\displaystyle \textbf{Exercise - 20(A)}

\displaystyle \textbf{Question 1: }\text{Find the area of a triangle whose sides are }18\text{ cm, }24\text{ cm and }30\text{ cm.}
\displaystyle \text{Also, find the length of altitude corresponding to the largest side of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }18^2+24^2=324+576=900=30^2,
\displaystyle \text{the triangle is right-angled, with perpendicular sides }18\text{ cm and }24\text{ cm.}
\displaystyle \text{Area of triangle}=\frac{1}{2}\times18\times24=216\text{ cm}^2.
\displaystyle \text{Let the altitude corresponding to the largest side }30\text{ cm be }h\text{ cm.}
\displaystyle \frac{1}{2}\times30\times h=216
\displaystyle 15h=216
\displaystyle h=\frac{216}{15}=14.4\text{ cm}.
\displaystyle \therefore \text{Area of the triangle}=216\text{ cm}^2\text{ and the required altitude}=14.4\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The lengths of the sides of a triangle are in the ratio }3:4:5.\text{ Find the area}
\displaystyle \text{of the triangle if its perimeter is }144\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the sides of the triangle be }3x,\ 4x\text{ and }5x.
\displaystyle 3x+4x+5x=144
\displaystyle 12x=144
\displaystyle x=12
\displaystyle \therefore \text{The sides are }36\text{ cm, }48\text{ cm and }60\text{ cm.}
\displaystyle \text{Since }36^2+48^2=1296+2304=3600=60^2,
\displaystyle \text{the triangle is right-angled, with perpendicular sides }36\text{ cm and }48\text{ cm.}
\displaystyle \text{Area of triangle}=\frac{1}{2}\times36\times48=864\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }864\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{ABC is a triangle in which }AB=AC=4\text{ cm and }\angle A=90^\circ.\text{ Calculate:}
\displaystyle \text{(i) the area of }\triangle ABC,
\displaystyle \text{(ii) the length of perpendicular from }A\text{ to }BC.
\displaystyle \text{Answer:} \displaystyle \text{(i) Since }\angle A=90^\circ,\ AB\text{ and }AC\text{ are perpendicular sides.}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times AB\times AC
\displaystyle =\frac{1}{2}\times4\times4=8\text{ cm}^2.
\displaystyle \therefore \text{Area of }\triangle ABC=8\text{ cm}^2.
\displaystyle \text{(ii) By Pythagoras Theorem,}
\displaystyle BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+4^2}=4\sqrt{2}\text{ cm}.
\displaystyle \text{Let the perpendicular from }A\text{ to }BC=h\text{ cm.}
\displaystyle \frac{1}{2}\times BC\times h=8
\displaystyle \frac{1}{2}\times4\sqrt{2}\times h=8
\displaystyle h=\frac{4}{\sqrt{2}}=2\sqrt{2}\text{ cm}.
\displaystyle \therefore \text{The required perpendicular is }2\sqrt{2}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The area of an equilateral triangle is }36\sqrt{3}\text{ sq. cm. Find its perimeter.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the equilateral triangle be }a\text{ cm.}
\displaystyle \text{Area of an equilateral triangle}=\frac{\sqrt{3}}{4}a^2
\displaystyle \frac{\sqrt{3}}{4}a^2=36\sqrt{3}
\displaystyle a^2=144
\displaystyle a=12\text{ cm}.
\displaystyle \text{Perimeter}=3a=3\times12=36\text{ cm}.
\displaystyle \therefore \text{The perimeter of the equilateral triangle is }36\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the area of an isosceles triangle with perimeter }36\text{ cm}
\displaystyle \text{and base }16\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Let each equal side be }a\text{ cm.}
\displaystyle 2a+16=36
\displaystyle 2a=20
\displaystyle a=10\text{ cm.}
\displaystyle \text{Draw the perpendicular from the vertex to the base.}
\displaystyle \text{It bisects the base into two segments of }8\text{ cm each.}
\displaystyle \text{Let the height be }h\text{ cm.}
\displaystyle h^2+8^2=10^2
\displaystyle h^2=100-64=36
\displaystyle h=6\text{ cm.}
\displaystyle \text{Area of triangle}=\frac{1}{2}\times16\times6=48\text{ cm}^2.
\displaystyle \therefore \text{The area of the isosceles triangle is }48\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The base of an isosceles triangle is }24\text{ cm and its area is }192\text{ sq. cm.}
\displaystyle \text{Find its perimeter.}
\displaystyle \text{Answer:}
\displaystyle \text{Let each equal side be }a\text{ cm.}
\displaystyle \frac{1}{2}\times24\times h=192
\displaystyle 12h=192
\displaystyle h=16\text{ cm.}
\displaystyle \text{The perpendicular bisects the base into two segments of }12\text{ cm each.}
\displaystyle a^2=16^2+12^2
\displaystyle a^2=256+144=400
\displaystyle a=20\text{ cm.}
\displaystyle \text{Perimeter}=20+20+24=64\text{ cm}.
\displaystyle \therefore \text{The perimeter of the isosceles triangle is }64\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The given figure shows a right-angled triangle }ABC\text{ and an equilateral triangle}
\displaystyle BCD.\text{ Find the area of the shaded portion, if }AC=16\text{ cm and }BC=8\text{ cm.} \displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle AB^2+BC^2=AC^2
\displaystyle AB^2+8^2=16^2
\displaystyle AB^2=256-64=192
\displaystyle AB=8\sqrt{3}\text{ cm.}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times BC\times AB
\displaystyle =\frac{1}{2}\times8\times8\sqrt{3}=32\sqrt{3}\text{ cm}^2.
\displaystyle \text{Since }\triangle BCD\text{ is equilateral, its side }BC=8\text{ cm.}
\displaystyle \text{Area of }\triangle BCD=\frac{\sqrt{3}}{4}\times8^2=16\sqrt{3}\text{ cm}^2.
\displaystyle \text{Area of shaded portion}=32\sqrt{3}-16\sqrt{3}=16\sqrt{3}\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded portion is }16\sqrt{3}\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the area and the perimeter of quadrilateral }ABCD,\text{ if }
\displaystyle AB=8\text{ cm,} \ AD=10\text{ cm, }BD=12\text{ cm, }DC=13\text{ cm and }\angle DBC=90^\circ. \displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle DBC,
\displaystyle BC^2+BD^2=DC^2
\displaystyle BC^2+12^2=13^2
\displaystyle BC^2=169-144=25
\displaystyle BC=5\text{ cm.}
\displaystyle \text{Area of }\triangle DBC=\frac{1}{2}\times12\times5=30\text{ cm}^2.
\displaystyle \text{For }\triangle ABD,\text{ the sides are }8\text{ cm, }10\text{ cm and }12\text{ cm.}
\displaystyle s=\frac{8+10+12}{2}=15\text{ cm.}
\displaystyle \text{Area of }\triangle ABD=\sqrt{s(s-8)(s-10)(s-12)}
\displaystyle =\sqrt{15\times7\times5\times3}
\displaystyle =\sqrt{225\times7}=15\sqrt{7}\text{ cm}^2.
\displaystyle \text{Area of quadrilateral }ABCD=30+15\sqrt{7}\text{ cm}^2.
\displaystyle \text{Perimeter of }ABCD=AB+BC+CD+DA
\displaystyle =8+5+13+10=36\text{ cm.}
\displaystyle \therefore \text{Area}=30+15\sqrt{7}\text{ cm}^2\text{ and perimeter}=36\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The base of a triangular field is three times its height. If the cost of}
\displaystyle \text{cultivating the field at Rs. }36.72\text{ per }100\text{ m}^2\text{ is Rs. }49,572,\text{ find its base and height.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of the field}=\frac{49,572\times100}{36.72}=135000\text{ m}^2.
\displaystyle \text{Let the height of the field}=h\text{ m.}
\displaystyle \therefore \text{Base}=3h\text{ m.}
\displaystyle \frac{1}{2}\times3h\times h=135000
\displaystyle \frac{3}{2}h^2=135000
\displaystyle h^2=90000
\displaystyle h=300\text{ m.}
\displaystyle \therefore \text{Base}=3\times300=900\text{ m.}
\displaystyle \therefore \text{The base is }900\text{ m and the height is }300\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The sides of a triangular field are in the ratio }5:3:4\text{ and its perimeter}
\displaystyle \text{is } 180\text{ m.} \ \text{Find: (i) its area, (ii) the altitude corresponding to its largest side,}
\displaystyle \text{(iii) the cost of levelling the field at the rate of Rs. }10\text{ per square metre.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the sides of the triangle be }5x,\ 3x\text{ and }4x.
\displaystyle 5x+3x+4x=180
\displaystyle 12x=180
\displaystyle x=15
\displaystyle \therefore \text{The sides are }75\text{ m, }45\text{ m and }60\text{ m.}
\displaystyle \text{Since }45^2+60^2=2025+3600=5625=75^2,
\displaystyle \text{the triangle is right-angled, with perpendicular sides }45\text{ m and }60\text{ m.}
\displaystyle \text{(i) Area}=\frac{1}{2}\times45\times60=1350\text{ m}^2.
\displaystyle \therefore \text{The area of the field is }1350\text{ m}^2.
\displaystyle \text{(ii) Let the altitude corresponding to the largest side }75\text{ m be }h\text{ m.}
\displaystyle \frac{1}{2}\times75\times h=1350
\displaystyle 75h=2700
\displaystyle h=36\text{ m.}
\displaystyle \therefore \text{The required altitude is }36\text{ m.}
\displaystyle \text{(iii) Cost of levelling}=1350\times10=\text{Rs. }13,500.
\displaystyle \therefore \text{The cost of levelling the field is Rs. }13,500.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Each of the equal sides of an isosceles triangle is }4\text{ cm greater than}
\displaystyle \text{its height. If the base of the triangle is }24\text{ cm, calculate its perimeter and area.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the triangle be }h\text{ cm.}
\displaystyle \therefore \text{Each equal side}=(h+4)\text{ cm.}
\displaystyle \text{The perpendicular from the vertex bisects the base into two parts of }12\text{ cm each.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle (h+4)^2=h^2+12^2
\displaystyle h^2+8h+16=h^2+144
\displaystyle 8h=128
\displaystyle h=16\text{ cm.}
\displaystyle \therefore \text{Each equal side}=16+4=20\text{ cm.}
\displaystyle \text{Perimeter}=20+20+24=64\text{ cm.}
\displaystyle \text{Area}=\frac{1}{2}\times24\times16=192\text{ cm}^2.
\displaystyle \therefore \text{The perimeter is }64\text{ cm and the area is }192\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Calculate the area and the height of an equilateral triangle} \\ \text{whose perimeter is }60\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Side of the equilateral triangle}=\frac{60}{3}=20\text{ cm.}
\displaystyle \text{Area}=\frac{\sqrt{3}}{4}\times20^2
\displaystyle =\frac{\sqrt{3}}{4}\times400=100\sqrt{3}\text{ cm}^2.
\displaystyle \text{Height}=\frac{\sqrt{3}}{2}\times20=10\sqrt{3}\text{ cm.}
\displaystyle \therefore \text{The area is }100\sqrt{3}\text{ cm}^2\text{ and the height is }10\sqrt{3}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In triangle }ABC,\ \angle A=90^\circ,\ AB=x\text{ cm, }AC=(x+5)\text{ cm and}
\displaystyle \text{area}=150\text{ cm}^2.\text{ Find the sides of the triangle.}
\displaystyle \text{Answer:} \displaystyle \text{Since }\angle A=90^\circ,\ AB\text{ and }AC\text{ are perpendicular sides.}
\displaystyle \frac{1}{2}\times x\times(x+5)=150
\displaystyle x(x+5)=300
\displaystyle x^2+5x-300=0
\displaystyle (x+20)(x-15)=0
\displaystyle x=15\text{ or }x=-20
\displaystyle \text{Rejecting }x=-20,\text{ since length cannot be negative, }x=15.
\displaystyle \therefore AB=15\text{ cm and }AC=20\text{ cm.}
\displaystyle BC=\sqrt{AB^2+AC^2}
\displaystyle =\sqrt{15^2+20^2}=\sqrt{625}=25\text{ cm.}
\displaystyle \therefore \text{The sides of the triangle are }15\text{ cm, }20\text{ cm and }25\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If the difference between the sides of a right-angled triangle is }
\displaystyle 3\text{ cm and its area} \ \text{is }54\text{ cm}^2,\text{ find its perimeter.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the perpendicular sides be }x\text{ cm and }(x+3)\text{ cm.}
\displaystyle \frac{1}{2}\times x\times(x+3)=54
\displaystyle x(x+3)=108
\displaystyle x^2+3x-108=0
\displaystyle (x+12)(x-9)=0
\displaystyle x=9\text{ or }x=-12
\displaystyle \text{Rejecting }x=-12,\text{ since length cannot be negative, }x=9.
\displaystyle \therefore \text{The perpendicular sides are }9\text{ cm and }12\text{ cm.}
\displaystyle \text{Hypotenuse}=\sqrt{9^2+12^2}=\sqrt{225}=15\text{ cm.}
\displaystyle \text{Perimeter}=9+12+15=36\text{ cm.}
\displaystyle \therefore \text{The perimeter of the triangle is }36\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{AD is altitude of an isosceles triangle }ABC\text{ in which }
\displaystyle AB=AC=30\text{ cm} \ \text{and }BC=36\text{ cm. A point }O\text{ is marked on }AD
\displaystyle \text{ such that }\angle BOC=90^\circ. \ \text{Find the area of quadrilateral }ABOC.
\displaystyle \text{Answer:} \displaystyle \text{Since }AB=AC,\text{ altitude }AD\text{ bisects }BC.
\displaystyle \therefore BD=DC=\frac{36}{2}=18\text{ cm.}
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AD^2=AB^2-BD^2
\displaystyle =30^2-18^2=900-324=576
\displaystyle \therefore AD=24\text{ cm.}
\displaystyle \text{Since }O\text{ lies on the perpendicular bisector of }BC,\ OB=OC.
\displaystyle \text{Also, }\angle BOC=90^\circ,\text{ so }\triangle BOC\text{ is an isosceles right-angled triangle.}
\displaystyle \therefore \angle BOD=45^\circ.
\displaystyle \text{In right-angled }\triangle BOD,\ \tan45^\circ=\frac{BD}{OD}
\displaystyle 1=\frac{18}{OD}
\displaystyle \therefore OD=18\text{ cm.}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times36\times24=432\text{ cm}^2.
\displaystyle \text{Area of }\triangle BOC=\frac{1}{2}\times36\times18=324\text{ cm}^2.
\displaystyle \text{Area of quadrilateral }ABOC=432-324=108\text{ cm}^2.
\displaystyle \therefore \text{The area of quadrilateral }ABOC\text{ is }108\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Exercise - 20(B)}

\displaystyle \textbf{Question 1: }\text{Find the area of a quadrilateral, one of whose diagonals is }30\text{ cm long}
\displaystyle \text{and the perpendiculars from the other two vertices are }19\text{ cm and }11\text{ cm respectively.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of quadrilateral}=\frac{1}{2}\times\text{diagonal}\times\text{sum of perpendiculars}
\displaystyle =\frac{1}{2}\times30\times(19+11)
\displaystyle =15\times30=450\text{ cm}^2.
\displaystyle \therefore \text{The area of the quadrilateral is }450\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The diagonals of a quadrilateral are }16\text{ cm and }13\text{ cm. If they intersect each}
\displaystyle \text{other at right angles, find the area of the quadrilateral.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of quadrilateral}=\frac{1}{2}\times\text{product of diagonals}
\displaystyle =\frac{1}{2}\times16\times13
\displaystyle =104\text{ cm}^2.
\displaystyle \therefore \text{The area of the quadrilateral is }104\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Calculate the area of quadrilateral }ABCD,\text{ in which }
\displaystyle \angle ABD=90^\circ,\text{ triangle }BCD \ \text{is an equilateral triangle of side }24\text{ cm and } \\ AD=26\text{ cm.}
\displaystyle \text{Answer:} \displaystyle \text{Since }\triangle BCD\text{ is equilateral, }BD=24\text{ cm.}
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AB^2+BD^2=AD^2
\displaystyle AB^2+24^2=26^2
\displaystyle AB^2=676-576=100
\displaystyle AB=10\text{ cm.}
\displaystyle \text{Area of }\triangle ABD=\frac{1}{2}\times10\times24=120\text{ cm}^2.
\displaystyle \text{Area of equilateral }\triangle BCD=\frac{\sqrt{3}}{4}\times24^2
\displaystyle =144\sqrt{3}\text{ cm}^2.
\displaystyle \text{Area of quadrilateral }ABCD=120+144\sqrt{3}\text{ cm}^2.
\displaystyle \therefore \text{The area of quadrilateral }ABCD\text{ is }(120+144\sqrt{3})\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Calculate the area of quadrilateral }ABCD\text{ in which }AB=32\text{ cm, }
\displaystyle AD=24\text{ cm,} \ \angle A=90^\circ\text{ and }BC=CD=52\text{ cm.}
\displaystyle \text{Answer:}

\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle BD=\sqrt{AB^2+AD^2}
\displaystyle =\sqrt{32^2+24^2}=\sqrt{1600}=40\text{ cm.}
\displaystyle \text{Area of }\triangle ABD=\frac{1}{2}\times32\times24=384\text{ cm}^2.
\displaystyle \text{In isosceles }\triangle BCD,\text{ draw }CE\perp BD.
\displaystyle \therefore BE=ED=\frac{40}{2}=20\text{ cm.}
\displaystyle CE^2=CD^2-DE^2
\displaystyle =52^2-20^2=2704-400=2304
\displaystyle CE=48\text{ cm.}
\displaystyle \text{Area of }\triangle BCD=\frac{1}{2}\times40\times48=960\text{ cm}^2.
\displaystyle \text{Area of quadrilateral }ABCD=384+960=1344\text{ cm}^2.
\displaystyle \therefore \text{The area of quadrilateral }ABCD\text{ is }1344\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The perimeter of a rectangular field is }\frac{3}{5}\text{ km. If the length of the field is twice}
\displaystyle \text{its width, find the area of the rectangle in square metres.}
\displaystyle \text{Answer:}
\displaystyle \text{Perimeter}=\frac{3}{5}\text{ km}=\frac{3}{5}\times1000=600\text{ m.}
\displaystyle \text{Let the width}=x\text{ m. Therefore, length}=2x\text{ m.}
\displaystyle 2(2x+x)=600
\displaystyle 6x=600
\displaystyle x=100\text{ m.}
\displaystyle \therefore \text{Width}=100\text{ m and length}=200\text{ m.}
\displaystyle \text{Area}=200\times100=20000\text{ m}^2.
\displaystyle \therefore \text{The area of the rectangular field is }20000\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A rectangular plot }85\text{ m long and }60\text{ m broad is to be covered with}
\displaystyle \text{grass leaving 5 m all around. Find the area to be laid with grass.}
\displaystyle \text{Answer:}
\displaystyle \text{Length of the grass-covered portion}=85-2(5)=75\text{ m.}
\displaystyle \text{Breadth of the grass-covered portion}=60-2(5)=50\text{ m.}
\displaystyle \text{Area to be laid with grass}=75\times50=3750\text{ m}^2.
\displaystyle \therefore \text{The area to be laid with grass is }3750\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The length and breadth of a rectangle are }6\text{ cm and }4\text{ cm respectively. Find}
\displaystyle \text{the height of a triangle whose base is }6\text{ cm and area is }3\text{ times that of the rectangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of rectangle}=6\times4=24\text{ cm}^2.
\displaystyle \text{Area of triangle}=3\times24=72\text{ cm}^2.
\displaystyle \text{Let the height of the triangle be }h\text{ cm.}
\displaystyle \frac{1}{2}\times6\times h=72
\displaystyle 3h=72
\displaystyle h=24\text{ cm.}
\displaystyle \therefore \text{The height of the triangle is }24\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{How many tiles, each of area }400\text{ cm}^2,\text{ will be needed to pave a}
\displaystyle \text{footpath which is }2\text{ m wide and surrounds a grass plot }25\text{ m long and }13\text{ m wide?}
\displaystyle \text{Answer:}
\displaystyle \text{Outer length}=25+2(2)=29\text{ m.}
\displaystyle \text{Outer breadth}=13+2(2)=17\text{ m.}
\displaystyle \text{Area including footpath}=29\times17=493\text{ m}^2.
\displaystyle \text{Area of grass plot}=25\times13=325\text{ m}^2.
\displaystyle \text{Area of footpath}=493-325=168\text{ m}^2.
\displaystyle \text{Area of each tile}=400\text{ cm}^2=\frac{400}{10000}=0.04\text{ m}^2.
\displaystyle \text{Number of tiles}=\frac{168}{0.04}=4200.
\displaystyle \therefore \text{The number of tiles required is }4200.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The cost of enclosing a rectangular garden with a fence all round, at the}
\displaystyle \text{rate of 75 paise per metre, is Rs. }300.\text{ If the length is }120\text{ m, find its area in square metres.}
\displaystyle \text{Answer:}
\displaystyle 75\text{ paise}=\text{Rs. }0.75.
\displaystyle \text{Perimeter of the garden}=\frac{300}{0.75}=400\text{ m.}
\displaystyle 2(\text{length}+\text{breadth})=400
\displaystyle 2(120+b)=400
\displaystyle 120+b=200
\displaystyle b=80\text{ m.}
\displaystyle \text{Area}=120\times80=9600\text{ m}^2.
\displaystyle \therefore \text{The area of the garden is }9600\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The width of a rectangular room is }\frac{4}{7}\text{ of its length, }x,\text{ and its perimeter is }y.
\displaystyle \text{Write an equation connecting }x\text{ and }y.\text{ Find the length of the room when the perimeter is }4400\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Length}=x\text{ and width}=\frac{4x}{7}.
\displaystyle y=2\left(x+\frac{4x}{7}\right)
\displaystyle y=2\left(\frac{11x}{7}\right)=\frac{22x}{7}
\displaystyle \therefore 7y=22x.
\displaystyle \text{When }y=4400,
\displaystyle 7(4400)=22x
\displaystyle x=\frac{7\times4400}{22}=1400\text{ cm.}
\displaystyle \therefore \text{The length of the room is }1400\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The length of a rectangular verandah is }3\text{ m more than its breadth. The numerical}
\displaystyle \text{value of its area is equal to the numerical value of its perimeter.}
\displaystyle \text{(i) Taking }x\text{ as the breadth, write an equation in }x\text{ that represents the above statement.}
\displaystyle \text{(ii) Solve the equation obtained in (i) and find the dimensions of the verandah.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the breadth}=x\text{ m. Therefore, length}=(x+3)\text{ m.}
\displaystyle \text{(i) Area}=x(x+3)
\displaystyle \text{Perimeter}=2\{x+(x+3)\}=4x+6
\displaystyle \text{Since their numerical values are equal,}
\displaystyle x(x+3)=4x+6
\displaystyle \therefore x^2-x-6=0.
\displaystyle \text{(ii) }x^2-x-6=0
\displaystyle (x-3)(x+2)=0
\displaystyle x=3\text{ or }x=-2
\displaystyle \text{Rejecting }x=-2,\text{ since breadth cannot be negative, }x=3.
\displaystyle \therefore \text{Breadth}=3\text{ m and length}=3+3=6\text{ m.}
\displaystyle \therefore \text{The dimensions of the verandah are }6\text{ m}\times3\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The diagram shows two paths drawn inside a rectangular field }80\text{ m long and }45\text{ m}
\displaystyle \text{wide. The widths of the two paths are }8\text{ m and }15\text{ m as shown. Find the area of}
\displaystyle \text{the shaded portion.} \displaystyle \text{Answer:}
\displaystyle \text{Area of horizontal path}=80\times8=640\text{ m}^2.
\displaystyle \text{Area of vertical path}=45\times15=675\text{ m}^2.
\displaystyle \text{Area of common portion}=15\times8=120\text{ m}^2.
\displaystyle \text{Area of shaded portion}=640+675-120
\displaystyle =1195\text{ m}^2.
\displaystyle \therefore \text{The area of the shaded portion is }1195\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The rate for a }1.20\text{ m wide carpet is Rs. }40\text{ per metre; find the}
\displaystyle \text{cost of covering a hall 45 m long and }32\text{ m wide with this carpet. Also, find the cost}
\displaystyle \text{of carpeting the same hall if the carpet, }80\text{ cm wide, is at Rs. }25\text{ per metre.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of hall}=45\times32=1440\text{ m}^2.
\displaystyle \text{For the }1.20\text{ m wide carpet,}
\displaystyle \text{Length of carpet required}=\frac{1440}{1.20}=1200\text{ m.}
\displaystyle \text{Cost}=1200\times40=\text{Rs. }48,000.
\displaystyle \text{For the }80\text{ cm wide carpet, }80\text{ cm}=0.80\text{ m.}
\displaystyle \text{Length of carpet required}=\frac{1440}{0.80}=1800\text{ m.}
\displaystyle \text{Cost}=1800\times25=\text{Rs. }45,000.
\displaystyle \therefore \text{The respective costs are Rs. }48,000\text{ and Rs. }45,000.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the area and perimeter of a square plot of land, the length}
\displaystyle \text{of whose diagonal is 15 m. Give your answer correct to }2\text{ decimal places.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }a\text{ m.}
\displaystyle a\sqrt{2}=15
\displaystyle a=\frac{15}{\sqrt{2}}=\frac{15\sqrt{2}}{2}\text{ m.}
\displaystyle \text{Area}=a^2=\left(\frac{15}{\sqrt{2}}\right)^2=\frac{225}{2}=112.50\text{ m}^2.
\displaystyle \text{Perimeter}=4a=4\left(\frac{15}{\sqrt{2}}\right)=30\sqrt{2}\text{ m}
\displaystyle \approx30\times1.4142=42.43\text{ m.}
\displaystyle \therefore \text{The area is }112.50\text{ m}^2\text{ and the perimeter is }42.43\text{ m (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The shaded region represents the lawn in the form of a house. On the}
\displaystyle \text{three sides of the lawn there are flower-beds having a uniform width of }2\text{ m.}
\displaystyle \text{(i) Find the length and the breadth of the lawn.}
\displaystyle \text{(ii) Hence, or otherwise, find the area of the flower-beds.} \displaystyle \text{Answer:}
\displaystyle \text{(i) Outer length}=30\text{ m and outer breadth}=12\text{ m.}
\displaystyle \text{Flower-beds are }2\text{ m wide on the left, right and top.}
\displaystyle \text{Length of lawn}=30-2-2=26\text{ m.}
\displaystyle \text{Breadth of lawn}=12-2=10\text{ m.}
\displaystyle \therefore \text{The lawn measures }26\text{ m}\times10\text{ m.}
\displaystyle \text{(ii) Area of outer rectangle}=30\times12=360\text{ m}^2.
\displaystyle \text{Area of lawn}=26\times10=260\text{ m}^2.
\displaystyle \text{Area of flower-beds}=360-260=100\text{ m}^2.
\displaystyle \therefore \text{The area of the flower-beds is }100\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{A floor which measures }15\text{ m}\times8\text{ m is to be laid with tiles measuring}
\displaystyle 50\text{ cm}\times25\text{ cm. Find the number of tiles required.}
\displaystyle \text{Further, if a carpet is laid on the floor so that a space of }1\text{ m exists between its edges}
\displaystyle \text{and the edges of the floor, what fraction of the floor is left uncovered?}
\displaystyle \text{Answer:}
\displaystyle \text{Area of floor}=15\times8=120\text{ m}^2.
\displaystyle 50\text{ cm}=0.5\text{ m and }25\text{ cm}=0.25\text{ m.}
\displaystyle \text{Area of each tile}=0.5\times0.25=0.125\text{ m}^2.
\displaystyle \text{Number of tiles required}=\frac{120}{0.125}=960.
\displaystyle \text{Length of carpet}=15-2(1)=13\text{ m.}
\displaystyle \text{Breadth of carpet}=8-2(1)=6\text{ m.}
\displaystyle \text{Area covered by carpet}=13\times6=78\text{ m}^2.
\displaystyle \text{Area left uncovered}=120-78=42\text{ m}^2.
\displaystyle \text{Fraction of floor left uncovered}=\frac{42}{120}=\frac{7}{20}.
\displaystyle \therefore \text{The number of tiles required is }960\text{ and the fraction left uncovered is }\frac{7}{20}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Two adjacent sides of a parallelogram are }24\text{ cm and }18\text{ cm. If the distance}
\displaystyle \text{between the longer sides is }12\text{ cm, find the distance between the shorter sides.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of parallelogram}=24\times12=288\text{ cm}^2.
\displaystyle \text{Let the distance between the shorter sides be }h\text{ cm.}
\displaystyle 18\times h=288
\displaystyle h=\frac{288}{18}=16\text{ cm.}
\displaystyle \therefore \text{The distance between the shorter sides is }16\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Two adjacent sides of a parallelogram are }10\text{ cm and }12\text{ cm. If one diagonal}
\displaystyle \text{is }16\text{ cm long, find the area of the parallelogram. Also, find the distance between its}
\displaystyle \text{shorter sides.}
\displaystyle \text{Answer:}
\displaystyle \text{The diagonal divides the parallelogram into two congruent triangles.}
\displaystyle \text{For one triangle, the sides are }10\text{ cm, }12\text{ cm and }16\text{ cm.}
\displaystyle s=\frac{10+12+16}{2}=19\text{ cm.}
\displaystyle \text{Area of one triangle}=\sqrt{s(s-10)(s-12)(s-16)}
\displaystyle =\sqrt{19\times9\times7\times3}
\displaystyle =3\sqrt{399}\text{ cm}^2.
\displaystyle \text{Area of parallelogram}=2\times3\sqrt{399}=6\sqrt{399}\text{ cm}^2
\displaystyle \approx119.85\text{ cm}^2.
\displaystyle \text{Let the distance between the shorter sides be }h\text{ cm.}
\displaystyle 10h=6\sqrt{399}
\displaystyle h=\frac{3\sqrt{399}}{5}\approx11.98\text{ cm.}
\displaystyle \therefore \text{The area is }119.85\text{ cm}^2\text{ and the required distance is }11.98\text{ cm (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The area of a rhombus is }216\text{ sq. cm. If one diagonal is }24\text{ cm, find:}
\displaystyle \text{(i) length of its other diagonal, (ii) length of its side, (iii) perimeter of the rhombus.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Area of rhombus}=\frac{1}{2}\times\text{product of diagonals}
\displaystyle 216=\frac{1}{2}\times24\times d
\displaystyle 216=12d
\displaystyle d=18\text{ cm.}
\displaystyle \therefore \text{The other diagonal is }18\text{ cm.}
\displaystyle \text{(ii) The diagonals of a rhombus bisect each other at right angles.}
\displaystyle \text{Half of the diagonals are }12\text{ cm and }9\text{ cm.}
\displaystyle \text{Side}=\sqrt{12^2+9^2}
\displaystyle =\sqrt{144+81}=\sqrt{225}=15\text{ cm.}
\displaystyle \therefore \text{The side of the rhombus is }15\text{ cm.}
\displaystyle \text{(iii) Perimeter}=4\times15=60\text{ cm.}
\displaystyle \therefore \text{The perimeter of the rhombus is }60\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The perimeter of a rhombus is }52\text{ cm. If one diagonal is }24\text{ cm, find:}
\displaystyle \text{(i) the length of its other diagonal, (ii) its area.}
\displaystyle \text{Answer:}
\displaystyle \text{Side of rhombus}=\frac{52}{4}=13\text{ cm.}
\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}
\displaystyle \text{Half of the given diagonal}=\frac{24}{2}=12\text{ cm.}
\displaystyle \text{(i) Let half of the other diagonal}=x\text{ cm.}
\displaystyle 13^2=12^2+x^2
\displaystyle 169=144+x^2
\displaystyle x^2=25
\displaystyle x=5\text{ cm.}
\displaystyle \therefore \text{Other diagonal}=2\times5=10\text{ cm.}
\displaystyle \text{(ii) Area of rhombus}=\frac{1}{2}\times24\times10=120\text{ cm}^2.
\displaystyle \therefore \text{The other diagonal is }10\text{ cm and the area is }120\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{The perimeter of a rhombus is }46\text{ cm. If the height}
\displaystyle \text{of the rhombus is 8 cm, find its area.}
\displaystyle \text{Answer:}
\displaystyle \text{Side of rhombus}=\frac{46}{4}=11.5\text{ cm.}
\displaystyle \text{Area of rhombus}=\text{base}\times\text{height}
\displaystyle =11.5\times8=92\text{ cm}^2.
\displaystyle \therefore \text{The area of the rhombus is }92\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The figure shows the cross-section of a concrete structure. Calculate}
\displaystyle \text{its area if } AB=1.8\text{ m, }CD=0.6\text{ m, }DE=0.8\text{ m, }EF=0.3\text{ m and } \\ AF=1.2\text{ m.} \displaystyle \text{Answer:}
\displaystyle \text{Total height}=AF+DE=1.2+0.8=2\text{ m.}
\displaystyle \text{Divide the figure into two rectangles and a right triangle.}
\displaystyle \text{Area of first rectangle}=0.3\times1.2=0.36\text{ m}^2.
\displaystyle \text{Area of second rectangle}=0.6\times2=1.20\text{ m}^2.
\displaystyle \text{Base of right triangle}=1.8-(0.3+0.6)=0.9\text{ m.}
\displaystyle \text{Height of right triangle}=2\text{ m.}
\displaystyle \text{Area of right triangle}=\frac{1}{2}\times0.9\times2=0.90\text{ m}^2.
\displaystyle \text{Area of cross-section}=0.36+1.20+0.90=2.46\text{ m}^2.
\displaystyle \therefore \text{The area of the cross-section is }2.46\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Calculate the area of the figure given below, which is not drawn to scale.}
\displaystyle \text{The given dimensions are }15\text{ cm, }25\text{ cm, }26\text{ cm and }12\text{ cm.} \displaystyle \text{Answer:}
\displaystyle \text{The left portion is a trapezium with parallel sides }15\text{ cm and }25\text{ cm.}
\displaystyle \text{Difference between the parallel sides}=25-15=10\text{ cm.}
\displaystyle \text{Let the distance between the parallel sides be }x\text{ cm.}
\displaystyle x^2+10^2=26^2
\displaystyle x^2=676-100=576
\displaystyle x=24\text{ cm.}
\displaystyle \text{Area of trapezium}=\frac{1}{2}(15+25)\times24
\displaystyle =480\text{ cm}^2.
\displaystyle \text{Area of triangular portion}=\frac{1}{2}\times25\times12
\displaystyle =150\text{ cm}^2.
\displaystyle \text{Total area}=480+150=630\text{ cm}^2.
\displaystyle \therefore \text{The area of the figure is }630\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The diagram shows a pentagonal field }ABCDE\text{ in which }AF,\ FG,\ GH\text{ and }HD
\displaystyle \text{are }50\text{ m, }40\text{ m, }15\text{ m and }25\text{ m respectively; and the lengths of perpendiculars}
\displaystyle BF,\ CH\text{ and }EG\text{ are }50\text{ m, }25\text{ m and }60\text{ m respectively. Determine the area of the field.} \displaystyle \text{Answer:}
\displaystyle AD=AF+FG+GH+HD
\displaystyle =50+40+15+25=130\text{ m.}
\displaystyle \text{Area of quadrilateral }ABCD=\frac{1}{2}\times AD\times(BF+CH)
\displaystyle =\frac{1}{2}\times130\times(50+25)
\displaystyle =4875\text{ m}^2.
\displaystyle \text{Area of }\triangle ADE=\frac{1}{2}\times AD\times EG
\displaystyle =\frac{1}{2}\times130\times60=3900\text{ m}^2.
\displaystyle \text{Area of pentagonal field }ABCDE=4875+3900
\displaystyle =8775\text{ m}^2.
\displaystyle \therefore \text{The area of the field is }8775\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{A footpath of uniform width runs all around the outside of a rectangular}
\displaystyle \text{field 30 m long and }24\text{ m wide. If the path occupies an area of }360\text{ m}^2,\text{ find its width.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the width of the footpath be }x\text{ m.}
\displaystyle \text{Outer length}=(30+2x)\text{ m and outer breadth}=(24+2x)\text{ m.}
\displaystyle (30+2x)(24+2x)-30\times24=360
\displaystyle 720+108x+4x^2-720=360
\displaystyle 4x^2+108x-360=0
\displaystyle x^2+27x-90=0
\displaystyle (x+30)(x-3)=0
\displaystyle x=-30\text{ or }x=3
\displaystyle \text{Rejecting }x=-30,\text{ since width cannot be negative, }x=3\text{ m.}
\displaystyle \therefore \text{The width of the footpath is }3\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{A wire when bent in the form of a square encloses an area of }484\text{ m}^2.
\displaystyle \text{Find the largest area enclosed by the same wire when bent to form:}
\displaystyle \text{(i) an equilateral triangle, (ii) a rectangle of breadth }16\text{ m.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of square}=484\text{ m}^2.
\displaystyle \therefore \text{Side of square}=\sqrt{484}=22\text{ m.}
\displaystyle \text{Length of wire}=\text{perimeter of square}=4\times22=88\text{ m.}
\displaystyle \text{(i) Let the side of the equilateral triangle be }a\text{ m.}
\displaystyle 3a=88
\displaystyle a=\frac{88}{3}\text{ m.}
\displaystyle \text{Area}=\frac{\sqrt{3}}{4}\left(\frac{88}{3}\right)^2
\displaystyle =\frac{1936\sqrt{3}}{9}\text{ m}^2.
\displaystyle \therefore \text{The area enclosed is }\frac{1936\sqrt{3}}{9}\text{ m}^2.
\displaystyle \text{(ii) Let the length of the rectangle be }l\text{ m.}
\displaystyle 2(l+16)=88
\displaystyle l+16=44
\displaystyle l=28\text{ m.}
\displaystyle \text{Area of rectangle}=28\times16=448\text{ m}^2.
\displaystyle \therefore \text{The area enclosed by the rectangle is }448\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{For each trapezium given below, find its area.}
\displaystyle \text{(i) Parallel sides are }20\text{ cm and }12\text{ cm, and the non-parallel sides are }10\text{ cm each.}
\displaystyle \text{(ii) Parallel sides are }14\text{ cm and }8\text{ cm, and the slant side is }10\text{ cm.}
\displaystyle \text{(iii) Parallel sides are }32\text{ cm and }20\text{ cm, and the non-parallel sides are }10\text{ cm and }16\text{ cm.}
\displaystyle \text{(iv) Parallel sides are }30\text{ cm and }18\text{ cm, and the non-parallel sides are }12\text{ cm each.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Difference of parallel sides}=20-12=8\text{ cm.}
\displaystyle \text{Horizontal projection on each side}=\frac{8}{2}=4\text{ cm.}
\displaystyle \text{Let the height be }h\text{ cm.}
\displaystyle h^2+4^2=10^2
\displaystyle h^2=100-16=84
\displaystyle h=2\sqrt{21}\text{ cm.}
\displaystyle \text{Area}=\frac{1}{2}(20+12)\times2\sqrt{21}
\displaystyle =32\sqrt{21}\text{ cm}^2.
\displaystyle \therefore \text{Area of the trapezium}=32\sqrt{21}\text{ cm}^2.
\displaystyle \text{(ii) Difference of parallel sides}=14-8=6\text{ cm.}
\displaystyle \text{Let the height be }h\text{ cm.}
\displaystyle h^2+6^2=10^2
\displaystyle h^2=100-36=64
\displaystyle h=8\text{ cm.}
\displaystyle \text{Area}=\frac{1}{2}(14+8)\times8=88\text{ cm}^2.
\displaystyle \therefore \text{Area of the trapezium}=88\text{ cm}^2.
\displaystyle \text{(iii) Let the horizontal projection of the }10\text{ cm side be }x\text{ cm.}
\displaystyle \text{Then the horizontal projection of the }16\text{ cm side}=(12-x)\text{ cm.}
\displaystyle \text{Let the height be }h\text{ cm.}
\displaystyle x^2+h^2=10^2
\displaystyle (12-x)^2+h^2=16^2
\displaystyle (12-x)^2-x^2=256-100
\displaystyle 144-24x=156
\displaystyle x=-\frac{1}{2}
\displaystyle h^2=100-\left(\frac{1}{2}\right)^2=\frac{399}{4}
\displaystyle h=\frac{\sqrt{399}}{2}\text{ cm.}
\displaystyle \text{Area}=\frac{1}{2}(32+20)\times\frac{\sqrt{399}}{2}
\displaystyle =13\sqrt{399}\text{ cm}^2\approx259.67\text{ cm}^2.
\displaystyle \therefore \text{Area of the trapezium}=13\sqrt{399}\text{ cm}^2\approx259.67\text{ cm}^2.
\displaystyle \text{(iv) Difference of parallel sides}=30-18=12\text{ cm.}
\displaystyle \text{Horizontal projection on each side}=\frac{12}{2}=6\text{ cm.}
\displaystyle \text{Let the height be }h\text{ cm.}
\displaystyle h^2+6^2=12^2
\displaystyle h^2=144-36=108
\displaystyle h=6\sqrt{3}\text{ cm.}
\displaystyle \text{Area}=\frac{1}{2}(30+18)\times6\sqrt{3}
\displaystyle =144\sqrt{3}\text{ cm}^2.
\displaystyle \therefore \text{Area of the trapezium}=144\sqrt{3}\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{The perimeter of a rectangular board is }70\text{ cm. Taking its length as }x\text{ cm, find}
\displaystyle \text{its width in terms of }x.\text{ If the area of the rectangular board is }300\text{ cm}^2,\text{ find its dimensions.}
\displaystyle \text{Answer:}
\displaystyle 2(\text{length}+\text{width})=70
\displaystyle x+\text{width}=35
\displaystyle \therefore \text{Width}=(35-x)\text{ cm.}
\displaystyle x(35-x)=300
\displaystyle 35x-x^2=300
\displaystyle x^2-35x+300=0
\displaystyle (x-20)(x-15)=0
\displaystyle x=20\text{ or }x=15.
\displaystyle \therefore \text{The dimensions of the board are }20\text{ cm}\times15\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{The area of a rectangle is }640\text{ m}^2.\text{ Taking its length as }x\text{ m, find in terms}
\displaystyle \text{of }x,\text{ the width of the rectangle. If its perimeter is }104\text{ m, find its dimensions.}
\displaystyle \text{Answer:}
\displaystyle \text{Area}=\text{length}\times\text{width}
\displaystyle 640=x\times\text{width}
\displaystyle \therefore \text{Width}=\frac{640}{x}\text{ m.}
\displaystyle 2\left(x+\frac{640}{x}\right)=104
\displaystyle x+\frac{640}{x}=52
\displaystyle x^2-52x+640=0
\displaystyle (x-32)(x-20)=0
\displaystyle x=32\text{ or }x=20.
\displaystyle \therefore \text{The dimensions of the rectangle are }32\text{ m}\times20\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{The length of a rectangle is twice the side of a square and its width is }
\displaystyle 6\text{ cm greater} \ \text{than the side of the square. If the area of the rectangle is three times}
\displaystyle \text{the area of the square, find the dimensions of each.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }x\text{ cm.}
\displaystyle \therefore \text{Length of rectangle}=2x\text{ cm and width}=(x+6)\text{ cm.}
\displaystyle \text{Area of square}=x^2
\displaystyle \text{Area of rectangle}=2x(x+6)
\displaystyle \text{Given, }2x(x+6)=3x^2
\displaystyle 2x^2+12x=3x^2
\displaystyle x^2-12x=0
\displaystyle x(x-12)=0
\displaystyle x=0\text{ or }x=12
\displaystyle \text{Rejecting }x=0,\text{ since a side cannot be zero, }x=12\text{ cm.}
\displaystyle \therefore \text{Square dimensions}=12\text{ cm}\times12\text{ cm.}
\displaystyle \text{Rectangle dimensions}=24\text{ cm}\times18\text{ cm.}
\displaystyle \therefore \text{The square is }12\text{ cm}\times12\text{ cm and the rectangle is }24\text{ cm}\times18\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{ABCD is a square with each side }12\text{ cm. }P\text{ is a point on }BC
\displaystyle \text{ such that area of} \ \triangle ABP:\text{ area of trapezium }APCD=1:5.\text{ Find the length of }CP.
\displaystyle \text{Answer:} \displaystyle \text{Area of square }ABCD=12\times12=144\text{ cm}^2.
\displaystyle \text{Since area of }\triangle ABP:\text{ area of trapezium }APCD=1:5,
\displaystyle \text{Area of }\triangle ABP=\frac{1}{1+5}\times144=24\text{ cm}^2.
\displaystyle \frac{1}{2}\times AB\times BP=24
\displaystyle \frac{1}{2}\times12\times BP=24
\displaystyle 6BP=24
\displaystyle BP=4\text{ cm.}
\displaystyle CP=BC-BP=12-4=8\text{ cm.}
\displaystyle \therefore CP=8\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{A rectangular plot of land measures }45\text{ m}\times30\text{ m. A boundary}
\displaystyle \text{wall of height 2.4 m is built all around the plot at a distance of }1\text{ m from the plot.}
\displaystyle \text{Find the area of the inner surface of the boundary wall.}
\displaystyle \text{Answer:}
\displaystyle \text{Inner length of boundary wall}=45+2(1)=47\text{ m.}
\displaystyle \text{Inner breadth of boundary wall}=30+2(1)=32\text{ m.}
\displaystyle \text{Inner perimeter}=2(47+32)=158\text{ m.}
\displaystyle \text{Area of inner surface}=\text{inner perimeter}\times\text{height}
\displaystyle =158\times2.4=379.2\text{ m}^2.
\displaystyle \therefore \text{The area of the inner surface of the boundary wall is }379.2\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{A wire when bent in the form of a square encloses an area of }
\displaystyle 576\text{ cm}^2.\text{ Find the largest} \ \text{area enclosed by the same wire when bent to form:}
\displaystyle \text{(i) an equilateral triangle, (ii) a rectangle whose adjacent sides differ by }4\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of square}=576\text{ cm}^2.
\displaystyle \therefore \text{Side of square}=\sqrt{576}=24\text{ cm.}
\displaystyle \text{Length of wire}=\text{perimeter of square}=4\times24=96\text{ cm.}
\displaystyle \text{(i) Let the side of the equilateral triangle be }a\text{ cm.}
\displaystyle 3a=96
\displaystyle a=32\text{ cm.}
\displaystyle \text{Area}=\frac{\sqrt{3}}{4}\times32^2
\displaystyle =256\sqrt{3}\text{ cm}^2.
\displaystyle \therefore \text{The area enclosed is }256\sqrt{3}\text{ cm}^2.
\displaystyle \text{(ii) Let the breadth of the rectangle be }x\text{ cm.}
\displaystyle \therefore \text{Length}=(x+4)\text{ cm.}
\displaystyle 2\{x+(x+4)\}=96
\displaystyle 4x+8=96
\displaystyle 4x=88
\displaystyle x=22\text{ cm.}
\displaystyle \therefore \text{Length}=26\text{ cm.}
\displaystyle \text{Area of rectangle}=26\times22=572\text{ cm}^2.
\displaystyle \therefore \text{The area enclosed by the rectangle is }572\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{The area of a parallelogram is }y\text{ cm}^2\text{ and its height is }h\text{ cm. The base of}
\displaystyle \text{another parallelogram is }x\text{ cm more than the base of the first parallelogram and its area is}
\displaystyle \text{twice the area of the first. Find, in terms of }y,\ h\text{ and }x,\text{ the height of the second parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the base of the first parallelogram be }b\text{ cm.}
\displaystyle y=b\times h
\displaystyle \therefore b=\frac{y}{h}\text{ cm.}
\displaystyle \text{Base of the second parallelogram}=\left(\frac{y}{h}+x\right)\text{ cm.}
\displaystyle \text{Let the height of the second parallelogram be }H\text{ cm.}
\displaystyle \left(\frac{y}{h}+x\right)H=2y
\displaystyle H=\frac{2y}{\frac{y}{h}+x}
\displaystyle H=\frac{2yh}{y+xh}\text{ cm.}
\displaystyle \therefore \text{The height of the second parallelogram is }\frac{2yh}{y+xh}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{The distance between parallel sides of a trapezium is }15\text{ cm and the length of the line}
\displaystyle \text{segment joining the mid-points of its non-parallel sides is }26\text{ cm. Find the area of the trapezium.}
\displaystyle \text{Answer:}
\displaystyle \text{The line segment joining the mid-points of the non-parallel sides}
\displaystyle =\frac{1}{2}\times\text{sum of parallel sides}.
\displaystyle \therefore \text{Area of trapezium}=\text{mid-point segment}\times\text{height}
\displaystyle =26\times15=390\text{ cm}^2.
\displaystyle \therefore \text{The area of the trapezium is }390\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{The diagonal of a rectangular plot is }34\text{ m and its perimeter is }92\text{ m. Find its area.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length and breadth be }l\text{ m and }b\text{ m respectively.}
\displaystyle 2(l+b)=92
\displaystyle l+b=46
\displaystyle l^2+b^2=34^2=1156
\displaystyle (l+b)^2=l^2+b^2+2lb
\displaystyle 46^2=1156+2lb
\displaystyle 2116=1156+2lb
\displaystyle 2lb=960
\displaystyle lb=480
\displaystyle \therefore \text{The area of the rectangular plot is }480\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Exercise - 20(C)}

\displaystyle \textbf{Question 1: }\text{The diameter of a circle is }28\text{ cm. Find its: (i) circumference (ii) area.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter}=28\text{ cm.}
\displaystyle \therefore \text{Radius}=\frac{28}{2}=14\text{ cm.}
\displaystyle \text{(i) Circumference}=\pi d
\displaystyle =\frac{22}{7}\times28=88\text{ cm.}
\displaystyle \therefore \text{The circumference of the circle is }88\text{ cm.}
\displaystyle \text{(ii) Area}=\pi r^2
\displaystyle =\frac{22}{7}\times14\times14=616\text{ cm}^2.
\displaystyle \therefore \text{The area of the circle is }616\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The circumference of a circular field is }308\text{ m. Find its: (i) radius (ii) area.}
\displaystyle \text{Answer:}
\displaystyle \text{Circumference}=2\pi r
\displaystyle 308=2\times\frac{22}{7}\times r
\displaystyle r=\frac{308\times7}{44}=49\text{ m.}
\displaystyle \text{(i) }\therefore \text{The radius of the field is }49\text{ m.}
\displaystyle \text{(ii) Area}=\pi r^2
\displaystyle =\frac{22}{7}\times49\times49
\displaystyle =7546\text{ m}^2.
\displaystyle \therefore \text{The area of the field is }7546\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The sum of the circumference and diameter of a circle is }116\text{ cm. Find its radius.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the circle be }r\text{ cm.}
\displaystyle \text{Circumference}=2\pi r\text{ and diameter}=2r.
\displaystyle 2\pi r+2r=116
\displaystyle 2r\left(\frac{22}{7}+1\right)=116
\displaystyle 2r\times\frac{29}{7}=116
\displaystyle \frac{58r}{7}=116
\displaystyle r=\frac{116\times7}{58}=14\text{ cm.}
\displaystyle \therefore \text{The radius of the circle is }14\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The radii of two circles are }25\text{ cm and }18\text{ cm. Find the radius of the circle whose}
\displaystyle \text{circumference is equal to the sum of the circumferences of these two circles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the required circle be }R\text{ cm.}
\displaystyle 2\pi R=2\pi(25)+2\pi(18)
\displaystyle 2\pi R=2\pi(25+18)
\displaystyle R=43\text{ cm.}
\displaystyle \therefore \text{The radius of the required circle is }43\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The radii of two circles are }48\text{ cm and }13\text{ cm. Find the area of the circle which}
\displaystyle \text{has its circumference equal to the difference of the circumferences of the given two circles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the required circle be }R\text{ cm.}
\displaystyle 2\pi R=2\pi(48)-2\pi(13)
\displaystyle 2\pi R=2\pi(48-13)
\displaystyle R=35\text{ cm.}
\displaystyle \text{Area}=\pi R^2
\displaystyle =\frac{22}{7}\times35\times35=3850\text{ cm}^2.
\displaystyle \therefore \text{The area of the required circle is }3850\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The diameters of two circles are }32\text{ cm and }24\text{ cm. Find the radius of the circle}
\displaystyle \text{having its area equal to the sum of the areas of the two given circles.}
\displaystyle \text{Answer:}
\displaystyle \text{Radii of the given circles}=\frac{32}{2}=16\text{ cm and }\frac{24}{2}=12\text{ cm.}
\displaystyle \text{Let the radius of the required circle be }R\text{ cm.}
\displaystyle \pi R^2=\pi(16)^2+\pi(12)^2
\displaystyle R^2=256+144
\displaystyle R^2=400
\displaystyle R=20\text{ cm.}
\displaystyle \therefore \text{The radius of the required circle is }20\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The radius of a circle is }5\text{ m. Find the circumference of the}
\displaystyle \text{circle whose area is } 49\text{ times the area of the given circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the required circle be }R\text{ m.}
\displaystyle \pi R^2=49\times\pi(5)^2
\displaystyle R^2=49\times25
\displaystyle R=35\text{ m.}
\displaystyle \text{Circumference}=2\pi R
\displaystyle =2\times\frac{22}{7}\times35=220\text{ m.}
\displaystyle \therefore \text{The circumference of the required circle is }220\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A circle of largest area is cut from a rectangular piece of card-board with}
\displaystyle \text{dimensions }55\text{ cm and }42\text{ cm. Find the ratio between the area of the circle cut}
\displaystyle \text{and the area of the remaining card-board.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the largest circle}=42\text{ cm.}
\displaystyle \therefore \text{Radius}=21\text{ cm.}
\displaystyle \text{Area of the rectangle}=55\times42=2310\text{ cm}^2.
\displaystyle \text{Area of the circle}=\pi r^2
\displaystyle =\frac{22}{7}\times21\times21=1386\text{ cm}^2.
\displaystyle \text{Area of the remaining card-board}=2310-1386=924\text{ cm}^2.
\displaystyle \text{Required ratio}=1386:924
\displaystyle =3:2.
\displaystyle \therefore \text{The required ratio is }3:2.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The following figure shows a square card-board }ABCD\text{ of side }
\displaystyle 28\text{ cm. Four identical} \ \text{circles of largest possible size are cut from this card as shown}
\displaystyle \text{below. Find the area of the remaining card-board.} \displaystyle \text{Answer:}
\displaystyle \text{Since two identical circles fit along each side of the square,}
\displaystyle \text{diameter of each circle}=\frac{28}{2}=14\text{ cm.}
\displaystyle \therefore \text{Radius of each circle}=7\text{ cm.}
\displaystyle \text{Area of square}=28\times28=784\text{ cm}^2.
\displaystyle \text{Area of four circles}=4\times\pi r^2
\displaystyle =4\times\frac{22}{7}\times7\times7=616\text{ cm}^2.
\displaystyle \text{Area of remaining card-board}=784-616=168\text{ cm}^2.
\displaystyle \therefore \text{The area of the remaining card-board is }168\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The radii of two circles are in the ratio }3:8.\text{ If the difference}
\displaystyle \text{between their areas is }2695\pi\text{ cm}^2,\text{ find the area of the smaller circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radii of the two circles be }3x\text{ cm and }8x\text{ cm.}
\displaystyle \pi(8x)^2-\pi(3x)^2=2695\pi
\displaystyle 64x^2-9x^2=2695
\displaystyle 55x^2=2695
\displaystyle x^2=49
\displaystyle x=7.
\displaystyle \therefore \text{Radius of the smaller circle}=3\times7=21\text{ cm.}
\displaystyle \text{Area of the smaller circle}=\pi(21)^2=441\pi\text{ cm}^2
\displaystyle =\frac{22}{7}\times441=1386\text{ cm}^2.
\displaystyle \therefore \text{The area of the smaller circle is }1386\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The diameters of three circles are in the ratio }3:5:6.\text{ If the sum }
\displaystyle \text{of the circumferences of these circles is }308\text{ cm, find the difference between the}
\displaystyle \text{areas of the largest and the smallest of these circles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the diameters of the three circles be }3x,\ 5x\text{ and }6x\text{ cm.}
\displaystyle \pi(3x)+\pi(5x)+\pi(6x)=308
\displaystyle 14\pi x=308
\displaystyle 14\times\frac{22}{7}\times x=308
\displaystyle 44x=308
\displaystyle x=7.
\displaystyle \therefore \text{Smallest diameter}=21\text{ cm and largest diameter}=42\text{ cm.}
\displaystyle \text{Smallest radius}=10.5\text{ cm and largest radius}=21\text{ cm.}
\displaystyle \text{Difference between their areas}=\pi\{21^2-(10.5)^2\}
\displaystyle =\frac{22}{7}(441-110.25)
\displaystyle =\frac{22}{7}\times330.75=1039.5\text{ cm}^2.
\displaystyle \therefore \text{The difference between their areas is }1039.5\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A wheel has diameter }84\text{ cm. Find how many complete revolutions}
\displaystyle \text{it must make to cover }3.168\text{ km.}
\displaystyle \text{Answer:}
\displaystyle \text{Circumference of the wheel}=\pi d
\displaystyle =\frac{22}{7}\times84=264\text{ cm.}
\displaystyle 3.168\text{ km}=3.168\times1000\times100=316800\text{ cm.}
\displaystyle \text{Number of revolutions}=\frac{316800}{264}=1200.
\displaystyle \therefore \text{The wheel must make }1200\text{ complete revolutions.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Each wheel of a car is of diameter }80\text{ cm. How many complete}
\displaystyle \text{revolutions does each wheel make in }10\text{ minutes when the car is travelling at a} \\ \text{speed of }66\text{ km per hour?}
\displaystyle \text{Answer:}
\displaystyle \text{Distance travelled in }10\text{ minutes}=66\times\frac{10}{60}=11\text{ km.}
\displaystyle 11\text{ km}=11\times1000\times100=1100000\text{ cm.}
\displaystyle \text{Circumference of each wheel}=\pi d
\displaystyle =\frac{22}{7}\times80=\frac{1760}{7}\text{ cm.}
\displaystyle \text{Number of revolutions}=\frac{1100000}{1760/7}
\displaystyle =\frac{1100000\times7}{1760}=4375.
\displaystyle \therefore \text{Each wheel makes }4375\text{ complete revolutions.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{An express train is running between two stations with a uniform speed. If the}
\displaystyle \text{diameter of each wheel is }42\text{ cm and each wheel makes }1200\text{ revolutions per minute,}
\displaystyle \text{find the speed of the train.}
\displaystyle \text{Answer:}
\displaystyle \text{Circumference of each wheel}=\pi d
\displaystyle =\frac{22}{7}\times42=132\text{ cm.}
\displaystyle \text{Distance covered in }1\text{ minute}=132\times1200=158400\text{ cm.}
\displaystyle 158400\text{ cm}=1.584\text{ km.}
\displaystyle \text{Distance covered in }60\text{ minutes}=1.584\times60=95.04\text{ km.}
\displaystyle \therefore \text{The speed of the train is }95.04\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The minute hand of a clock is }8\text{ cm long. Find the area swept}
\displaystyle \text{by the minute hand between }8.30\text{ a.m. and }9.05\text{ a.m.}
\displaystyle \text{Answer:}
\displaystyle \text{Time elapsed from }8.30\text{ a.m. to }9.05\text{ a.m.}=35\text{ minutes.}
\displaystyle \text{Angle swept by the minute hand in }1\text{ minute}=6^\circ.
\displaystyle \therefore \text{Angle swept in }35\text{ minutes}=35\times6^\circ=210^\circ.
\displaystyle \text{Area swept}=\frac{210}{360}\times\pi\times8^2
\displaystyle =\frac{7}{12}\times\frac{22}{7}\times64
\displaystyle =\frac{352}{3}\text{ cm}^2\approx117.33\text{ cm}^2.
\displaystyle \therefore \text{The area swept by the minute hand is }117.33\text{ cm}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The shaded portion of the figure shows two concentric circles. If the}
\displaystyle \text{circumferences of the two circles are }396\text{ cm and }374\text{ cm, find the area of the shaded portion.} \displaystyle \text{Answer:}
\displaystyle \text{Let the radii of the outer and inner circles be }R\text{ cm and }r\text{ cm respectively.}
\displaystyle 2\pi R=396
\displaystyle R=\frac{396\times7}{44}=63\text{ cm.}
\displaystyle 2\pi r=374
\displaystyle r=\frac{374\times7}{44}=59.5\text{ cm.}
\displaystyle \text{Area of shaded portion}=\pi(R^2-r^2)
\displaystyle =\frac{22}{7}\{63^2-(59.5)^2\}
\displaystyle =\frac{22}{7}(63-59.5)(63+59.5)
\displaystyle =\frac{22}{7}\times3.5\times122.5
\displaystyle =1347.5\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded portion is }1347.5\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In the figure given above for Question 19, the area of the shaded portion is }770\text{ cm}^2.
\displaystyle \text{If the circumference of the outer circle is }132\text{ cm, find the width of the shaded portion.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the outer and inner radii be }R\text{ cm and }r\text{ cm respectively.}
\displaystyle 2\pi R=132
\displaystyle 2\times\frac{22}{7}\times R=132
\displaystyle R=21\text{ cm.}
\displaystyle \text{Area of shaded portion}=\pi(R^2-r^2)
\displaystyle 770=\frac{22}{7}(21^2-r^2)
\displaystyle 245=441-r^2
\displaystyle r^2=196
\displaystyle r=14\text{ cm.}
\displaystyle \text{Width of shaded portion}=R-r=21-14=7\text{ cm.}
\displaystyle \therefore \text{The width of the shaded portion is }7\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{The cost of fencing a circular field at the rate of Rs. }240\text{ per metre is Rs. }52,800.
\displaystyle \text{The field is to be ploughed at the rate of Rs. }12.50\text{ per m}^2.\text{ Find the cost of ploughing the field.}
\displaystyle \text{Answer:}
\displaystyle \text{Circumference of the field}=\frac{52,800}{240}=220\text{ m.}
\displaystyle 2\pi r=220
\displaystyle 2\times\frac{22}{7}\times r=220
\displaystyle r=35\text{ m.}
\displaystyle \text{Area of the field}=\pi r^2
\displaystyle =\frac{22}{7}\times35\times35=3850\text{ m}^2.
\displaystyle \text{Cost of ploughing}=3850\times12.50
\displaystyle =\text{Rs. }48,125.
\displaystyle \therefore \text{The cost of ploughing the field is Rs. }48,125.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Two circles touch each other externally. The sum of their areas is }58\pi\text{ cm}^2
\displaystyle \text{and the distance between their centres is }10\text{ cm. Find the radii of the two circles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radii of the two circles be }R\text{ cm and }r\text{ cm.}
\displaystyle \text{Since the circles touch externally, }R+r=10.
\displaystyle \pi R^2+\pi r^2=58\pi
\displaystyle R^2+r^2=58.
\displaystyle (R+r)^2=R^2+r^2+2Rr
\displaystyle 10^2=58+2Rr
\displaystyle 2Rr=42
\displaystyle Rr=21.
\displaystyle \text{Thus, the radii have sum }10\text{ and product }21.
\displaystyle x^2-10x+21=0
\displaystyle (x-7)(x-3)=0
\displaystyle x=7\text{ or }x=3.
\displaystyle \therefore \text{The radii of the two circles are }7\text{ cm and }3\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{The given figure shows a rectangle }ABCD\text{ inscribed in a circle as shown alongside.}
\displaystyle \text{If }AB=28\text{ cm and }BC=21\text{ cm, find the area of the shaded portion of the given figure.} \displaystyle \text{Answer:}
\displaystyle \text{Since }ABCD\text{ is a rectangle, its diagonal is the diameter of the circle.}
\displaystyle AC=\sqrt{AB^2+BC^2}
\displaystyle =\sqrt{28^2+21^2}
\displaystyle =\sqrt{784+441}=\sqrt{1225}=35\text{ cm.}
\displaystyle \therefore \text{Radius of the circle}=\frac{35}{2}=17.5\text{ cm.}
\displaystyle \text{Area of circle}=\pi r^2
\displaystyle =\frac{22}{7}\times17.5\times17.5=962.5\text{ cm}^2.
\displaystyle \text{Area of rectangle}=28\times21=588\text{ cm}^2.
\displaystyle \text{Area of shaded portion}=962.5-588=374.5\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded portion is }374.5\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{A square is inscribed in a circle of radius }7\text{ cm. Find the area of the square.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the circle}=2\times7=14\text{ cm.}
\displaystyle \text{The diagonal of the inscribed square equals the diameter of the circle.}
\displaystyle \therefore \text{Diagonal of the square}=14\text{ cm.}
\displaystyle \text{Let the side of the square be }a\text{ cm.}
\displaystyle a\sqrt{2}=14
\displaystyle a=\frac{14}{\sqrt{2}}=7\sqrt{2}\text{ cm.}
\displaystyle \text{Area of square}=a^2=(7\sqrt{2})^2=98\text{ cm}^2.
\displaystyle \therefore \text{The area of the square is }98\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{A metal wire, when bent in the form of an equilateral triangle of largest}
\displaystyle \text{area, encloses an area of }484\sqrt{3}\text{ cm}^2.\text{ If the same wire is bent into the form of a circle}
\displaystyle \text{of largest area, find the area of this circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the equilateral triangle be }a\text{ cm.}
\displaystyle \frac{\sqrt{3}}{4}a^2=484\sqrt{3}
\displaystyle a^2=1936
\displaystyle a=44\text{ cm.}
\displaystyle \text{Length of wire}=3\times44=132\text{ cm.}
\displaystyle \text{Let the radius of the circle be }r\text{ cm.}
\displaystyle 2\pi r=132
\displaystyle 2\times\frac{22}{7}\times r=132
\displaystyle r=21\text{ cm.}
\displaystyle \text{Area of circle}=\pi r^2
\displaystyle =\frac{22}{7}\times21\times21=1386\text{ cm}^2.
\displaystyle \therefore \text{The area of the circle is }1386\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{The diameters of the front and rear wheels of a tractor are }63\text{ cm and }
\displaystyle 1.54\text{ m} \ \text{respectively. The rear wheel is rotating at }24\frac{6}{11}\text{ revolutions per minute. Find:}
\displaystyle \text{(i) the revolutions per minute made by the front wheel,}
\displaystyle \text{(ii) the distance travelled by the tractor in }40\text{ minutes.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of rear wheel}=1.54\text{ m}=154\text{ cm.}
\displaystyle 24\frac{6}{11}=\frac{270}{11}\text{ revolutions per minute.}
\displaystyle \text{(i) Let the front wheel make }n\text{ revolutions per minute.}
\displaystyle \pi\times63\times n=\pi\times154\times\frac{270}{11}
\displaystyle n=\frac{154\times270}{11\times63}
\displaystyle =60.
\displaystyle \therefore \text{The front wheel makes }60\text{ revolutions per minute.}

\displaystyle \text{(ii) Circumference of the front wheel}=\pi d
\displaystyle =\frac{22}{7}\times63=198\text{ cm}=1.98\text{ m.}
\displaystyle \text{Distance travelled in }1\text{ minute}=1.98\times60=118.8\text{ m.}
\displaystyle \text{Distance travelled in }40\text{ minutes}=118.8\times40=4752\text{ m.}
\displaystyle =4.752\text{ km.}
\displaystyle \therefore \text{The tractor travels }4.752\text{ km in }40\text{ minutes.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If a square is inscribed in a circle, find the ratio of the}
\displaystyle \text{areas of the circle and the square.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the circle be }r.
\displaystyle \text{Since the square is inscribed in the circle, its diagonal}=2r.
\displaystyle \text{Let the side of the square be }a.
\displaystyle a\sqrt{2}=2r
\displaystyle \therefore a=r\sqrt{2}.
\displaystyle \text{Area of the circle}=\pi r^2.
\displaystyle \text{Area of the square}=a^2=(r\sqrt{2})^2=2r^2.
\displaystyle \therefore \text{Required ratio}=\pi r^2:2r^2=\pi:2.
\displaystyle =\frac{22}{7}:2=11:7.
\displaystyle \therefore \text{The required ratio is }11:7.
\displaystyle \\

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