Maths / Areas Of Parallelograms And Triangles / Area of Triangle

QUESTION

Find the area of the following quadrilateral.

EXPLANATION
Explain TypeExplanation Content
Text

$AP = \sqrt{10^2-8^2}=\sqrt{36}=6\ cm$

$QD = \sqrt{10^2-8^2}=\sqrt{36}=6\ cm$

Area of Trapezium ABCD =$\frac{h(a+b)}{2}$ $=\frac{8(6+18)}{2}=96\ cm^2$

We can also calculate by finding area of triangle APB, triangle BCQ, rectangle PQCB.

traingle APB = $\frac{1}{2} \times 6\times 8=24 \ cm^2$

triangle BCQ = $\frac{1}{2} \times 6\times 8=24 \ cm^2$

rectangle PQCB = $6\times 8=48 \ cm^2$

Total Area = $24 +24+48=96\ cm^2$

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