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The lengths of the sides of a right triangle are (2x – 1) m, (4x) m and (4x + 1) m, where x > 0. Find : i. the value of x, ii. the area of the triangle.

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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 :

  1. the value of x, 
  2. 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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Chapter 6: Problems on Quadratic Equations - EXERCISE 6 [Page 82]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 6 Problems on Quadratic Equations
EXERCISE 6 | Q 24. | Page 82
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