Chapter 9: Areas of Parallelograms and Triangles

False

Explanation:

In ΔABC, X is the midpoint of AB

Considering the given question,

Hence ar(ΔAXC) = ar(ΔBXC) = ½ ar(ΔABC)

Therefore,

ar(ΔAXC) = ar(ΔBXC) = 12 cm²

Area of ||gm ABCD = 2 x ar(ΔABC)

= 2 x 24 = 48 cm²

But, according to the question,

area of ||gm ABCD = ar(AXCD) + ar(ΔBXC)

= 24 + 12 = 36 cm²

Hence, we find a contradiction here.

So, If ar (AXCD) = 24 cm², then ar (ABC) ≠ 24 cm²

True. Since A is any point on PQ, then ar (PAS) ≠ 30 cm²

But, the statement can be true if PA is equal to PS.

Justification:

According to the question,

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm.

Given,

A is any point on PQ

PA<PQ

ar (△PQR) = ½ ×PQ×QR

= ½ ×12×5

= 30cm²

PS=5 cm

Suppose PA<PQ,

ar(△PAS) < ar(△PQR)

ar(△PAS) < 30 cm²

Suppose PA=PQ,

ar (△PAS) = ½ ×PQ×PS

= ½ ×12×5

= 30cm²