Question

in trapezium abcd diagonals ac and bd intersect at o ar(COD) = 24cm² ar( AOB) = 96 cm² find the area of trapezium​

Answer 1

Answer:

The area of the trapezium is 216 cm²

Step-by-step explanation:

The parameters of the trapezium are;

The point of intersection of the diagonals ac and bd = point o

The area of ΔCOD = 24 cm²

The area of ΔAOB = 96 cm²

Let h represent the height of the trapezium and let x represent the height of triangle ΔCOD, we have;

The height of ΔAOB = h - x

Therefore;

The area of ΔCOD = (1/2) × The length of CD × x = 24

∴ The length of CD = 24/((1/2) × x) = 48/x

The area of ΔAOB = (1/2) × The length of AB × (h - x) = 96

The length of AB = 96/((1/2) × (h - x)) = 192/(h - x)

The area of the trapezium, A = ((48/x + 192/(h - x))/2) × h

We note that ΔCOD ~ ΔAOB, therefore;

(AB)²/(CD)² = (h - x)²/x² = (arΔAOB)/(arΔCOD) = 96/24 = 4

∴ √((h - x)²/x²) = (h - x)/x = √4 = 2

∴ h - x = 2·x

h = 2·x + x = 3·x

A = ((48/x + 192/(h - x))/2) × h

∴ A = ((48/x + 192/(2·x))/2) × 3·x = ((48 + 192/(2))/2) × 3 = 216

The area of the trapezium, A = 216 cm².