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Suppose the medians PP′ and QQ′ of ΔPQR intersect at right angles. If QR = 3 and PR = 4, then the length of side PQ is

(a) √3 

(b) √2 

(c) √5 

(d) √6

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Best answer

(c) √5.

Join P′Q′. 

P′, Q′ being the mid-points of QR and PR respectively, we have P′Q′ || PQ and P′Q′ = \(\frac{1}{2}\) PQ 

(By the mid-point theoram) 

Let OP′ = a, OQ′ = b, OP = c, OQ = d and PQ = x

Then, P′Q′ = \(\frac{1}{2}\) \(x\)

∴ In rt. Δ OP′Q′, 

a2 + b2 = \(\frac{2}{4}\) \(x\)             …(i) 

In rt. Δ OP′Q, 

a2 + d2 = \(\frac{9}{4}\)              …(ii) 

In rt. Δ OQP, 

c2 + d2 = x2            …(iii) 

In rt. Δ OQ′P, 

b2 + c2 = 4            …(iv) 

∴ Eq (i) – Eq (ii) + Eq (iii) – Eq (iv) 

⇒ a2 + b2 – (a2 + d2) + (c2 + d2) – (b2 + c2) = \(\frac{x^2}{4}\) - \(\frac{9}{4}\) + x- 4 

⇒ 0 = \(\frac{5x^2}{4}-\frac{25}{4}\) ⇒ \(\frac{5x^2}{4}=\frac{25}{4}\) ⇒ x2 = 5 ⇒ x = √5.

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