Question

Diagonals of an isosceles trapezoid are perpendicular to each other. The length of its leg is 26 cm. An altitude from the vertex of an obtuse angle to the base divides the longer base into two parts, such that the shorter part is 10 cm long. What is the area of this trapezoid? PLEASE HELP!

Answer 1

Answer:

the area of this trapezoid = 576  cm²

Step-by-step explanation:

Leg length  = 26 cm

Shorter  Part = 10 cm

Hence Altitude = √26² - 10² = √576 = 24

now as isosceles trapezoid

Let say diagonal cut length = a & b

a² + b² = 26²

=> a² + b² = 676

let say c is shorter parallel side

then c + 10 + 10 = c + 20  longer paralle side

a² + a² = c² => 2a² = c²

b² + b² = (c + 20)² => 2b² = c² + 40c + 400

Adding both

2a² + 2b² = c² + c² + 40c + 400

=> 2( a² + b²) = 2c² + 40c + 400

=> a² + b² = c² + 20c + 200

=> 676 = c² + 20c + 200

=> c² + 20c - 476 = 0

=> c² + 34c - 14c - 476 = 0

=> (c + 34)(c - 14) = 0

=> c = 14

Area = (1/2)(c + c + 20) * Altitude

= (1/2) ( 48) * 24

= 576  cm²

the area of this trapezoid = 576  cm²