SSC CGL 2024 Tier-I · 17 September 2024, Shift 2 · Quantitative Aptitude · Question 11 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

In right-angled triangle ABC, ∠ C=90, CM is perpendicular on AB. If AB = 18 cm and BM = 6 cm, then find the length of CM.

(a)2√2 cm
(b)6√2 cm
(c)7√2 cm
(d)4√2 cm

Answer

Why

Correct — B.

M lies on AB, so AM = AB − BM = 18 − 6 = 12 cm.

Triangles AMC and CMB are each similar to ABC — a right angle plus a shared acute angle — so AM ⁄ CM = CM ⁄ BM. That is the altitude rule CM² = AM × BM.

CM² = 12 × 6 = 72
CM = √72 = √(36 × 2) = 6√2 cm → option (b).

Why the others are wrong

(a)
2√2 cm — 2√2 ≈ 2.83 cm squares to 8. The two pieces of the hypotenuse are 12 and 6, and the altitude rule fixes CM² at 72, so 8 cannot arise from these lengths.
(c)
7√2 cm — 7√2 ≈ 9.90 cm is geometrically impossible here. C lies on the circle with AB as diameter, so the altitude to AB can never exceed the radius, AB ⁄ 2 = 9 cm.
(d)
4√2 cm — 4√2 squares to 32, which would need the two hypotenuse pieces to multiply to 32. They are AM = 12 and BM = 6, and 12 × 6 = 72.

Concept

Dropping the altitude from the right angle to the hypotenuse cuts a right triangle into two smaller triangles, both similar to the original and to each other.

That single fact produces three relations. The altitude is the geometric mean of the two hypotenuse pieces, CM² = AM × MB, and each leg is the geometric mean of its adjacent piece and the whole hypotenuse: AC² = AM × AB and BC² = BM × AB.

So one perpendicular turns a length hunt into a multiplication. Here AB = 18 and BM = 6 give AM = 12 immediately, and the rest is arithmetic.

The stem writes the right angle as '∠ C=90' without the degree sign; read it as 90°. It also never says M is the midpoint — M is only the foot of the perpendicular, and here it is not the midpoint, since AM = 12 and BM = 6.

Key facts

In a right triangle, the altitude to the hypotenuse squared equals the product of the two pieces it cuts the hypotenuse into.
Each leg squared equals its adjacent piece times the whole hypotenuse: here AC² = 12 × 18 = 216 and BC² = 6 × 18 = 108.
Check: 216 + 108 = 324 = 18², so the right angle at C is consistent with CM = 6√2.
The altitude to the hypotenuse can never exceed half the hypotenuse, because the right-angled vertex lies on the circle with the hypotenuse as diameter.

Study next

The three geometric-mean relations produced by the altitude on the hypotenuseSimilar triangles created by a perpendicular from a right angle, and how to pair corresponding sidesFinding the altitude from the area identity AC × BC = AB × CM

Common traps

Pairing BM = 6 with AB = 18 instead of with AM = 12, which gives the wrong 108.
Reading M as the midpoint of AB, so that BM is taken as 9.
Stopping at CM² = 72 and not simplifying √72 to 6√2.

SSC asks this as a two-line computation: give the hypotenuse and one piece, ask for the altitude. The whole item is AM = AB − BM followed by one multiplication and one surd.

A neighbouring shift tests the same habit of reading a perpendicular into a pair of right triangles: 17 Sep 2024, 16:00, Quant Q.16 drops DL ⊥ AB and DM ⊥ AC from the midpoint of BC and asks what kind of triangle results.

Related PYQs

No directly related past PYQ was found.