The length of the base of a triangle is 1 cm more than the corresponding height. If the area of the triangle is 45 cm²​, find the length of the base of the triangle.

Correct option is D

Given:

The length of the base of the triangle is 1 cm more than the height.

The area of the triangle is 45 cm².

Formula Used:

The formula for the area of a triangle is:

Area =$$\frac{1}{2} \times \text{base} \times \text{height}$$​

Solution:

Let the height of the triangle be h cm.
The base of the triangle will be h + 1 cm (since it is 1 cm more than the height).

Using the given information, the area of the triangle is:

45 =$$\frac{1}{2} \times (h + 1) \times$$​h

90 = (h+1) × h

90 =$$h^2$$​+ h

​$$h^2 + h - 9$$​0 = 0

Solve this quadratic equation using the quadratic formula:

h =$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$​

Here, a = 1, b = 1, and c = -90.

h =$$\frac{-1 \pm \sqrt{1^2 - 4(1)(-90)}}{2(1)}$$​

h =$$\frac{-1 \pm \sqrt{1 + 360}}{2}$$​

h $$= \frac{-1 \pm \sqrt{361}}{2}$$​

h $$= \frac{-1 \pm 19}{2}$$​

We have two possible solutions for h:

h $$= \frac{-1 + 19}{2} = \frac{18}{2} = 9$$​

h = $$\frac{-1 - 19}{2} = \frac{-20}{2} = -10$$​

Since height can not be negative, we take h = 9 cm.
The base of the triangle is h + 1 = 9 + 1 = 10 cm.
The length of the base of the triangle is 10 cm.