Two straight poles of unequal length stand upright on a ground. The length of the shorter pole is 10 m. A pole joins the top of the two vertical poles. The distance between the two tops is 5m. The distance between the poles along the ground is 4 m. The area thus formed by the three poles with the ground is:

c m^{2}

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c m^{2}

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c m^{2}

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c m^{2}

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Solution

The correct option is B 46 $c m^{2}$
$\begin{matrix} n o w x^{2} + 4^{2} = 5^{2} \Rightarrow x = 3 A r e a o f Δ A B C = \frac{1}{2} \times 4 \times x = \frac{1}{2} \times 4 \times 3 = 6 m^{2} A r e a o f R e c t a n g l e B C \in D = 10 \times 4 = 40 m^{2} t o t a l A r e a = 40 + 6 = 46 m^{2} \end{matrix}$

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The correct option is B 46 cm2nowx2+42=52⇒x=3AreaofΔABC=12×4×x=12×4×3=6m2AreaofRectangleBC∈D=10×4=40m2totalArea=40+6=46m2