The height of a triangle is 7 m more than its base. If the area of the triangle is 39 sq.m. , then find its height. [ 3 marks]

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Solution

Let $x$ be the height of the triangle, then $(x - 7)$ will be the base.

Area of a triangle $= \frac{1}{2} \times b a s e \times h e i g h t$

$\Rightarrow \frac{1}{2} \times (x - 7) \times x = 39$
$\Rightarrow (x - 7) \times x = 39 \times 2$
$\Rightarrow x^{2} - 7 x = 78$
$\Rightarrow x^{2} - 7 x - 78 = 0$ [ 1 mark]

For a quadratic equation $a x^{2} + b x + c$ ,
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Here $a = 1, b = - 7$ and $c = - 78$

So, $x = \frac{- (- 7) \pm \sqrt{(- 7)^{2} - 4 \times 1 \times - 78}}{2}$ [ 1mark]

$x = \frac{7 \pm \sqrt{361}}{2} = \frac{7 \pm 19}{2}$

$x = 13 o r x = - 6$

$x$ cannot be a negative value. So $x = 13$

$∴$ The height of the triangle is 13 m. [ 1 mark]

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