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.
Answer: S = 32.