Question

The diagram consists of a right-angled Δ ABC. D is the mid-point of AB and F is the mid-point of AD. Given that AG // DE // BC and the area of the triangle is 36 cm2, find the area of the trapezium FGED.
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Answer 1

$\huge{ \green{ \mathfrak{ \underline{Answer}}}}$

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$\boxed{\red{\bold{Given:}}}$

$FG \: || \: DE \:||\: BC$

$\boxed{\red{\bold{In\: \triangle ABC\: and\: \triangle ADE :}}}$

$\angle ADE = \angle ABC (each\: 90^o)$

$\angle AED = \angle ACB (corresponding\; angles )$

$\tt{\blue{By \:AA\: criterion, \triangle ABC \sim \triangle ADE}}$

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$\displaystyle{\frac{ar(ADE)}{ar(ABC)}} = \left( {\frac{AD}{AB}} \right)^2$$(By\; theorem)$

$\displaystyle{\frac{ar(ADE)}{ar(ABC)}} =\left( {\frac{AD}{2AD}} \right) ^2$$(Because \:D\: is \\ the\: midpoint)$

$\displaystyle{\frac{ar(ADE)}{36}} = \left( {\frac{1}{2}}\right)^2$

$ar(ADE) = \displaystyle{\frac{36}{4}}$

$ar(ADE) = 9 cm^2$

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$\boxed{\red{\bold{In\: \triangle ADE and \triangle AFG :}}}$

$\angle AFG = \angle ADE (each \:90^o)$

$\angle AGF = \angle AED (corresponding \;angles)$

$\tt{\blue{By \:AA\: criterion, \triangle AFG \sim \triangle ADE}}$

$\displaystyle{\frac{ar(AFG)}{ar(ADE)}} =\left( {\frac{AF}{AD}} \right)^2$$(By\; theorem)$

$\displaystyle{\frac{ar(AFG)}{ar(ADE)}} = \left( {\frac{AF}{2AF}} \right)^2$$(Because\: F\: is \\ the\: midpoint)$

$\displaystyle{\frac{ar(AFG)}{9}} = \left( {\frac{1}{2}} \right)^2$

$ar(AFG) = \displaystyle{\frac{9}{4}}$

$ar(AFG) = 2.25 cm^2$

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$\boxed{\red{\bold{ar(FGED):}}}$

$ar(FGED) = ar(ADE) - ar(AFG)$

$ar(FGED) = 9 cm^2 - 2.25 cm^2$

$ar(FGED) = 6.75 cm^2$

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