Question

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The Area of a trapezium is 405
cm2, whose height of is 9cm.If one of the parallel sides of trapezium is
48 cm. Find the other parallel sides​

Answer 1

Answer:

$let \: the \: parallel \: side \: be \: 4x \: . \: 5x \: \\ respectively$

$height = 18cm$

$the \: area \: of \: trapezium \: = 405 {cm}^{2}$

$\frac{1}{2} h(a + b) = 405$

$\frac{1}{2} \times 18 \times (4x + 5x) = 405 \\ \\ 9 \times 9x = 405$

$81x = 405$

$x = \frac{405}{81} \\ \\ x = 5$

$the \: length \: of \: each \: of \: parallel \: \\ sides \: 20cm \: and \: 25cm \: \\ respecively$

Answer 2

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$\huge{\bf{\green{\mathfrak{Question:-}}}}$

• The Area of a trapezium is 405 cm², whose height of is 9cm. If one of the parallel sides of trapezium is 48 cm. Find the other parallel sides.

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$\huge {\bf{\orange{\mathfrak{Answer:-}}}}$

• $\sf{Area \: of \: trapezium\: = \: 405 \: cm^{2}.}$

• $\sf{Height \:of \:trapezium\: = \: 9 \: cm.}$

• $\sf{Length \: of \: 1 \: parallel \: side \: = \: a \: = \: 48 \: cm.}$

• $\sf{Area \: of \: trapezium \: = \: \dfrac{a \: + \: b}{2} \: * \: h}$

• $\sf{Substituting \: the \: value,}$

• $\sf{405\: = \: \dfrac{48 \: + \: b}{2} \: * \: 9}$

• $\sf{\dfrac{405 \: * \: 2}{9} \: = \: 48 \: + \: b}$

• $\sf{\dfrac{810}{9} \: = \: 48 \: + \: b}$

• $\sf{90 \: = \: 48 \: + \: b}$

• $\sf{b \: = \: 90 \: - \: 48}$

• $\boxed{\therefore{\sf{b \: = \: 42 \: cm.}}}$

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$\huge{\bf{\red{\mathfrak{Conclusion:-}}}}$

• $\boxed{\therefore{\sf{Length \: of \: other \: parallel \: side \: = \: b \: = \: 42 \: cm.}}}$

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