Question

In the given trapezium ABCD in which AD//BC//MN, MN is a median. Prove that:

i. Area of AABN = area of ADCM.

A. True

B. False

ii. Area of AABN = area of 1/2 of the trapezium ABCD.

A. True

B. False

Answer 1

It is true that the area of triangle ABN is equal to the area of triangle DCM, and the area of triangle ABN is equal to half of the area of trapezium ABCD because MN is a median.

Step-by-step explanation:

The question requires the proof of two statements regarding areas in a trapezium ABCD, where MN is a median.

Statement i: Area of ΔABN = area of ΔDCM

This statement is True. Since MN is a median and AD is parallel to BC, MN is also parallel to both AD and BC. This means the triangles ΔABN and ΔDCM have the same height and their bases (AB and DC) are equal as MN bisects the area of trapezium. Therefore, the areas of ΔABN and ΔDCM are equal.

Statement ii: Area of ΔABN = area of 1/2 of trapezium ABCD

This statement is True. The median of a trapezium divides it into two trapeziums of equal area. Triangle ΔABN occupies half of the upper trapezium, thus its area is equal to 1/2 of the total area of trapezium ABCD.