Word problems on Quadratic Equations (3)

One leg of a right triangle is 7 cm more than the other. Given that the hypotenuse is 17 cm, find the measure of the other two sides of the triangle.

Solution:

  • Let the length of the shorter leg be `x`
  • Therefore the length of the other leg is `7 + x`
Given that hypotenuse is 17cm, using the Pythagoras theorem:
`x^2 + (7 + x)^2 = 17^2`
`x^2 + 49 + 14x + x^2 = 289`
`2x^2 + 14x + 49- 289 = 0`
`2x^2 + 14x - 240 = 0`
`2x^2 + 30x - 16x - 240 = 0`
`2x(x + 15) -16(x + 15)= 0`
`(x + 15) (2x - 16) = 0`
`x + 15 = 0` and `2x - 16 = 0`
`x = -15` and `2x = 16`
`x = -15` and `x = 8` 
Since length can't be negative, so x = 8 is the length of the shorter leg. Since one side measures 8 cm therefore the other side measures 8 + 7 = 15 cm. So the two sides of the right triangle are 8 and 15 cm respectively.

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