In a trapezium ABCD, AB parallel CD. If AB = | vedclass

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(B) $1$. Draw a line $CE$ parallel to $AD$ such that $E$ lies on $AB$. Since $AD \parallel CE$ and $AE \parallel CD$, $AECD$ is a parallelogram.$2$. T

Vedclass · Accepted Answer

$180$

Vedclass · Answer

(B) $1$. Draw a line $CE$ parallel to $AD$ such that $E$ lies on $AB$. Since $AD \parallel CE$ and $AE \parallel CD$, $AECD$ is a parallelogram.$2$. Thus, $AE = CD = 20 \, cm$ and $CE = AD = 9 \, cm$. $3$. The length $EB = AB - AE = 30 - 20 = 10 \, cm$.$4$. In $	riangle CEB$, the sides are $17 \, cm$, $9 \, cm$, and $10 \, cm$. The semi-perimeter $s = (17 + 9 + 10) / 2 = 18 \, cm$.$5$. Using Heron's formula, Area of $	riangle CEB = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18(18-17)(18-9)(18-10)} = \sqrt{18 	imes 1 	imes 9 	imes 8} = \sqrt{1296} = 36 \, cm^2$.$6$. The area of $	riangle CEB$ is also given by $(1/2) 	imes 	ext{base} 	imes 	ext{height} = (1/2) 	imes 10 	imes h = 36$, so $h = 7.2 \, cm$.$7$. The area of trapezium $ABCD = (1/2) 	imes (AB + CD) 	imes h = (1/2) 	imes (30 + 20) 	imes 7.2 = 25 	imes 7.2 = 180 \, cm^2$.

In a trapezium $ABCD$, $AB \parallel CD$. If $AB = 30 \, cm$, $BC = 17 \, cm$, $CD = 20 \, cm$, and $DA = 9 \, cm$, then find its area. (in $cm^2$)

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