Question

In the given figure, P and Q are the mid-points of AC and AB. Also, PG = GR and HQ = HR. What is the ratio of the Area of ΔPQR : Area of ΔABC

(a) 1 : 2 

(b) 2 : 3 

(c) 3 : 4 

(d) 3 : 5

Answer 1

(a) 1 : 2

P and Q being the mid-points of AC and AB respectively, PQ || BC and PQ = \(\frac{1}{2}\) BC 

Let AF ⊥ BC be drawn such that it intersects PQ and BC in E and F respectively. 

PQ || BC ⇒ \(\frac{AE}{EF}=\frac{AP}{PC} = 1\) ⇒ AE = EF 

Also let RI ⊥ PQ be drawn such that it intersect BC and PQ in J and I respectively. G and H being the mid-points of sides PR and RQ of ΔPQR, GH || PQ and GH = \(\frac{1}{2}\) PQ 

(By midpoint Theorem)

Also, GH || PQ ⇒ \(\frac{RJ}{JI}=\frac{RG}{GP}=1\) ⇒ RJ = RG 

(By Basic Proportionality Theorem) 

But EF = JI 

∴ AE = EF = RJ = JI 

∴ AF = AE + EF = RJ + JI = RI = h (say)

Then, \(\frac{\text{Area (ΔPQR)}}{\text Area (ΔABC)}\) = \(\frac{\frac{1}{2}\times{PQ}\times{h}}{\frac{1}{2}\times{BC}\times{h}}=\frac{PQ}{BC}=\frac{1}{2}.\)