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Find the lengths of the sides of the triangle?

Find the lengths of the sides of the triangle?-example-1

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Explanation:

it is a right-angled triangle.

so, Pythagoras applies.

c² = a² + b²

c is the Hypotenuse (the side opposite of the 90° angle), a and b are the legs.

so, in our case

(x + 4)² = x² + (x + 1)²

x² + 8x + 16 = x² + x² + 2x + 1 = 2x² + 2x + 1

6x + 15 = x²

0 = x² - 6x - 15

a quadratic equation

ax² + bx + c = 0

has the general solution

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = -6

c = -15

x = (6 ± sqrt((-6)² - 4×1×-15))/(2×1) =

= (6 ± sqrt(36 + 60))/2 =

= (6 ± sqrt(96))/2 =

= (6 ± sqrt(16×6))/2 =

= (6 ± 4×sqrt(6))/2 = 3 ± 2×sqrt(6)

x1 = 3 + 2×sqrt(6) = 7.898979486... ≈ 7.9

x2 = 3 - 2×sqrt(6) = -1.898979486... ≈ -1.9

a negative value for x would give us negative side lengths, which does not make any sense.

so, x1 is our only solution.

that means

x = 7.9

x + 1 = 8.9

x + 4 = 11.9

User Tandi
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