Question

In a triangle abc;a=90degree and side ab=x cm and ac=(x+5)cm and area=150cm2.Find sides of the triangle

Answer 1

Step-by-step explanation:

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Answer 2

ANSWER:-

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$area = 150 {cm}^{2}$

$= > area = \frac{1}{2} \times b \times h = \frac{1}{2} \times AC \times AB$

$= > 150 \times \frac{1}{2} \times (x + 5) \times x$

$= > 300 = {x}^{2} + 5x$

$= > {x}^{2} + 5x - 300 = 0$

$= > {x}^{2} (x + 20) - 15(x + 20) = 0$

$= > (x + 20)(x - 15) = 0$

$= > x = 15$

$∴AB = 15cm, AC=x+5=15 + 5 = 20cm,BC = \sqrt{ {15}^{2} + {20}^{2} } = 25cm$