Class 9: Perimeter and Areas of Plane Figures

$\displaystyle \textbf{Question 1: }\text{Find the area of a parallelogram whose base is }32\text{ cm and the}$
$\displaystyle \text{corresponding altitude is }4\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of a parallelogram}=\text{Base}\times\text{Height}$
$\displaystyle =32\times4=128\text{ cm}^2.$
$\displaystyle \therefore \text{The area of the parallelogram is }128\text{ cm}^2.$
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$\displaystyle \textbf{Question 2: }\text{Find the area of a rhombus whose diagonals are }10\text{ cm and }8\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of a rhombus}=\frac{1}{2}d_1d_2$
$\displaystyle =\frac{1}{2}\times10\times8=40\text{ cm}^2.$
$\displaystyle \therefore \text{The area of the rhombus is }40\text{ cm}^2.$
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$\displaystyle \textbf{Question 3: }\text{If the lengths of the diagonals of a rhombus are }(a+b)\text{ and }$
$\displaystyle (a-b),\text{ what is} \ \text{the area of the rhombus?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of a rhombus}=\frac{1}{2}d_1d_2$
$\displaystyle =\frac{1}{2}(a+b)(a-b)$
$\displaystyle =\frac{a^2-b^2}{2}.$
$\displaystyle \therefore \text{The area of the rhombus is }\frac{a^2-b^2}{2}.$
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$\displaystyle \textbf{Question 4: }\text{The area of a rhombus is }72\text{ cm}^2.\text{ If one of the diagonals is }18\text{ cm}$
$\displaystyle \text{long, find the length of the other diagonal.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the rhombus}=72\text{ cm}^2.$
$\displaystyle \text{Area of a rhombus}=\frac{1}{2}d_1d_2$
$\displaystyle 72=\frac{1}{2}\times18\times d_2$
$\displaystyle 72=9d_2$
$\displaystyle d_2=8\text{ cm.}$
$\displaystyle \therefore \text{The length of the other diagonal is }8\text{ cm.}$
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$\displaystyle \textbf{Question 5: }\text{The area of a parallelogram is }338\text{ m}^2.\text{ If its altitude is twice}$
$\displaystyle \text{the corresponding base, determine the base and the altitude.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the base be }x\text{ m.}$
$\displaystyle \therefore \text{Altitude}=2x\text{ m.}$
$\displaystyle \text{Area of parallelogram}=\text{Base}\times\text{Altitude}$
$\displaystyle 338=x\times2x$
$\displaystyle 2x^2=338$
$\displaystyle x^2=169$
$\displaystyle x=13\text{ m.}$
$\displaystyle \therefore \text{Base}=13\text{ m and Altitude}=26\text{ m.}$
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$\displaystyle \textbf{Question 6: }\text{The adjacent sides of a parallelogram are }10\text{ m and }8\text{ m. If the}$
$\displaystyle \text{distance between the longer sides is }4\text{ m, find the distance between the shorter sides.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the parallelogram}=10\times4=40\text{ m}^2.$
$\displaystyle \text{Let the distance between the shorter sides be }h\text{ m.}$
$\displaystyle 8\times h=40$
$\displaystyle h=5\text{ m.}$
$\displaystyle \therefore \text{The distance between the shorter sides is }5\text{ m.}$
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$\displaystyle \textbf{Question 7: }\text{The area of a triangle is equal to the area of a parallelogram whose base is }$
$\displaystyle 15\text{ cm} \text{and altitude is }8\text{ cm. If the base of the triangle is }20\text{ cm, find its altitude.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Base of the triangle}=20\text{ cm and Height}=h\text{ cm.}$
$\displaystyle \text{Base of the parallelogram}=15\text{ cm and Height}=8\text{ cm.}$
$\displaystyle \text{Area of the triangle}=\text{Area of the parallelogram}$
$\displaystyle \frac{1}{2}\times20\times h=15\times8$
$\displaystyle 10h=120$
$\displaystyle h=12\text{ cm.}$
$\displaystyle \therefore \text{The altitude of the triangle is }12\text{ cm.}$
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$\displaystyle \textbf{Question 8: }\text{The diagonals of a rhombus are }15\text{ cm and }36\text{ cm long. Find its perimeter.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}$
$\displaystyle \text{Half of the diagonals are }\frac{15}{2}\text{ cm and }18\text{ cm.}$
$\displaystyle \text{Side of the rhombus}=\sqrt{\left(\frac{15}{2}\right)^2+18^2}$
$\displaystyle =\sqrt{\frac{225}{4}+324}$
$\displaystyle =\sqrt{\frac{1521}{4}}=\frac{39}{2}=19.5\text{ cm.}$
$\displaystyle \text{Perimeter}=4\times19.5=78\text{ cm.}$
$\displaystyle \therefore \text{The perimeter of the rhombus is }78\text{ cm.}$
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$\displaystyle \textbf{Question 9: }\text{In quadrilateral }ABCD,\text{ diagonal }AC=44\text{ cm and the lengths of the perpendiculars}$
$\displaystyle \text{from }B\text{ and }D\text{ to }AC\text{ are }20\text{ cm and }15\text{ cm respectively. Find the area of the quadrilateral.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of }ABCD=\text{Area of }\triangle ABC+\text{Area of }\triangle ACD$
$\displaystyle =\frac{1}{2}\times44\times20+\frac{1}{2}\times44\times15$
$\displaystyle =440+330=770\text{ cm}^2.$
$\displaystyle \therefore \text{The area of quadrilateral }ABCD\text{ is }770\text{ cm}^2.$
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$\displaystyle \textbf{Question 10: }\text{Find the diagonal of a quadrilateral whose area is }495\text{ dm}^2\text{ and whose}$
$\displaystyle \text{offsets are }19\text{ dm and }11\text{ dm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area}=495\text{ dm}^2.$
$\displaystyle \text{Offsets}=19\text{ dm and }11\text{ dm.}$
$\displaystyle \text{Area of a quadrilateral}=\frac{1}{2}\times\text{Diagonal}\times(\text{Sum of offsets})$
$\displaystyle \therefore \text{Diagonal}=\frac{2\times\text{Area}}{\text{Sum of offsets}}$
$\displaystyle =\frac{2\times495}{19+11}$
$\displaystyle =\frac{990}{30}=33\text{ dm.}$
$\displaystyle \therefore \text{The diagonal of the quadrilateral is }33\text{ dm.}$
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$\displaystyle \textbf{Question 11: }\text{Find the cost of levelling a plot of ground in the form of a quadrilateral at}$
$\displaystyle \text{Rs. }250\text{ per square metre whose diagonal measures }75\text{ m and whose offsets are }50\text{ m}$
$\displaystyle \text{and }40\text{ m respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the quadrilateral}=\frac{1}{2}\times\text{Diagonal}\times(\text{Sum of offsets})$
$\displaystyle =\frac{1}{2}\times75\times(50+40)$
$\displaystyle =3375\text{ m}^2.$
$\displaystyle \text{Cost of levelling}=3375\times250=\text{Rs. }843750.$
$\displaystyle \therefore \text{The cost of levelling the plot is Rs. }843750.$
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$\displaystyle \textbf{Question 12: }\text{Find the area of a quadrangular field whose diagonals measure }48\text{ m and }32\text{ m}$
$\displaystyle \text{and bisect each other at right angles. Find also the cost of land at the rate of Rs. }70\text{ per}$
$\displaystyle \text{square metre.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the diagonals bisect each other at right angles, the quadrilateral is a rhombus.}$
$\displaystyle \text{Area of a rhombus}=\frac{1}{2}d_1d_2$
$\displaystyle =\frac{1}{2}\times48\times32=768\text{ m}^2.$
$\displaystyle \text{Cost of land}=768\times70=\text{Rs. }53760.$
$\displaystyle \therefore \text{The area of the field is }768\text{ m}^2\text{ and the cost of land is Rs. }53760.$
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$\displaystyle \textbf{Question 13: }\text{The parallel sides of a trapezium are }85\text{ cm and }63\text{ cm and its altitude is}$
$\displaystyle 36\text{ cm. Find its area.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of a trapezium}=\frac{1}{2}(a+b)h$
$\displaystyle =\frac{1}{2}(85+63)\times36$
$\displaystyle =\frac{1}{2}\times148\times36$
$\displaystyle =2664\text{ cm}^2.$
$\displaystyle \therefore \text{The area of the trapezium is }2664\text{ cm}^2.$
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$\displaystyle \textbf{Question 14: }\text{Two parallel sides of a trapezium are }60\text{ cm and }77\text{ cm, and the other}$
$\displaystyle \text{sides are }25\text{ cm and }26\text{ cm. Find the area of the trapezium.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the parallel sides}=77-60=17\text{ cm.}$
$\displaystyle \text{For }\triangle BCE,\text{ the sides are }25\text{ cm, }26\text{ cm and }17\text{ cm.}$
$\displaystyle s=\frac{25+26+17}{2}=34.$
$\displaystyle \text{By Heron's Formula,}$
$\displaystyle \text{Area of }\triangle BCE=\sqrt{s(s-a)(s-b)(s-c)}$
$\displaystyle =\sqrt{34(34-25)(34-26)(34-17)}$
$\displaystyle =\sqrt{34\times9\times8\times17}$
$\displaystyle =\sqrt{41616}=204\text{ cm}^2.$
$\displaystyle \text{Let the height of the trapezium be }h\text{ cm.}$
$\displaystyle 204=\frac{1}{2}\times17\times h$
$\displaystyle h=24\text{ cm.}$
$\displaystyle \text{Area of the trapezium}=\frac{1}{2}(60+77)\times24$
$\displaystyle =1644\text{ cm}^2.$
$\displaystyle \therefore \text{The area of the trapezium is }1644\text{ cm}^2.$
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$\displaystyle \textbf{Question 15: }\text{The cross-section of a canal is a trapezium in shape. If the canal is }$
$\displaystyle 10\text{ m wide at} \ \text{the top, }6\text{ m wide at the bottom and the area of the cross-section is }$
$\displaystyle 64\text{ m}^2,\text{ find the} \ \text{depth of the canal.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of a trapezium}=\frac{1}{2}(a+b)h.$
$\displaystyle 64=\frac{1}{2}(10+6)h$
$\displaystyle 64=\frac{1}{2}\times16\times h$
$\displaystyle 64=8h$
$\displaystyle h=8\text{ m.}$
$\displaystyle \therefore \text{The depth of the canal is }8\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{The parallel sides }AB\text{ and }DC\text{ of trapezium }ABCD\text{ are }51\text{ cm and}$
$\displaystyle 30\text{ cm respectively. If the sides }AD\text{ and }BC\text{ are }20\text{ cm and }13\text{ cm respectively, find}$
$\displaystyle \text{the distance between the parallel sides and the area of trapezium }ABCD.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the parallel sides}=51-30=21\text{ cm.}$
$\displaystyle \text{For the auxiliary triangle, the sides are }20\text{ cm, }13\text{ cm and }21\text{ cm.}$
$\displaystyle s=\frac{20+13+21}{2}=27.$
$\displaystyle \text{By Heron's Formula,}$
$\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$
$\displaystyle =\sqrt{27(27-20)(27-13)(27-21)}$
$\displaystyle =\sqrt{27\times7\times14\times6}$
$\displaystyle =\sqrt{15876}=126\text{ cm}^2.$
$\displaystyle \text{Let the distance between the parallel sides be }h\text{ cm.}$
$\displaystyle 126=\frac{1}{2}\times21\times h$
$\displaystyle h=12\text{ cm.}$
$\displaystyle \text{Area of trapezium }ABCD=\frac{1}{2}(51+30)\times12$
$\displaystyle =486\text{ cm}^2.$
$\displaystyle \therefore \text{The distance between the parallel sides is }12\text{ cm and the area is }486\text{ cm}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{The area of a trapezium is }540\text{ cm}^2.\text{ If the ratio of its parallel sides is}$
$\displaystyle 7:5\text{ and the distance between them is }18\text{ cm, find the lengths of the parallel sides.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the parallel sides be }7x\text{ cm and }5x\text{ cm.}$
$\displaystyle \text{Area of a trapezium}=\frac{1}{2}(a+b)h$
$\displaystyle 540=\frac{1}{2}(7x+5x)\times18$
$\displaystyle 540=\frac{1}{2}\times12x\times18$
$\displaystyle 540=108x$
$\displaystyle x=5.$
$\displaystyle \therefore \text{The parallel sides are }7\times5=35\text{ cm and }5\times5=25\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{The parallel sides of an isosceles trapezium are in the ratio }$
$\displaystyle 2:3.\text{ If its height is} \ 4\text{ cm and its area is }60\text{ cm}^2,\text{ find the perimeter.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the parallel sides be }2x\text{ cm and }3x\text{ cm.}$
$\displaystyle \text{Area of a trapezium}=\frac{1}{2}(a+b)h$
$\displaystyle 60=\frac{1}{2}(2x+3x)\times4$
$\displaystyle 60=10x$
$\displaystyle x=6.$
$\displaystyle \therefore \text{The parallel sides are }12\text{ cm and }18\text{ cm.}$
$\displaystyle \text{Difference between the parallel sides}=18-12=6\text{ cm.}$
$\displaystyle \text{Since the trapezium is isosceles, each horizontal offset}=\frac{6}{2}=3\text{ cm.}$
$\displaystyle \text{Each non-parallel side}=\sqrt{3^2+4^2}$
$\displaystyle =\sqrt{9+16}=5\text{ cm.}$
$\displaystyle \text{Perimeter}=12+18+5+5=40\text{ cm.}$
$\displaystyle \therefore \text{The perimeter of the trapezium is }40\text{ cm.}$
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$\displaystyle \textbf{Question 19: }\text{The area of a parallelogram is }x\text{ cm}^2\text{ and its height is }y\text{ cm. A second}$
$\displaystyle \text{parallelogram has equal area but its base is }z\text{ cm more than that of the first. Obtain an}$
$\displaystyle \text{expression in terms of }x,\ y\text{ and }z\text{ for the height of the second parallelogram.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the base of the first parallelogram be }b\text{ cm.}$
$\displaystyle x=b\times y$
$\displaystyle b=\frac{x}{y}\text{ cm.}$
$\displaystyle \text{Let the height of the second parallelogram be }h\text{ cm.}$
$\displaystyle \text{Base of the second parallelogram}=\left(\frac{x}{y}+z\right)\text{ cm.}$
$\displaystyle x=\left(\frac{x}{y}+z\right)h$
$\displaystyle h=\frac{x}{\frac{x}{y}+z}$
$\displaystyle =\frac{xy}{x+yz}\text{ cm.}$
$\displaystyle \therefore \text{The height of the second parallelogram is }\frac{xy}{x+yz}\text{ cm.}$
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$\displaystyle \textbf{Question 20: }\text{The area of a parallelogram is }98\text{ cm}^2.\text{ If one altitude is half the corresponding}$
$\displaystyle \text{base, determine the base and altitude of the parallelogram.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the altitude be }h\text{ cm.}$
$\displaystyle \therefore \text{Corresponding base}=2h\text{ cm.}$
$\displaystyle \text{Area of parallelogram}=\text{Base}\times\text{Altitude}$
$\displaystyle 98=2h\times h$
$\displaystyle 2h^2=98$
$\displaystyle h^2=49$
$\displaystyle h=7\text{ cm.}$
$\displaystyle \therefore \text{Base}=2\times7=14\text{ cm.}$
$\displaystyle \therefore \text{The base is }14\text{ cm and the altitude is }7\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{A triangle and a parallelogram have the same base and same area.}$
$\displaystyle \text{If the sides of the triangle are }26\text{ cm, }28\text{ cm and }30\text{ cm, and the parallelogram}$
$\displaystyle \text{stands on the base 28 cm, find the height of the parallelogram.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the triangle, }a=26,\ b=28\text{ and }c=30.$
$\displaystyle s=\frac{26+28+30}{2}=42.$
$\displaystyle \text{By Heron's Formula,}$
$\displaystyle \text{Area of the triangle}=\sqrt{s(s-a)(s-b)(s-c)}$
$\displaystyle =\sqrt{42(42-26)(42-28)(42-30)}$
$\displaystyle =\sqrt{42\times16\times14\times12}$
$\displaystyle =\sqrt{112896}=336\text{ cm}^2.$
$\displaystyle \text{Area of the parallelogram}=336\text{ cm}^2.$
$\displaystyle \text{Let its height be }h\text{ cm.}$
$\displaystyle 336=28\times h$
$\displaystyle h=12\text{ cm.}$
$\displaystyle \therefore \text{The height of the parallelogram is }12\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{The cross-section of a canal is in the form of a trapezium whose parallel}$
$\displaystyle \text{sides are along the top and bottom of the canal. If the canal is }8\text{ m wide at the top and }$
$\displaystyle 6\text{ m wide at the} \ \text{bottom and the area of the cross-section is }16.8\text{ m}^2,\text{ find its depth.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the depth of the canal be }h\text{ m.}$
$\displaystyle \text{Area of a trapezium}=\frac{1}{2}(a+b)h$
$\displaystyle 16.8=\frac{1}{2}(8+6)h$
$\displaystyle 16.8=7h$
$\displaystyle h=2.4\text{ m.}$
$\displaystyle \therefore \text{The depth of the canal is }2.4\text{ m.}$
$\displaystyle \\$