Question

The sides of a triangle have length x, x 4, and 20. If the length of the longest side is 20, which value of x would make the triangle acute? 8 10 12 14.

Answer 1

The correct option is b, 10.

Given to us,

Sides of the triangle x, (x+4) and, 20.

According to the Triangle Inequality Theorem, the sum of any 2 sides of a triangle must be greater than the third side.

Thus,

$x+(x+4)>20\\2x+4>20\\2x> 16\\x>8$

Now, let us assume it as a right-angle triangle with the longest side as 20.

Thus,

$x^2+(x+4)^2=20^2$

using the expression, (a+b)² = a² + b² + 2ab;

$x^2+x^2+16+8x=400$

Diving both sides by 2,

$x^2+8+4x=200$

subtracting 200 on both sides,

$x^2+8+4x-200=200-200\\2x^2+4x-192=0$

Solving further, using ;

x = 12, -16

As negative value can be neglected.

for an Acute angle triangle,

X should be 12 >x >8. Therefore the only feasible option is 10.

Hence, the correct option is b, 10.

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