Question 34

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waecmaths question: 

In the diagram, $\left| EF \right|=8cm$ $\left| FG \right|=xcm$ $\left| GH \right|=(x+2)cm$ , $\angle EFG={{90}^{\circ }}$ If the area of the shaded portion is 40cm2. Find the area of $\Delta EFG$

Option A: 

128cm2

Option B: 

72cm2

Option C: 

64cm2

Option D: 

32cm2

waecmaths solution: 

$\begin{align}  & Area\,\text{of triangle }=\frac{1}{2}base\times height \\ & \text{Area of triangle }EFG=\frac{1}{2}\times 8\times xc{{m}^{2}}=4xc{{m}^{2}} \\ & \text{Area of}\vartriangle \text{ }EFG=4xc{{m}^{2}} \\ & \text{Area of }\vartriangle EFH=\frac{1}{2}\times 8\times (x+x+2)c{{m}^{2}} \\ & \text{Area of}\vartriangle EFH=4(2x+2)c{{m}^{2}} \\ & \text{Area of }\vartriangle EGH=\text{Area of}\vartriangle EFH-\text{Area of}\vartriangle \text{ }EFG \\ & \text{Area of }\vartriangle EGH=[4(2x+2)-4x]c{{m}^{2}} \\ & \text{Area of }\vartriangle EGH=(8x+8-4x)c{{m}^{2}}=40c{{m}^{2}} \\ & 4x-8=40 \\ & 4x=32 \\ & x=8 \\ & \text{Area of }\vartriangle EFG=4xc{{m}^{2}}=4\times 8c{{m}^{2}}=32c{{m}^{2}} \\\end{align}$

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