The area of a right triangle is $28 cm^2$. If one of its perpendicular sides exceeds the other by 10 cm . then the length of the longest of the perpendicular is
A
16 cm
B
14 cm
C$6 \sqrt{5} cm$
D
18 cm
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B
14 cm
(b) 14 cm Let the perpendicular sides be of length $x cm$ and $(x+10) cm$ respectively. Then, $\text { Area }=28 cm^2 \Rightarrow \frac{1}{2} x(x+10)=28 \Rightarrow x^2+10 x-56=0 \Rightarrow(x+14)(x-4)=0\Rightarrow x=4$ Thus, the length of the longest perpendicular side is $(x+10) cm =14 cm$.
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