IPMAT Indore 2024 · Quantitative Ability · Question 22

Step 1 — exploit the median. BD is the median to AC, so D is the midpoint of AC. A median splits a triangle into two of equal area:

\left[ \triangle ABC \right]=2\,\left[ \triangle ABD \right]

Step 2 — work inside triangle ABD. Here AB=c, BD=k, \angle BDA=\theta, and AD is unknown. Apply the cosine rule at D:

{c}^{2}=A{D}^{2}+{k}^{2}-2\,\left( AD \right)\,k\cos \theta

A{D}^{2}-\left( 2k\cos \theta \right)AD+\left({k}^{2}-{c}^{2}\right)=0

Step 3 — solve for AD.

AD=k\cos \theta \pm \sqrt{{k}^{2}{\cos}^{2}\theta -{k}^{2}+{c}^{2}}=k\cos \theta \pm \sqrt{{c}^{2}-{k}^{2}{\sin}^{2}\theta}

Since AD>0 and \theta <{90}^{∘}, take the + sign.

Step 4 — area of ABD. Using two sides and the included angle at D:

\left[ \triangle ABD \right]=\frac{1}{2}\cdot AD\cdot BD\cdot \sin \theta =\frac{1}{2}\,k\sin \theta \left( k\cos \theta +\sqrt{{c}^{2}-{k}^{2}{\sin}^{2}\theta}\right)

Step 5 — double it.

\left[ \triangle ABC \right]=k\sin \theta \left( k\cos \theta +\sqrt{{c}^{2}-{k}^{2}{\sin}^{2}\theta}\right)={k}^{2}\sin \theta \cos \theta +k\sin \theta \sqrt{{c}^{2}-{k}^{2}{\sin}^{2}\theta}

=\frac{{k}^{2}\sin 2\theta}{2}+k\sin \theta \sqrt{{c}^{2}-{k}^{2}{\sin}^{2}\theta}

Answer: Option 2

Shortcut. Don’t derive — eliminate on structure. Two observations kill three options instantly:

Area must come from 1/2ab\sin C, so the leading term carries \sin \theta \cos \theta =1/2\sin 2\theta — not \cos 2\theta and not a bare \cos \theta. That removes (a), (c) and (d).

The surd came from a discriminant {c}^{2}-{k}^{2}{\sin}^{2}\theta; a +{k}^{2}{\sin}^{2}\theta inside the root (option c) can never arise from the cosine rule here.

Only option (b) survives. On “messy expression” questions, match the trigonometric shape rather than redoing the algebra.