Note: If \displaystyle a, b, c are the sides of the triangle and s is the semi perimeter, then its area if given by \displaystyle A = \sqrt{s(s-a)(s-b)(s-c)}  \text{ Where } s = \frac{a+b+c}{2} . This is Heron’s Formula.

\displaystyle \textbf{Question 1: }\text{Find the area of a triangle whose sides are }150\text{ cm, }120\text{ cm and }200\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Here }a=150,\ b=120\text{ and }c=200.
\displaystyle s=\frac{a+b+c}{2}=\frac{150+120+200}{2}=235.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{235(235-150)(235-120)(235-200)}
\displaystyle =\sqrt{235\times85\times115\times35}
\displaystyle \approx8966.57\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }8966.57\text{ cm}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the area of a triangle whose sides are }9\text{ cm, }12\text{ cm and }15\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Here }a=9,\ b=12\text{ and }c=15.
\displaystyle s=\frac{a+b+c}{2}=\frac{9+12+15}{2}=18.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{18(18-9)(18-12)(18-15)}
\displaystyle =\sqrt{18\times9\times6\times3}
\displaystyle =54\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }54\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the area of a triangle two sides of which are }18\text{ cm and }10\text{ cm,}
\displaystyle \text{and the perimeter is }42\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Perimeter}=42\text{ cm.}
\displaystyle \therefore \text{Third side}=42-18-10=14\text{ cm.}
\displaystyle \text{Here }a=18,\ b=10\text{ and }c=14.
\displaystyle s=\frac{a+b+c}{2}=\frac{18+10+14}{2}=21.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{21(21-18)(21-10)(21-14)}
\displaystyle =\sqrt{21\times3\times11\times7}
\displaystyle =\sqrt{4851}\approx69.65\text{ cm}^2.
\displaystyle \therefore \text{The area of the triangle is }69.65\text{ cm}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In }\triangle ABC,\ AB=15\text{ cm, }BC=13\text{ cm and }AC=14\text{ cm. Find the area}
\displaystyle \text{of }\triangle ABC\text{ and hence its altitude on }AC.
\displaystyle \text{Answer:}
\displaystyle \text{Here }a=15,\ b=13\text{ and }c=14.
\displaystyle s=\frac{a+b+c}{2}=\frac{15+13+14}{2}=21.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{21(21-15)(21-13)(21-14)}
\displaystyle =\sqrt{21\times6\times8\times7}
\displaystyle =84\text{ cm}^2.
\displaystyle \text{Let the altitude on }AC\text{ be }h\text{ cm.}
\displaystyle \frac{1}{2}\times14\times h=84
\displaystyle 7h=84
\displaystyle h=12\text{ cm.}
\displaystyle \therefore \text{The area of }\triangle ABC\text{ is }84\text{ cm}^2\text{ and the altitude on }AC\text{ is }12\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The perimeter of a triangular field is }540\text{ m and its sides are in the ratio }25:17:12.
\displaystyle \text{Find the area of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Ratio of sides}=25:17:12.
\displaystyle \text{Let the sides be }25x,\ 17x\text{ and }12x\text{ m.}
\displaystyle 25x+17x+12x=540
\displaystyle 54x=540
\displaystyle x=10.
\displaystyle \therefore \text{The sides are }250\text{ m, }170\text{ m and }120\text{ m.}
\displaystyle s=\frac{250+170+120}{2}=270\text{ m.}
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{270(270-250)(270-170)(270-120)}
\displaystyle =\sqrt{270\times20\times100\times150}
\displaystyle =\sqrt{81000000}=9000\text{ m}^2.
\displaystyle \therefore \text{The area of the triangular field is }9000\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The perimeter of a triangle is }300\text{ m. If its sides are in the ratio }3:5:7,
\displaystyle \text{find the area of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Ratio of sides}=3:5:7.
\displaystyle \text{Let the sides be }3x,\ 5x\text{ and }7x\text{ m.}
\displaystyle 3x+5x+7x=300
\displaystyle 15x=300
\displaystyle x=20.
\displaystyle \therefore \text{The sides are }60\text{ m, }100\text{ m and }140\text{ m.}
\displaystyle s=\frac{60+100+140}{2}=150\text{ m.}
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{150(150-60)(150-100)(150-140)}
\displaystyle =\sqrt{150\times90\times50\times10}
\displaystyle =\sqrt{6750000}=1500\sqrt{3}\text{ m}^2.
\displaystyle \therefore \text{The area of the triangle is }1500\sqrt{3}\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The perimeter of a triangular field is }240\text{ dm. If two of its sides are }78\text{ dm}
\displaystyle \text{and }50\text{ dm, find the length of the perpendicular on the side of length }50\text{ dm from the}
\displaystyle \text{opposite vertex.}
\displaystyle \text{Answer:}
\displaystyle \text{Perimeter}=240\text{ dm.}
\displaystyle \therefore \text{Third side}=240-78-50=112\text{ dm.}
\displaystyle \text{Here }a=112,\ b=78\text{ and }c=50.
\displaystyle s=\frac{a+b+c}{2}=\frac{112+78+50}{2}=120\text{ dm.}
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{120(120-112)(120-78)(120-50)}
\displaystyle =\sqrt{120\times8\times42\times70}
\displaystyle =1680\text{ dm}^2.
\displaystyle \text{Let the perpendicular on the side }50\text{ dm be }h\text{ dm.}
\displaystyle \frac{1}{2}\times50\times h=1680
\displaystyle h=\frac{2\times1680}{50}=67.2\text{ dm.}
\displaystyle \therefore \text{The required perpendicular is }67.2\text{ dm.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A triangle has sides }35\text{ cm, }54\text{ cm and }61\text{ cm. Find its area. Also, find}
\displaystyle \text{the smallest of its altitudes.}
\displaystyle \text{Answer:}
\displaystyle \text{Here }a=35,\ b=54\text{ and }c=61.
\displaystyle s=\frac{a+b+c}{2}=\frac{35+54+61}{2}=75\text{ cm.}
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{75(75-35)(75-54)(75-61)}
\displaystyle =\sqrt{75\times40\times21\times14}
\displaystyle =\sqrt{882000}=420\sqrt{5}\text{ cm}^2
\displaystyle \approx939.15\text{ cm}^2.
\displaystyle \text{The smallest altitude corresponds to the largest side, }61\text{ cm.}
\displaystyle \text{Let the smallest altitude be }h\text{ cm.}
\displaystyle \frac{1}{2}\times61\times h=420\sqrt{5}
\displaystyle h=\frac{840\sqrt{5}}{61}\approx30.79\text{ cm.}
\displaystyle \therefore \text{The area is }420\sqrt{5}\text{ cm}^2\approx939.15\text{ cm}^2\text{ and the smallest altitude is }30.79\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The lengths of the sides of a triangle are in the ratio }3:4:5\text{ and its perimeter}
\displaystyle \text{is }144\text{ cm. Find the area of the triangle and the height corresponding to the longest side.}
\displaystyle \text{Answer:}
\displaystyle \text{Ratio of sides}=3:4:5.
\displaystyle \text{Let the sides be }3x,\ 4x\text{ and }5x\text{ cm.}
\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 s=\frac{36+48+60}{2}=72\text{ cm.}
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{72(72-36)(72-48)(72-60)}
\displaystyle =\sqrt{72\times36\times24\times12}
\displaystyle =864\text{ cm}^2.
\displaystyle \text{Let the altitude corresponding to the longest side }60\text{ cm be }h\text{ cm.}
\displaystyle \frac{1}{2}\times60\times h=864
\displaystyle h=\frac{2\times864}{60}=28.8\text{ cm.}
\displaystyle \therefore \text{The area is }864\text{ cm}^2\text{ and the required height is }28.8\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The perimeter of an isosceles triangle is }42\text{ cm and its base is }\frac{3}{2}\text{ times}
\displaystyle \text{each of the equal sides. Find the length of each side, area and height of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let each equal side be }x\text{ cm.}
\displaystyle \therefore \text{Base}=\frac{3x}{2}\text{ cm.}
\displaystyle x+x+\frac{3x}{2}=42
\displaystyle \frac{7x}{2}=42
\displaystyle x=12.
\displaystyle \therefore \text{The sides are }12\text{ cm, }12\text{ cm and }18\text{ cm.}
\displaystyle s=\frac{12+12+18}{2}=21\text{ cm.}
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{21(21-12)(21-12)(21-18)}
\displaystyle =\sqrt{21\times9\times9\times3}
\displaystyle =27\sqrt{7}\text{ cm}^2\approx71.44\text{ cm}^2.
\displaystyle \text{Let the height corresponding to the base }18\text{ cm be }h\text{ cm.}
\displaystyle \frac{1}{2}\times18\times h=27\sqrt{7}
\displaystyle 9h=27\sqrt{7}
\displaystyle h=3\sqrt{7}\text{ cm}\approx7.94\text{ cm.}
\displaystyle \therefore \text{The sides are }12\text{ cm, }12\text{ cm and }18\text{ cm, the area is }27\sqrt{7}\text{ cm}^2
\displaystyle \text{and the height is }3\sqrt{7}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the area of quadrilateral }ABCD\text{ in which }AB=3\text{ cm, }BC=4\text{ cm,}
\displaystyle CD=4\text{ cm, }DA=5\text{ cm and }AC=5\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of }ABCD=\text{Area of }\triangle ABC+\text{Area of }\triangle ACD.
\displaystyle \text{For }\triangle ABC,\ a=3,\ b=4\text{ and }c=5.
\displaystyle s=\frac{a+b+c}{2}=\frac{3+4+5}{2}=6.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{6(6-3)(6-4)(6-5)}
\displaystyle =\sqrt{6\times3\times2\times1}=6\text{ cm}^2.
\displaystyle \text{For }\triangle ACD,\ a=5,\ b=4\text{ and }c=5.
\displaystyle s=\frac{a+b+c}{2}=\frac{5+4+5}{2}=7.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ACD=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{7(7-5)(7-4)(7-5)}
\displaystyle =\sqrt{7\times2\times3\times2}
\displaystyle =\sqrt{84}=2\sqrt{21}\text{ cm}^2\approx9.165\text{ cm}^2.
\displaystyle \text{Area of }ABCD=6+2\sqrt{21}\text{ cm}^2
\displaystyle \approx15.165\text{ cm}^2.
\displaystyle \therefore \text{The area of quadrilateral }ABCD\text{ is }6+2\sqrt{21}\text{ cm}^2\approx15.165\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The sides of a quadrangular field, taken in order, are }26\text{ m, }27\text{ m, }7\text{ m}
\displaystyle \text{and }24\text{ m respectively. The angle contained by the last two sides is a right angle. Find its area.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the angle between the sides }7\text{ m and }24\text{ m is }90^\circ,
\displaystyle AC=\sqrt{7^2+24^2}
\displaystyle =\sqrt{49+576}=\sqrt{625}=25\text{ m.}
\displaystyle \text{Area of }ABCD=\text{Area of }\triangle ABC+\text{Area of }\triangle ACD.
\displaystyle \text{For }\triangle ABC,\ a=26,\ b=27\text{ and }c=25.
\displaystyle s=\frac{a+b+c}{2}=\frac{26+27+25}{2}=39.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{39(39-26)(39-27)(39-25)}
\displaystyle =\sqrt{39\times13\times12\times14}
\displaystyle \approx291.85\text{ m}^2.
\displaystyle \text{For }\triangle ACD,\ a=25,\ b=7\text{ and }c=24.
\displaystyle s=\frac{25+7+24}{2}=28.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ACD=\sqrt{28(28-25)(28-7)(28-24)}
\displaystyle =\sqrt{28\times3\times21\times4}
\displaystyle =84\text{ m}^2.
\displaystyle \text{Area of }ABCD=291.85+84=375.85\text{ m}^2.
\displaystyle \therefore \text{The area of the quadrangular field is }375.85\text{ m}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The sides of a quadrilateral, taken in order, are }5\text{ m, }12\text{ m, }14\text{ m}
\displaystyle \text{and }15\text{ m respectively, and the angle contained by the first two sides is a right angle. Find its area.}
\displaystyle \text{Answer:}
\displaystyle AC=\sqrt{5^2+12^2}
\displaystyle =\sqrt{25+144}=\sqrt{169}=13\text{ m.}
\displaystyle \text{Area of }ABCD=\text{Area of }\triangle ABC+\text{Area of }\triangle ACD.
\displaystyle \text{For }\triangle ABC,\ a=5,\ b=12\text{ and }c=13.
\displaystyle s=\frac{a+b+c}{2}=\frac{5+12+13}{2}=15.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{15(15-5)(15-12)(15-13)}
\displaystyle =\sqrt{15\times10\times3\times2}
\displaystyle =30\text{ m}^2.
\displaystyle \text{For }\triangle ACD,\ a=13,\ b=14\text{ and }c=15.
\displaystyle s=\frac{13+14+15}{2}=21.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ACD=\sqrt{21(21-13)(21-14)(21-15)}
\displaystyle =\sqrt{21\times8\times7\times6}
\displaystyle =84\text{ m}^2.
\displaystyle \text{Area of }ABCD=30+84=114\text{ m}^2.
\displaystyle \therefore \text{The area of the quadrilateral is }114\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{A park, in the shape of a quadrilateral }ABCD,\text{ has }\angle C=90^\circ,\ AB=9\text{ m,}
\displaystyle BC=12\text{ m, }CD=5\text{ m and }AD=8\text{ m. How much area does it occupy?}
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle BCD,
\displaystyle BD=\sqrt{BC^2+CD^2}
\displaystyle =\sqrt{12^2+5^2}=\sqrt{169}=13\text{ m.}
\displaystyle \text{Area of }ABCD=\text{Area of }\triangle ABD+\text{Area of }\triangle BCD.
\displaystyle \text{For }\triangle ABD,\ a=9,\ b=13\text{ and }c=8.
\displaystyle s=\frac{a+b+c}{2}=\frac{9+13+8}{2}=15.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ABD=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{15(15-9)(15-13)(15-8)}
\displaystyle =\sqrt{15\times6\times2\times7}
\displaystyle =6\sqrt{35}\text{ m}^2\approx35.50\text{ m}^2.
\displaystyle \text{For }\triangle BCD,\ a=12,\ b=5\text{ and }c=13.
\displaystyle s=\frac{12+5+13}{2}=15.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle BCD=\sqrt{15(15-12)(15-5)(15-13)}
\displaystyle =\sqrt{15\times3\times10\times2}
\displaystyle =30\text{ m}^2.
\displaystyle \text{Area of }ABCD=6\sqrt{35}+30\text{ m}^2
\displaystyle \approx65.50\text{ m}^2.
\displaystyle \therefore \text{The area occupied by the park is }65.50\text{ m}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The two parallel sides of a trapezium are }60\text{ cm and }77\text{ cm, and its}
\displaystyle \text{non-parallel sides are }25\text{ cm and }26\text{ cm. Find the area of the trapezium.}
\displaystyle \text{Answer:}
\displaystyle \text{Difference between the parallel sides}=77-60=17\text{ cm.}
\displaystyle \text{Area of }ABCD=\text{Area of parallelogram }AECD+\text{Area of }\triangle BCE.
\displaystyle \text{For }\triangle BCE,\ a=17,\ b=26\text{ and }c=25.
\displaystyle s=\frac{a+b+c}{2}=\frac{17+26+25}{2}=34.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle BCE=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{34(34-17)(34-26)(34-25)}
\displaystyle =\sqrt{34\times17\times8\times9}
\displaystyle =204\text{ cm}^2.
\displaystyle \text{Let }CF=h\text{ be the height of the trapezium.}
\displaystyle \frac{1}{2}\times17\times h=204
\displaystyle h=\frac{2\times204}{17}=24\text{ cm.}
\displaystyle \text{Area of parallelogram }AECD=60\times24=1440\text{ cm}^2.
\displaystyle \text{Area of }ABCD=1440+204=1644\text{ cm}^2.
\displaystyle \therefore \text{The area of the trapezium is }1644\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Find the area of a rhombus whose perimeter is }80\text{ m and one of whose diagonals is}
\displaystyle 24\text{ m.}
\displaystyle \text{Answer:}
\displaystyle \text{Perimeter}=80\text{ m.}
\displaystyle \therefore \text{Side of the rhombus}=\frac{80}{4}=20\text{ m.}
\displaystyle \text{Diagonal }BD=24\text{ m divides the rhombus into two congruent triangles.}
\displaystyle \text{For }\triangle ABD,\ a=20,\ b=20\text{ and }c=24.
\displaystyle s=\frac{a+b+c}{2}=\frac{20+20+24}{2}=32.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ABD=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{32(32-20)(32-20)(32-24)}
\displaystyle =\sqrt{32\times12\times12\times8}
\displaystyle =192\text{ m}^2.
\displaystyle \text{Since }\triangle ABD\cong\triangle BCD,
\displaystyle \text{Area of }\triangle BCD=192\text{ m}^2.
\displaystyle \text{Area of rhombus }ABCD=192+192=384\text{ m}^2.
\displaystyle \therefore \text{The area of the rhombus is }384\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{A rhombus sheet, whose perimeter is }32\text{ m and whose one diagonal is }10\text{ m}
\displaystyle \text{long, is painted on both sides at the rate of Rs. }5\text{ per m}^2.\text{ Find the cost of painting.}
\displaystyle \text{Answer:}
\displaystyle \text{Perimeter of the rhombus}=32\text{ m.}
\displaystyle \therefore \text{Side of the rhombus}=\frac{32}{4}=8\text{ m.}
\displaystyle \text{The diagonal divides the rhombus into two congruent triangles.}
\displaystyle \text{For one triangle, }a=8,\ b=8\text{ and }c=10.
\displaystyle s=\frac{a+b+c}{2}=\frac{8+8+10}{2}=13.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of one triangle}=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{13(13-8)(13-8)(13-10)}
\displaystyle =\sqrt{13\times5\times5\times3}
\displaystyle =5\sqrt{39}\text{ m}^2.
\displaystyle \text{Area of rhombus}=2\times5\sqrt{39}=10\sqrt{39}\text{ m}^2.
\displaystyle \text{Area painted on both sides}=2\times10\sqrt{39}=20\sqrt{39}\text{ m}^2.
\displaystyle \text{Cost of painting}=20\sqrt{39}\times5
\displaystyle =100\sqrt{39}\text{ Rs.}\approx624.50\text{ Rs.}
\displaystyle \therefore \text{The cost of painting is Rs. }624.50\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Find the area of quadrilateral }ABCD\text{ in which }AD=24\text{ cm, }\angle BAD=90^\circ
\displaystyle \text{and }\triangle BCD\text{ is an equilateral triangle with each side equal to }26\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }\triangle BCD\text{ is equilateral, }BD=26\text{ cm.}
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AB=\sqrt{BD^2-AD^2}
\displaystyle =\sqrt{26^2-24^2}=\sqrt{100}=10\text{ cm.}
\displaystyle \text{Area of }ABCD=\text{Area of }\triangle ABD+\text{Area of }\triangle BCD.
\displaystyle \text{For }\triangle ABD,\ a=10,\ b=26\text{ and }c=24.
\displaystyle s=\frac{10+26+24}{2}=30.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ABD=\sqrt{30(30-10)(30-26)(30-24)}
\displaystyle =\sqrt{30\times20\times4\times6}
\displaystyle =120\text{ cm}^2.
\displaystyle \text{For }\triangle BCD,\ a=26,\ b=26\text{ and }c=26.
\displaystyle s=\frac{26+26+26}{2}=39.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle BCD=\sqrt{39(39-26)(39-26)(39-26)}
\displaystyle =\sqrt{39\times13\times13\times13}
\displaystyle =169\sqrt{3}\text{ cm}^2\approx292.72\text{ cm}^2.
\displaystyle \text{Area of }ABCD=120+169\sqrt{3}\text{ cm}^2
\displaystyle \approx412.72\text{ cm}^2.
\displaystyle \therefore \text{The area of quadrilateral }ABCD\text{ is }412.72\text{ cm}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Find the area of quadrilateral }ABCD\text{ in which }AB=42\text{ cm, }BC=21\text{ cm,}
\displaystyle CD=29\text{ cm, }DA=34\text{ cm and diagonal }BD=20\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of }ABCD=\text{Area of }\triangle ABD+\text{Area of }\triangle BCD.
\displaystyle \text{For }\triangle ABD,\ a=42,\ b=20\text{ and }c=34.
\displaystyle s=\frac{a+b+c}{2}=\frac{42+20+34}{2}=48.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ABD=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{48(48-42)(48-20)(48-34)}
\displaystyle =\sqrt{48\times6\times28\times14}
\displaystyle =336\text{ cm}^2.
\displaystyle \text{For }\triangle BCD,\ a=21,\ b=29\text{ and }c=20.
\displaystyle s=\frac{a+b+c}{2}=\frac{21+29+20}{2}=35.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle BCD=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{35(35-21)(35-29)(35-20)}
\displaystyle =\sqrt{35\times14\times6\times15}
\displaystyle =210\text{ cm}^2.
\displaystyle \text{Area of }ABCD=336+210=546\text{ cm}^2.
\displaystyle \therefore \text{The area of quadrilateral }ABCD\text{ is }546\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Find the perimeter and area of quadrilateral }ABCD\text{ in which }AB=17\text{ cm,}
\displaystyle AD=9\text{ cm, }CD=12\text{ cm, }\angle ACB=90^\circ\text{ and }AC=15\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle BC=\sqrt{AB^2-AC^2}
\displaystyle =\sqrt{17^2-15^2}=\sqrt{64}=8\text{ cm.}
\displaystyle \text{Perimeter of }ABCD=17+8+12+9=46\text{ cm.}
\displaystyle \text{Area of }ABCD=\text{Area of }\triangle ABC+\text{Area of }\triangle ACD.
\displaystyle \text{For }\triangle ABC,\ a=17,\ b=8\text{ and }c=15.
\displaystyle s=\frac{17+8+15}{2}=20.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ABC=\sqrt{20(20-17)(20-8)(20-15)}
\displaystyle =\sqrt{20\times3\times12\times5}=60\text{ cm}^2.
\displaystyle \text{For }\triangle ACD,\ a=15,\ b=12\text{ and }c=9.
\displaystyle s=\frac{15+12+9}{2}=18.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ACD=\sqrt{18(18-15)(18-12)(18-9)}
\displaystyle =\sqrt{18\times3\times6\times9}=54\text{ cm}^2.
\displaystyle \text{Area of }ABCD=60+54=114\text{ cm}^2.
\displaystyle \therefore \text{The perimeter is }46\text{ cm and the area is }114\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{The adjacent sides of a parallelogram }ABCD\text{ measure }34\text{ cm and }20\text{ cm,}
\displaystyle \text{and the diagonal }AC\text{ measures }42\text{ cm. Find the area of the parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of parallelogram }ABCD=\text{Area of }\triangle ABC+\text{Area of }\triangle ACD.
\displaystyle \text{For }\triangle ABC,\ a=20,\ b=34\text{ and }c=42.
\displaystyle s=\frac{20+34+42}{2}=48.
\displaystyle \text{By Heron's Formula,}
\displaystyle \text{Area of }\triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}
\displaystyle =\sqrt{48(48-20)(48-34)(48-42)}
\displaystyle =\sqrt{48\times28\times14\times6}
\displaystyle =336\text{ cm}^2.
\displaystyle \text{Since a diagonal divides a parallelogram into two congruent triangles,}
\displaystyle \text{Area of }\triangle ACD=336\text{ cm}^2.
\displaystyle \text{Area of parallelogram }ABCD=336+336=672\text{ cm}^2.
\displaystyle \therefore \text{The area of the parallelogram is }672\text{ cm}^2.
\displaystyle \\

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