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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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
= 465 cm²
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Thanks Karthik.
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see the attachment for solution..
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