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Question
The lengths of the sides of a right triangle are (2x – 1) m, (4x) m and (4x + 1) m, where x > 0. Find :
- the value of x,
- the area of the triangle.
Sum
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Solution
The hypotenuse is the longest side, which is $$(4x + 1)\text{ m}$$.
By Pythagoras theorem: $$(2x - 1)^2 + (4x)^2 = (4x + 1)^2$$
$$(4x^2 - 4x + 1) + 16x^2 = 16x^2 + 8x + 1$$
$$4x^2 - 4x + 1 = 8x + 1$$
$$4x^2 - 12x = 0$$
$$4x(x - 3) = 0$$
Since $$x > 0$$, we have $$x = 3$$.
i. The value of $$x = 3$$.
ii. Sides containing the right angle are:
Side 1 $$= 2(3) - 1 = 5\text{ m}$$
Side 2 $$= 4(3) = 12\text{ m}$$
Area of the triangle $$= \frac{1}{2} \times \text{base} \times \text{height}$$
$$= \frac{1}{2} \times 5 \times 12$$
$$= 30\text{ m}^2$$
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