Let A, B, C be the mid-points of sides PQ, QR PR, respectively, of ΔPQR. If the area of ΔPQR is 32 cm² then find the area of ΔABC. 

Given:

Area of △PQR = 32 cm2

A, B, and C be the mid-points of sides PQ, QR and PR.

Concept used:

The line segment in a triangle joining the midpoint of any two sides of the triangle,

is said to be parallel to its third side and is also half of the length of the third side.

Heron's formula:

Area = √{S(S - a)(S - b)(S - c)}

Where, S = (a + b + c)/2

And a, b, and c are the sides of the triangle.

Calculation:

Semi-perimeter (S) =  (PQ + QR + PR)/2

Area of △XYZ = √S(S - p)(S - q)(S - r)

⇒ √[(PQ + QR + PR)/2 × {(PQ + QR + PR)/2 - QR} × {(PQ + QR + PR)/2 - PR} × {(PQ + QR + PR)/2 - PQ}]

⇒ √[(PQ + QR + PR)/2 × {(PQ + QR + PR) - 2QR}/2 × {(PQ + QR + PR) - 2PR}/2 × {(PQ + QR + PR) - 2PQ}/2]

⇒ √[(PQ + QR + PR)/2 × (PQ + QR + PR)/2 × (PQ + QR + PR)/2 × (PQ + QR + PR)/2]

⇒ (PQ + QR + PR)2/4 = 32 cm2 -------- (1)

With the midpoint theorem:

AC = QR/2 ; AB = PR/2 ; BC = PQ/2

Semi-perimeter (S) = (AB + BC + AC)/2

⇒ {(PR/2) + (PQ/2) + (QR/2)}/2

⇒ (PR + PQ + QR)/4

Area of △ABC = √S(S - p)(S - q)(S - r)

⇒ √[(PR + PQ + QR)/4 × {(PR + PQ + QR)/4 - QR/2} × {(PR + PQ + QR)/4 - PR/2} × {(PR + PQ + QR)/4 - PQ/2}]

⇒ √[(PR + PQ + QR)/4 × {(PR + PQ + QR) - 2QR}/4 × {(PR + PQ + QR) - 2PR}/4 × {(PR + PQ + QR) - 2PQ}/4]

⇒ √[(PR + PQ + QR)/4 × (PR + PQ + QR)/4 × (PR + PQ + QR)/4 × (PR + PQ + QR)/4]

⇒ (PR + PQ + QR)2/(4 ×4)

⇒ 32/4 = 8 cm2  ------ From (1)

∴ The correct answer is 8 cm2.

Shortcut TrickConcept used:

Area of △ABC = (1/4) × Area of △PQR

Calculation:

Area of △ABC = (1/4) × Area of △PQR

⇒ 32/4 = 8 cm2

∴ The correct answer is 8 cm2.