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Answer 1

$\sf \large \underbrace{ \underline{Understanding \: the \: Question}}$

Here we could see that the quadrilateral is divided into 2 right angled triangle, where base of both triangles are equal in length. So we could find the area of both the triangles separately and then have to add both area to find area of quadrilateral.

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GIVEN:

⊙ BM and LD are perpendicular on side AC.

⊙ Length of base of both triangles is 32cm.

⊙Height of triangle ABC=8cm.

⊙Height of triangle ADC=6cm.

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TO FIND:

⊙ Area of quadrilateral

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SOLUTION:

So let's start!

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~For area of triangle ABC

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$\small \star \boxed{ \sf Area \: of \: triangle \: ABC=\dfrac{1}{2}×Base×Height}$

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$\sf Area \: of \: triangle \: ABC=\dfrac{1}{2}×32cm×8cm$

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$\sf Area \: of \: triangle \: ABC= \cancel{\dfrac{1}{2}×32cm×8cm}$

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$\sf \green{Area \: of \: triangle \: ABC={128cm }^{2} }$

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~For area of triangle ADC

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$\small \star \boxed{ \sf Area \: of \: triangle \: ADC=\dfrac{1}{2}×Base×Height}$

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$\sf Area \: of \: triangle \: ADC=\dfrac{1}{2}×32cm×6cm$

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$\sf Area \: of \: triangle \: ADC= \cancel{\dfrac{1}{2}×32cm×6cm}$

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$\sf \blue{Area \: of \: triangle \: ADC=96cm {}^{2} }$

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~For area of quadrilateral ABCD

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$\star \boxed{ \sf \small Area \: of \: quadrilateral=Area \: of \: \triangle ABC \: and \: \triangle ADC}$

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$\sf : \implies \: 128cm {}^{2} + 96cm {}^{2}$

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$\sf : \implies \: 224cm {}^{2}$

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[•°•Ar. ABCD =224Cm²]

Answer 2

$\color{red}\huge{\underline{\underline{\bf GIVEN\dag}}}$

• $\tt BM=8cm$
• $\tt AC=32cm$
• $\tt LD=6cm$
• $\tt BM \: ⊥ \: AC(perpendicular)$
• $\tt and \: \: DL \: ⊥ \: AC$

$\color{red}\huge{\underline{\underline{\bf SOLUTION\dag}}}$

$\purple{ \sf consider \: \triangle ABC : }$

$\boxed{ \blue{ \tt Area \: of \: \triangle = \frac{1}{2} bh}}$

$\sf \color{darkblue}{where : }$

• $\tt b : base$
• $\tt h : height$

$\sf \color{darkblue}{here : }$

• $\tt BM=h = 8cm$
• $\tt AC=b = 32cm$

$\implies \tt Area(\triangle ABC ) = \frac{1}{ \cancel2} (32cm)( \cancel8cm)$

$\implies \tt Ar.(\triangle ABC ) = (32cm)(4cm)$

$\color{crimson}{\therefore \tt Ar.(\triangle ABC ) = 128cm ^{2} }$

$\purple{ \sf \: now \: consider \: \triangle ACD : }$

$\implies \tt Ar.( \triangle ACD) = \frac{1}{ \cancel2} (32cm)( \cancel6cm)$

$\implies \tt Ar.( \triangle ACD) = (32cm)(3cm)$

$\color{crimson}{ \therefore \tt Ar.( \triangle ACD) = 96cm^{2}}$

$\sf \purple{ but}$

$\tt Ar.(ABCD) =Ar.( \triangle ABC ) + Ar.( \triangle ACD )$

$\implies \tt Ar.(ABCD) =128 {cm}^{2} + 96 {cm}^{2}$

$\color{darkorange}{ \boxed{\therefore\tt Ar.(ABCD) =224 {cm}^{2} }}$

HOPE THIS HELPS!!