# In a right-angled triangle, the hypotenuse is 5 and one leg is 1 more than the other. find the area of the triangle.

Given:

right-angled triangle ABC;

angle C = 90;

AB – hypotenuse;

AB = 5;

CA = CB + 1.

Find the area ABC, that is

Solution:

Consider a right-angled triangle ABC. Let x, then CA is equal to x + 1. Then by the Pythagorean theorem (the square of the hypotenuse is equal to the sum of the squares of the legs):

AC ^ 2 + BC ^ 2 = AB ^ 2:

(x + 1) ^ 2 + x ^ 2 = (5) ^ 2;

x ^ 2 + 2x + 1 + x ^ 2 = 25;

2x ^ 2 + 2x + 1 – 25 = 0;

2x ^ 2 + 2x – 24 = 0;

a = 2, b = 2, c = -24;

D = b ^ 2 – 4 * a * c = 4 – 4 * 2 * (-24) = 196 (the discriminant is greater than zero, then this quadratic equation has two roots);

x = (-b + √ D) / 2 * a = (-2 + √ 196) / 2 * 2 = (-2 + 14) / 2 * 2 = 12/4 = 3;

x = (-b – √ D) / 2 * a = (-2 – √ 196) / 2 * 2 = (-2 – 14) / 4 = -4 does not satisfy the condition of the problem.

AC = 3 and BC = 4.

2) Saavs = 1/2 * CA * BC;

Sav = 1/2 * 3 * 4;

Savs = 6.