abcd is a rectangle ab=5 ad=2 e is any point on cd such that de=4 prove that ab substend a right angle at e
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Area of the rectangle ABCD=12×7=84sqcm.
Let AP=x cm. So, PB=(12−x) cm.
Area of shaded region =0.4×84=33.6 sq cm.
Area of triangle EAP=0.5×5×x=2.5x sq cm
Area of triangle CPB=0.5×7×(12−x)=42−3.5x sq cm
Sum of the area of triangles EAP and CPB = area of shaded region
2.5x+42−3.5x=33.6
42−x=33.6 or x=8.4cm
Point P should at 8.4 cm from vertex A.
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Let AP=x cm. So, PB=(12−x) cm.
Area of shaded region =0.4×84=33.6 sq cm.
Area of triangle EAP=0.5×5×x=2.5x sq cm
Area of triangle CPB=0.5×7×(12−x)=42−3.5x sq cm
Sum of the area of triangles EAP and CPB = area of shaded region
2.5x+42−3.5x=33.6
42−x=33.6 or x=8.4cm
Point P should at 8.4 cm from vertex A.
MARK ME AS A BRAINLIEST
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