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.
Answer: 6.