Question

In the figure, APCB is a trapezium. AP 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 square units. What is the area of the trapezium in square units?

• A. 90
• B. 120
• C. 82.5
• D. 80

Answer 1

Answer:

D. 80

Based on the provided image, the total area of the trapezium is 80 square units.

Let the intersection point of the diagonals be O. From the image, we can identify the areas of the given sections:

• Area of the left triangle $\Delta AOB$ = 15

• Area of the bottom triangle $\Delta AOP$ = 45

By applying the geometric properties of a trapezium's intersecting diagonals:

1. Equal Side Areas:
   The triangles adjacent to the non-parallel sides always have equal areas. Therefore, the area of the right triangle $\Delta COP$ is equal to the area of the left triangle $\Delta AOB$.
   $\text{Area\ of\ Right\ Triangle\ }(\Delta COP)=\mathbf{15}$

2. Cross-Product Property of Areas:
   The product of the areas of the opposite triangles formed by the diagonals is always equal:
   $\text{Left\ Area}\times \text{Right\ Area}=\text{Top\ Area}\times \text{Bottom\ Area}$
   $15\times 15=\text{Area\ of\ Top\ Triangle\ }(\Delta BOC)\times 45$
   $225=\text{Area\ of\ Top\ Triangle\ }(\Delta BOC)\times 45$
   $\text{Area\ of\ Top\ Triangle\ }(\Delta BOC)=\frac{225}{45}=\mathbf{5}$

Total Area Calculation

Summing up the areas of all four individual triangles gives the total area of the trapezium:
$\text{Total\ Area}=45\text{\ (Bottom)}+5\text{\ (Top)}+15\text{\ (Left)}+15\text{\ (Right)}=\mathbf{80}\text{\ square\ units}$