Question

The altitude of triangle is 6cm more than its base if the area is 108 cm2 then find the base and height with respect to altitude

Answer 1

GIVEN:

• The altitude of triangle is 6cm more than its base
• The area of triangle is 108 cm²

TO FIND:

• What is the base and height of the triangle ?

SOLUTION:

Let the base of the triangle be 'x' cm

➾ Altitude of triangle is 6cm more than its base

• ALTITUDE = (x + 6) cm

We know that the formula for finding the area of the triangle is:-

$\large\bf{\star \: AREA = \dfrac{1}{2} \times base \times height\: \star}$

According to question:-

On putting the given values in the formula, we get

$\rm{\dashrightarrow 108 = \dfrac{1}{2} \times x \times(x +6)}$

$\rm{\dashrightarrow 108 \times 2 = x(x+6) }$

$\rm{\dashrightarrow 216 = x^2 + 6x}$

$\rm{\dashrightarrow 0 = x^2 + 6x -216 }$

$\rm{\dashrightarrow 0 = x^2 +(18-12)x -216}$

$\rm{\dashrightarrow 0 = x^2 + 18x -12x - 216}$

$\rm{\dashrightarrow 0 = x(x + 18) -12(x + 18) }$

$\rm{\dashrightarrow 0 = (x+18) (x-12) }$

$\rm{\dashrightarrow x = -18 \: (Neglected)}$

$\bf{\dashrightarrow x = 12 }$

❝ The value of 'x' is 12 ❞

• BASE = x = 12 cm
• HEIGHT = (x+6) = 12+6 = 18 cm

______________________

Answer 2

GIVEN:-

b=12cm

area=${48cm^2}$

TO FIND:-

h=?

solution:-

WE KNOW THE AREA OF TRIANGLE AS

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

now, put the given values.....

$\frac{1}{2} \times 12 \times h = {48cm}^{2}$

$= > \frac{1}{ \cancel2} \times \cancel{12} \times h = 48$

$= > 6 \times h = 48$

$= > h = \frac{ \cancel{48}}{ \cancel 6}$

$= > h = 8cm$

so, the height is

$\boxed{8cm}$