Quadratic Equation Solution18





In this page quadratic equation solution18 we are going to see solution of the word problems of the topic quadratic equation.

Question 25

The area of the right angled triangle is 500 cm². If the base of the triangle exceeds the altitude by 15 cm,find the dimensions of the triangle.

Solution:

The area of the right angled triangle = 500 cm²

here base of the triangle is compared by using altitude

Let "x" be the altitude of the triangle

Let "y" be the base of the triangle

base of the triangle exceeds the altitude by 15 cm

 y = x + 15

Area of the triangle = 500

(1/2) x b x h = 500

(1/2) x  (x + 15) x x = 500

  x (x + 15) = 500 x 2

 x² + 15 x = 1000

 x² + 15 x - 1000 = 0

 x² + 40 x - 25 x - 1000 = 0

 x (x + 40) - 25 (x + 40) = 0

(x - 25) (x + 40) = 0

x - 25 = 0                  x + 40 = 0

x = 25                          x = - 40

Here x represents length of altitude. So negative value is not possible.

Now we are going to apply the value of x in the equation y = x + 15 to find the value of base length.

y = 25 + 15

y = 40 cm

Therefore the length of altitude = 25 cm

          value of base length = 40 cm     

Verification:

altitude = 25 cm

base = 40 cm    

Area of the triangle = 500

(1/2) x 40 x 25 = 500

      20 x 25 = 500

           500 = 500

quadratic equation solution18