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?

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:

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