Question

The base of a triangle is 5 meters longer than the height of a triangle. If the area of the triangle is 102 square meters, what are the base and height of the triangle ?

Answer 1

$area=\frac{1}{2} base*height \\

102=\frac{1}{2}(x+5)(x)\\

102=\frac{x^2+5x}{2} \\

204=x^2+5x\\

0=x^2+5x-204\\

0=(x+17)(x-12)\\

x=-17, x=12$

x cannot be a negative length, therefore x=12

$base=x+5=12+5=17\\

height=x=12$

therefore base = 17 meters and height = 12 meters.

Answer 2

The base and height of the triangle is 17 meters and 12 meters respectively.

• Let the base of the triangle be b.

• Let the height of the triangle be h.

Given the following data:

• Area of triangle = 102 square meters.

To find the base and height of the triangle:

Translating the word problem into an algebraic expression, we have;

$b = 5 + h$

Mathematically, the area of a triangle is given by the formula:

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

Substituting the given parameters into the formula, we have;

$102 = \frac{1}{2} \times (5+h) \times h\\\\102 = \frac{1}{2} \times 5h+h^2\\\\204 = 5h + h^2\\\\h^2 + 5h - 204 =0$

Solving the quadratic equation by factorization, we have:

$h^2 +17h -12h -204 = 0\\\\h(h +17)-12(h+17)=0\\\\(h+17)(h-12)=0$

Height, h = 12 meters

To find the base:

$b = 5 + h\\\\b = 5 + 12$

Base, b = 17 meters.

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