Question

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15. The parallel sides of a trapezium are 52 cm. and 24 cm. and distance between them is 24 cm. Find its area,
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Answer 1

$\huge{\underline{\bf{Given}}}$

• AD = 24cm
• BC = 52cm
• Height = 24cm

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$\huge{\underline{\bf{Figure}}}$

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 24\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 52\ cm }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 24\ cm }\end{picture}$

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$\huge{\underline{\bf{To\: find}}}$

• Area of the trapezium.

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$\huge{\underline{\bf{Solution}}}$

★ We know that

$\underline{\boxed{\tt{\orange{Area\: of\: trapezium = \dfrac{1}{2}(Sum\: of\: parallel\: sides) \times h}}}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {Area = \dfrac{1}{2}(52 + 24) \times 24}$

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$\tt:\implies\: \: \: \: \: \: \: \: {Area = \dfrac{1}{\cancel{2}}(\cancel{76}) \times 24 }$

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$\tt:\implies\: \: \: \: \: \: \: \: {Area = 38 \times 24}$

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$\tt:\implies\: \: \: \: \: \: \: \: {Area = 912cm^2}$

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Hence, the area of the trapezium is 912cm².

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