Areas of Parallelograms & Triangles — Assertion-Reason Type Questions

Assertion (A) : In the figure, ABCD is a parallelogram. Area of ΔABD = $\dfrac{1}{2}$ Area of ∥ gm ABCD.

Reason (R) : If a triangle and a parallelogram are on the same base and between the same parallels, then area of the triangle is equal to half of the area of the parallelogram.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

In a parallelogram ABCD, the diagonal BD divides it into two triangles ΔABD and ΔBCD of equal area.

Area of △ABD = $\dfrac{1}{2}$ × Area of parallelogram ABCD

Assertion (A) is true.

The statement is a standard area theorem:

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Reason (R) is true.

Both A and R are true.

Hence, option 3 is the correct option.

Assertion (A) : In ΔABC, if D is the mid-point of side AB, then area of ΔBCD = area of ΔACD.

Reason (R) : A triangle and a parallelogram on the same base and between the same parallels are equal in area.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

In ΔABC, point D is the midpoint of AB.

Thus, CD is the median of trinagle ABC.

A median of a triangle divides it into two triangles of equal area.

Thus, area of ΔBCD = area of ΔACD.

Assertion (A) is true.

We know that,

If a triangle and a parallelogram lie on the same base and between the same parallels then area of triangle is equal to half the area of parallelogram.

Area of the triangle = $\dfrac{1}{2}$ area of the parallelogram.

Reason (R) is false.

A is true, R is false.

Hence, option 1 is the correct option.