In this page quadratic equation solution17 we are going to see solution of the word problems of the topic quadratic equation.

**Question 24**

The length of the rectangle exceeds its width by 2 cm and the area of the rectangle is 195 sq.cm. Find the dimensions of the rectangle.

**Solution:**

Here length is compared by its width

Let "x" be the width of the rectangle

Let "y" be the length of the rectangle

The length of the rectangle exceeds its width by 2 cm

So,Length (y) = x + 2

Area of the rectangle = 195 sq.cm

Length **x** width = 195

(x + 2) x = 195

x² + 2 x - 195 = 0

x² + 15 x - 13 x - 195 = 0

x (x + 15) - 13 (x + 15) = 0

(x + 15) (x - 13) = 0

x + 15 = 0 x - 13 = 0

x = -15 x = 13 cm

Here x represents width of the rectangle. So the negative value is not possible.

To find the value of y we have to apply the value of x in the equation y = x + 2

y = 13 + 2

y = 15 cm

Therefore length of rectangle = 15 cm

Width of the rectangle = 13 cm

**Verification:**

area of the rectangle = 195 sq.cm

15 **x** 13 = 195

195 = 195

quadratic equation solution17

- Back to worksheet
- Factoring a Quadratic Equation
- Factoring Worksheets
- Framing Quadratic Equation From Roots
- Framing Quadratic Equation Worksheet
- Remainder Theorem
- Relationship Between Coefficients and roots
- Roots of Cubic equation
- Roots of Polynomial of Degree4
- Roots of Polynomial of Degree5