Assertion (A): In the given figure, if the area of the parallelogram ABEF is 120 cm², then area of rectangle ABCD is 120 cm².

Reason (R): Parallelogram and rectangle 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.

Both A and R are true.

Explanation

In Δ ADF and Δ BCE,

AD = BC (Opposite sides of rectangle ABCD)

∠ ADF = ∠ BCE (Corresponding angles)

∠ AFD = ∠ BEC (Corresponding angles)

∠ DAF = ∠ CBE (Since, two angles of triangles are equal, therefore their third angle will also be equal)

By ASA Congruency Criterion,

Δ ADF ≅ Δ BCE

⇒ Area (Δ ADF) = Area (Δ BCE) (Congruent triangles are equal in area)

Adding Area (Δ ABED) on both sides, we get

Area (Δ ADF) + Area (Δ ABED) = Area (Δ BCE) + Area (Δ ABED)

Area (||gm ABEF) = Area (Rectangle ABCD)

Hence, the area of parallelogram ABEF = area of rectangle ABCD = 120 cm².

∴ Assertion (A) is true.

As it is proved above, the area of a parallelogram is equal to the area of a rectangle on the same base and between the same parallels.

∴ Reason (R) is true.

Hence, both Assertion (A) and Reason (R) are true.