# In the figure AD PARALLEL TO CD AD PERPENDICULAR TO AB .CB:17cm, AB:35cm, CD:27cm . find the area of shaded region .

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by ultrab30

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by ultrab30

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AB parallel to CD.

AB perpendicular to AD.

It is a trapezium.

Length of parallel sides are :

AB = 35cm, CD = 27cm

Height of trapezium = AD = CE

CE is perpendicular to AB.

So CE,EB and BC form a right-angled triangle.

So BC² = CE² + EB²

⇒BC² = CE² + (AB-AE²)

⇒BC² = CE² + (AB-CD²) (AE=CD)

⇒17² = CE² + (35-27)²

⇒17² = CE² + 8²

⇒CE² = 17²-8²=289-64=225

⇒CE = 15 =AD

Area of trapezium = (1/2)(sum of parallel sides)×(height)

= (1/2)(35+27)×(15) = 31×15

=**465 cm²**

AB perpendicular to AD.

It is a trapezium.

Length of parallel sides are :

AB = 35cm, CD = 27cm

Height of trapezium = AD = CE

CE is perpendicular to AB.

So CE,EB and BC form a right-angled triangle.

So BC² = CE² + EB²

⇒BC² = CE² + (AB-AE²)

⇒BC² = CE² + (AB-CD²) (AE=CD)

⇒17² = CE² + (35-27)²

⇒17² = CE² + 8²

⇒CE² = 17²-8²=289-64=225

⇒CE = 15 =AD

Area of trapezium = (1/2)(sum of parallel sides)×(height)

= (1/2)(35+27)×(15) = 31×15

=

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