Question

in the following figure angle abc = 90° ,AB = x + 8 cm, BC= x + 1 cm and ac = x + 15 cm find the lengths of the side of the triangle

Answer 1

Given:

In ΔABC,

∠B = 90°

AB = x + 8 cm

BC = x + 1 cm

AC = x + 15 cm

To find:

The lengths of the sides of the triangle

Solution:

Let's assume,

In Δ ABC, side AC is the hypotenuse and the sides AB & BC are the other two sides

So using the Pythagoras theorem, we get

$AC^2 = AB ^2 + BC^2$

substituting the given values of AB, BC & AC

$\implies (x + 15)^2 = (x +8)^2+(x + 1)^2$

$\implies x^2 + 30x + 225 = x^2 + 16x + 64 + x^2 + 2x + 1$

$\implies 30x + 225 = 16x + 64 + x^2 + 2x + 1$

$\implies x^2 + 18x - 30x 65 -225 = 0$

$\implies x^2 -12x - 160 = 0$

$\implies x^2 -20x +8x - 160 = 0$

$\implies x(x - 20) + 8(x - 20) = 0$

$\implies (x - 20)(x + 8)= 0$

$\implies x = 20\: or\: -8$

since the side of a triangle cannot be negative so we cannot take the negative value of x

$\implies \bold{x = 20}$

Thus,

The lengths of the sides of the triangle are:

AB = x + 8 = 20 + 8 = 28 cm

BC = x + 1 = 20 + 1 = 21 cm

AC = x + 15 = 20 + 15 = 35 cm

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Answer 2

Answer:

above one is correct

Step-by-step explanation:

do follow it...