The difference between the sides at right angles in a right-angled triangle is 7 cm, The area of the triangle is 60 $c m^{2}$ . Find its perimeter.

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Solution

Let base = x and altitude be x + 7

Now area = 60 $c m^{2}$

i.e, $\frac{1}{2} \times b a s e \times a l t i t u d e = 60 \frac{1}{2} \times x \times (x + 7) = 60 x^{2} + 7 x = 120 x^{2} + 7 x - 120 = 0 x^{2} - 8 x + 15 x - 120 = 0 x (x - 8) + 15 (x - 8) = 0 (x - 8) (x + 15) = 0 x = 8 o r - 15$

Length annot be negative.

So base = 8 cm and altitude = 8 + 7 = 15 cm

Hypotenuse $= \sqrt{8^{2} + 15^{2}} = \sqrt{64 + 225} = \sqrt{289} = 17 c m$

Perimeter $= 8 + 15 + 17 = 40 c m$

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