\displaystyle \textbf{Question 1: }\text{Find the area of a triangle with base }16\text{ cm and height }7\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of a triangle}=\frac{1}{2}\times\text{Base}\times\text{Height}
\displaystyle =\frac{1}{2}\times16\times7=56\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }56\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The area of a triangle is }48\text{ cm}^2.\text{ Its base is }12\text{ cm. What is its altitude?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the altitude of the triangle be }h\text{ cm.}
\displaystyle \text{Area of a triangle}=\frac{1}{2}\times\text{Base}\times\text{Height}
\displaystyle 48=\frac{1}{2}\times12\times h
\displaystyle 48=6h
\displaystyle h=\frac{48}{6}=8\text{ cm.}
\displaystyle \therefore \text{The altitude of the triangle is }8\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the area of an equilateral triangle whose perimeter is }12\text{ m.}
\displaystyle \text{Answer:}
\displaystyle \text{Perimeter}=12\text{ m.}
\displaystyle \therefore \text{Side of the equilateral triangle}=\frac{12}{3}=4\text{ m.}
\displaystyle \text{Let }AD\perp BC.
\displaystyle \text{Then }BD=\frac{1}{2}BC=2\text{ m.}
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AD^2+BD^2=AB^2
\displaystyle AD^2+2^2=4^2
\displaystyle AD^2=12
\displaystyle AD=2\sqrt{3}\text{ m.}
\displaystyle \text{Area of the triangle}=\frac{1}{2}\times4\times2\sqrt{3}=4\sqrt{3}\text{ m}^2.
\displaystyle \therefore \text{The area of the equilateral triangle is }4\sqrt{3}\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the area of a right-angled triangle if the radius of its circumcircle is }3\text{ cm}
\displaystyle \text{and the altitude drawn to the hypotenuse is }2\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the circumcircle}=3\text{ cm.}
\displaystyle \therefore \text{Diameter}=2\times3=6\text{ cm.}
\displaystyle \text{Since the triangle is right-angled, the hypotenuse is the diameter of its circumcircle.}
\displaystyle \therefore \text{Hypotenuse}=6\text{ cm.}
\displaystyle \text{Area of the triangle}=\frac{1}{2}\times\text{Hypotenuse}\times\text{Altitude}
\displaystyle =\frac{1}{2}\times6\times2=6\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }6\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the area of a right-angled triangle if the diameter of its circumcircle is }10\text{ cm}
\displaystyle \text{and the altitude drawn to the hypotenuse is }4.5\text{ cm long.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the circumcircle}=10\text{ cm.}
\displaystyle \text{Since the triangle is right-angled, the hypotenuse is the diameter of its circumcircle.}
\displaystyle \therefore \text{Hypotenuse}=10\text{ cm.}
\displaystyle \text{Area of the triangle}=\frac{1}{2}\times\text{Hypotenuse}\times\text{Altitude}
\displaystyle =\frac{1}{2}\times10\times4.5=22.5\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }22.5\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The perimeter of a right triangle is }60\text{ cm. Its hypotenuse is }26\text{ cm. Find}
\displaystyle \text{the other two sides and the area of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Perimeter of the triangle}=60\text{ cm.}
\displaystyle \text{Sum of the other two sides}=60-26=34\text{ cm.}
\displaystyle \text{Let one side be }x\text{ cm. Then the other side is }(34-x)\text{ cm.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle x^2+(34-x)^2=26^2
\displaystyle x^2+1156-68x+x^2=676
\displaystyle 2x^2-68x+480=0
\displaystyle x^2-34x+240=0
\displaystyle (x-24)(x-10)=0
\displaystyle x=24\text{ or }x=10.
\displaystyle \therefore \text{The other two sides are }10\text{ cm and }24\text{ cm.}
\displaystyle \text{Area of the triangle}=\frac{1}{2}\times10\times24=120\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }120\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The cost of turfing a triangular field at the rate of Rs. }45\text{ per }100\text{ m}^2\text{ is}
\displaystyle \text{Rs. }900.\text{ Find the height, if double the base of the triangle is }5\text{ times the height.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the base of the triangle be }x\text{ m.}
\displaystyle \text{Then, }2\times\text{Base}=5\times\text{Height}
\displaystyle \therefore \text{Height}=\frac{2}{5}x\text{ m.}
\displaystyle \text{Rate of turfing}=\text{Rs. }\frac{45}{100}\text{ per m}^2.
\displaystyle \text{Area of the triangle}\times\frac{45}{100}=900
\displaystyle \frac{1}{2}\times x\times\frac{2}{5}x\times\frac{45}{100}=900
\displaystyle \frac{9x^2}{100}=900
\displaystyle x^2=10000
\displaystyle x=100\text{ m.}
\displaystyle \therefore \text{Height}=\frac{2}{5}\times100=40\text{ m.}
\displaystyle \therefore \text{The height of the triangle is }40\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The base of a triangular field is three times its height. If the cost of cultivating the}
\displaystyle \text{field at Rs. }36\text{ per hectare is Rs. }486,\text{ find its base and height.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the triangle be }h\text{ m.}
\displaystyle \therefore \text{Base}=3h\text{ m.}
\displaystyle \text{Area of the triangle}=\frac{1}{2}\times3h\times h=\frac{3}{2}h^2\text{ m}^2.
\displaystyle 1\text{ hectare}=10000\text{ m}^2.
\displaystyle \text{Rate of cultivation}=\text{Rs. }\frac{36}{10000}\text{ per m}^2.
\displaystyle \frac{3}{2}h^2\times\frac{36}{10000}=486
\displaystyle h^2=90000
\displaystyle h=300\text{ m.}
\displaystyle \therefore \text{Base}=3\times300=900\text{ m.}
\displaystyle \therefore \text{The height is }300\text{ m and the base is }900\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{An isosceles right triangle has area }200\text{ cm}^2.\text{ What is the length of its}
\displaystyle \text{hypotenuse?}
\displaystyle \text{Answer:}
\displaystyle \text{Let each perpendicular side of the isosceles right triangle be }x\text{ cm.}
\displaystyle \frac{1}{2}\times x\times x=200
\displaystyle x^2=400
\displaystyle x=20\text{ cm.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{20^2+20^2}
\displaystyle =\sqrt{800}=20\sqrt{2}\text{ cm}
\displaystyle \approx28.28\text{ cm.}
\displaystyle \therefore \text{The length of the hypotenuse is }20\sqrt{2}\text{ cm}\approx28.28\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the base of an isosceles triangle whose area is }60\text{ cm}^2\text{ and the length}
\displaystyle \text{of each of its equal sides is }13\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Let half the base be }a\text{ cm. Therefore, the base is }2a\text{ cm.}
\displaystyle \text{The altitude of an isosceles triangle bisects its base.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle h=\sqrt{13^2-a^2}=\sqrt{169-a^2}.
\displaystyle \text{Area}=\frac{1}{2}\times2a\times\sqrt{169-a^2}=60
\displaystyle a\sqrt{169-a^2}=60
\displaystyle a^2(169-a^2)=3600
\displaystyle a^4-169a^2+3600=0
\displaystyle (a^2-144)(a^2-25)=0
\displaystyle a^2=144\text{ or }a^2=25
\displaystyle a=12\text{ or }a=5.
\displaystyle \therefore \text{Base}=2a=24\text{ cm or }10\text{ cm.}
\displaystyle \therefore \text{The possible lengths of the base are }24\text{ cm and }10\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The perimeter of an isosceles triangle is }40\text{ cm. The base is two-third of the sum}
\displaystyle \text{of its equal sides. Find the length of each side.}
\displaystyle \text{Answer:}
\displaystyle \text{Let each of the two equal sides be }x\text{ cm.}
\displaystyle \text{Given, base}=\frac{2}{3}(2x)=\frac{4x}{3}\text{ cm.}
\displaystyle x+x+\frac{4x}{3}=40
\displaystyle \frac{10x}{3}=40
\displaystyle 10x=120
\displaystyle x=12\text{ cm.}
\displaystyle \therefore \text{Base}=\frac{4}{3}\times12=16\text{ cm.}
\displaystyle \therefore \text{The sides of the triangle are }12\text{ cm, }12\text{ cm and }16\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the area of an isosceles triangle whose equal sides are }12\text{ cm each and}
\displaystyle \text{the perimeter is }30\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Base}=30-12-12=6\text{ cm.}
\displaystyle \text{The altitude of an isosceles triangle bisects its base.}
\displaystyle \therefore \text{Half the base}=3\text{ cm.}
\displaystyle \text{Let the height be }h\text{ cm.}
\displaystyle h=\sqrt{12^2-3^2}
\displaystyle =\sqrt{144-9}=\sqrt{135}=3\sqrt{15}\text{ cm.}
\displaystyle \text{Area}=\frac{1}{2}\times6\times3\sqrt{15}
\displaystyle =9\sqrt{15}\text{ cm}^2\approx34.86\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }9\sqrt{15}\text{ cm}^2\approx34.86\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the area of an isosceles triangle whose base is }6\text{ cm and perimeter is }16\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Let each equal side be }x\text{ cm.}
\displaystyle 2x+6=16
\displaystyle 2x=10
\displaystyle x=5\text{ cm.}
\displaystyle \text{The altitude of an isosceles triangle bisects its base.}
\displaystyle \therefore \text{Half the base}=3\text{ cm.}
\displaystyle \text{Let the height be }h\text{ cm.}
\displaystyle h=\sqrt{5^2-3^2}
\displaystyle =\sqrt{25-9}=4\text{ cm.}
\displaystyle \text{Area}=\frac{1}{2}\times6\times4=12\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }12\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The sides of a right triangle containing the right angle are }5x\text{ cm and}
\displaystyle (3x-1)\text{ cm. Calculate the length of the hypotenuse if its area is }60\text{ cm}^2.
\displaystyle \text{Answer:}
\displaystyle \text{Area of the triangle}=60\text{ cm}^2.
\displaystyle \frac{1}{2}\times5x\times(3x-1)=60
\displaystyle 5x(3x-1)=120
\displaystyle x(3x-1)=24
\displaystyle 3x^2-x-24=0
\displaystyle (3x+8)(x-3)=0
\displaystyle x=3\text{ or }x=-\frac{8}{3}.
\displaystyle \text{Rejecting }x=-\frac{8}{3},\text{ since }x\text{ must be positive, }x=3.
\displaystyle \therefore \text{The perpendicular sides are }15\text{ cm and }8\text{ cm.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{15^2+8^2}
\displaystyle =\sqrt{225+64}=\sqrt{289}=17\text{ cm.}
\displaystyle \therefore \text{The length of the hypotenuse is }17\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{ABC is a right triangle right angled at }B.\text{ If }AB=(2x+1)\text{ cm,}
\displaystyle \text{BC}=(x+1)\text{ cm and area of }\triangle ABC\text{ is }60\text{ cm}^2,\text{ find its perimeter.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle ABC=60\text{ cm}^2.
\displaystyle \frac{1}{2}(x+1)(2x+1)=60
\displaystyle (x+1)(2x+1)=120
\displaystyle 2x^2+3x-119=0
\displaystyle (2x+17)(x-7)=0
\displaystyle x=7\text{ or }x=-\frac{17}{2}.
\displaystyle \text{Rejecting }x=-\frac{17}{2},\text{ since }x\text{ must be positive, }x=7.
\displaystyle \therefore AB=2(7)+1=15\text{ cm and }BC=7+1=8\text{ cm.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle AC=\sqrt{15^2+8^2}=\sqrt{289}=17\text{ cm.}
\displaystyle \text{Perimeter}=15+8+17=40\text{ cm.}
\displaystyle \therefore \text{The perimeter of the triangle is }40\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In the adjoining figure, }ABC\text{ is an equilateral triangle with each side of length}
\displaystyle 10\text{ cm. }D\text{ is a point inside }\triangle ABC\text{ such that }\angle BDC=90^\circ\text{ and }CD=6\text{ cm.}
\displaystyle \text{Find the area of the shaded region.}
\displaystyle \text{Answer:}
\displaystyle \text{Height of equilateral }\triangle ABC=\sqrt{10^2-5^2}
\displaystyle =\sqrt{75}=5\sqrt{3}\text{ cm.}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times10\times5\sqrt{3}
\displaystyle =25\sqrt{3}\text{ cm}^2.
\displaystyle \text{In right-angled }\triangle BDC,\ BC=10\text{ cm and }CD=6\text{ cm.}
\displaystyle BD=\sqrt{10^2-6^2}=\sqrt{64}=8\text{ cm.}
\displaystyle \text{Area of }\triangle BDC=\frac{1}{2}\times6\times8=24\text{ cm}^2.
\displaystyle \text{Area of shaded region}=25\sqrt{3}-24\text{ cm}^2
\displaystyle \approx19.30\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }25\sqrt{3}-24\text{ cm}^2\approx19.30\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If the difference between the two sides of a right-angled triangle is }2\text{ cm and}
\displaystyle \text{the area of the triangle is }24\text{ cm}^2,\text{ find the perimeter of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the perpendicular sides be }x\text{ cm and }(x+2)\text{ cm.}
\displaystyle \frac{1}{2}\times x\times(x+2)=24
\displaystyle x(x+2)=48
\displaystyle x^2+2x-48=0
\displaystyle (x+8)(x-6)=0
\displaystyle x=6\text{ or }x=-8.
\displaystyle \text{Rejecting }x=-8,\text{ since length cannot be negative, }x=6.
\displaystyle \therefore \text{The perpendicular sides are }6\text{ cm and }8\text{ cm.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{6^2+8^2}
\displaystyle =\sqrt{36+64}=\sqrt{100}=10\text{ cm.}
\displaystyle \text{Perimeter}=6+8+10=24\text{ cm.}
\displaystyle \therefore \text{The perimeter of the triangle is }24\text{ cm.}
\displaystyle \\

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