Question

The base of a triangle is 5 meters greater than the height. If the area of the triangle is 102 m2, which are the dimensions of the base and the height?

Answer 1

Answer: The base is 17 meters long, and the height is 12 meters.

Step-by-step explanation:

The area of a triangle is:

A = b*h/2

where:

b = base

h = height.

For this particular triangle, we have that:

b = 5m + h

A = 102 m^2 = b*h/2

We can replace the first equation into the second one:

b*h/2 = 102 m^2

(5m + h)*h/2 = 102 m^2

(5m + h)*h = 2*102 m^2 = 204m^2

5m*h + h^2 = 204m^2

We can rewrite this as a quadratic equation:

h^2 + 5m*h - 204m^2 = 0

The solutions are given by the Bhaskara formula, the solutions are:

$h = \frac{-5m +- \sqrt{(+5m)^2 -4*1*(-204m^2)} }{2*1} = \frac{-5m+-29m}{2}$

Then the two solutions are:

h = (-5m - 29m)/2 = -17 m

This is a negative height, that is not really defined, so this will be discarded.

The other solution is:

h = (-5m + 29m)/2 = 12m

This is positive, so this is the option we will use.

Knowing h, we can find the value of b.

b = 5m + h = 5m + 12m = 17m

Then the area is:

A = 12m*17m/2 = 102m^2