Question

In a triangle ABC;angle A=90degrees,side AB=xcm,AC=(x+5)cm and area =150cm2.find the sides of the triangle.

Answer 1

ABC is a right triangle
Area of right triangle = 1/2 × base × height
 
                                  = 1/2 × AB × AC
 
1/2 × (x) × (x+5) = 150

x(x+5) = 300

x² + 5x - 300 = 0

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

x = 15 

Sides AB = 15 cm and AC = 15 + 5 = 20 cm

Hypotenuse BC = $\sqrt{20^2 + 15^2}$

                          = $\sqrt{400 + 225}$

                          = $\sqrt{625}$

                          = $25$

Answer 2

ANSWER:-

R.E.F IMAGE

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

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

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

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

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

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

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

$= > x = 15$

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