Question

The height of a triangle is 6m more than its base. The area of the triangle is 56m^2.
What is the length of the base?
A: 7m
B: 8m
C: 14m
D: 15m

Answer 1

The answer would be B. 8m.,
because if the base is 8, the height would be 6+8=14, and 14 x 8=112 x 1/2=56 m^2.

Answer 2

The length of the base of the considered triangle whose height is 6m meter more than its base with area 56 sq. meters is given by: Option B: 8 m

How to find the area of a triangle?

If we have:

• Length of its base = b units
• Its height = h units long,

Then we get:

Area of a triangle = $\dfrac{b \times h}{2} \: \rm unit^2$

How to find the roots of a quadratic equation?

Suppose that the given quadratic equation is

$ax^2 + bx + c = 0$

Then its roots are given as:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

We're given that:

• The height of a triangle is 6m more than its base.
• The area of the triangle is 56m²

For this case, we can assume that:

• Length of the base of the considered triangle = b metres
• Length of the height of this triangle = b + 6 metres

Then its area would be:

$A = \dfrac{b(b+6)}{2} \: \rm m^2\\\\56 = \dfrac{b^2 + 6b}{2}\\\\b^2 + 6b - 112 = 0$

Using the formula for finding the roots of this quadratic equation, we get;

$b = \dfrac{-6 \pm \sqrt{6^2 + 4(1)(-112)}}{2 \times 1}\\\\b = \dfrac{-6 \pm \sqrt{36 + 448}}{2} = \dfrac{-6 \pm \sqrt{484}}{2}\\\\b = \dfrac{-6 \pm 22}{2} = -3 \pm 11\\\\b = -14, 8$

Since 'b' represents the length of the base, and length is a non-negative quantity, so we get: b = 8 meters.

Thus, the length of the base of the considered triangle whose height is 6m meter more than its base with area 56 sq. meters is given by: Option B: 8 m

Learn more about finding the solutions of a quadratic equation here:

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