Question

In the figure, APCB is a trapezium. AF is parallel to BC. The diagonals of the trapezium divide it into four parts. The areas of two parts are given as 45 and 15 sq. units. The area of the trapezium in sq units is:

• A. 90
• B. 120
• C. 82.5
• D. 80

Answer 1

Answer:

D. 80

Area Calculation

The diagonals of a trapezium divide it into four triangles. Let's denote the areas of these triangles as A1​, A2​, A3​, and A4​. There are two key properties of these triangles:

1. The areas of the two triangles sharing the non-parallel sides (the side triangles) are equal.

2. The product of the areas of the two triangles with the parallel sides as bases is equal to the product of the areas of the two side triangles.

Let the given areas be 45 and 15 sq. units. Since the two side triangles must have equal areas, the areas 45 and 15 must correspond to one of the triangles on a parallel base and one of the side triangles.

• Let the area of one triangle with a parallel base be 45 sq. units.

• Let the area of one side triangle be 15 sq. units.

Based on the properties above:

• The area of the other side triangle is also 15 sq. units.

• Let the area of the remaining triangle (on the other parallel base) be x.

Using the second property: (Area of first parallel-base triangle) × (Area of second parallel-base triangle) = (Area of first side triangle) × (Area of second side triangle)

45×x=15×15 45x=225 x=45225​ x=5

So, the areas of the four triangles are 45, 15, 15, and 5.

Total Area

To find the total area of the trapezium, we sum the areas of the four triangles:

Total Area = 45+15+15+5=80 sq. units.