In an isosceles trapezoid abcd ad = 10, bc = 6, angle abd = 90 degrees. find the area of the trapezoid.

Given a trapezoid ABCD: AB = CD, AD = 10, BC = 6, ∠ABD = 90 ° (thus BD is the diagonal).

1. The length of the diagonal of an isosceles trapezoid is found by the formula:

d = √ (a * b + c²),

where a and b are the lengths of the bases of the trapezoid, c is the length of the lateral side.

Substitute the known values ​​into the formula:

BD = √ (AD * BC + AB²);

BD = √ (10 * 6 + AB²);

BD = √ (60 + AB²).

2. Consider a rectangular △ ABD: AD = 10 – hypotenuse (since it lies opposite the right angle), AB and BD – legs.

By the Pythagorean theorem:

BD = √ (AD² – AB²);

BD = √ (10² – AB²);

BD = √ (100 – AB²).

3. Received a system of equations with two unknowns:

BD = √ (60 + AB²);

BD = √ (100 – AB²).

Let’s equate the right-hand sides of the equations:

√ (60 + AB²) = √ (100 – AB²).

Let’s square both sides of the resulting equation with one variable:

(√ (60 + AB²)) ² = (√ (100 – AB²)) ²;

60 + AB² = 100 – AB²;

AB² + AB² = 100 – 60;

2 * AB² = 40;

AB² = 40/2;

AB² = 20;

AB = √20;

AB = 2√5.

4. The area of ​​an isosceles trapezoid is equal to:

S = (p – c) * √ ((p – a) * (p – b)),

where p is a semi-perimeter.

The semi-perimeter of an isosceles trapezoid is:

p = (a + b + 2 * c) / 2.

Semi-perimeter ABCD is:

p = (AD + BC + 2 * AB) / 2 = (10 + 6 + 2 * 2√5) / 2 = (16 + 4√5) / 2 = 8 + 2√5.

Area ABCD is:

S = (8 + 2√5 – 2√5) * √ ((8 + 2√5 – 10) * (8 + 2√5 – 6)) = 8 * √ ((2√5 – 2) * ( 2√5 + 2)) = 8 * √ ((2√5) ² – 2²) = 8 * √ (20 – 4) = 8 * √16 = 8 * 4 = 32. 