SSC CGL 2024 Tier-I · 25 September 2024, Shift 3 · Quantitative Aptitude · Question 25 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 10 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 ∆PQR, S and T are the midpoints of the sides PQ and PR, respectively. The length of the side QR is 12 cm. If ST is parallel to QR, then find the length (in cm) of ST.

(a)4
(b)8
(c)6
(d)10

Answer

Why

Correct — C. S and T are the midpoints of PQ and PR, so ST is the midsegment of ∆PQR.

Midpoint theorem: the segment joining the midpoints of two sides is parallel to the third side and half its length.

ST = ½ × QR
= ½ × 12 = 6 cm → option (c).

The stem's remark that ST is parallel to QR is not extra data; it is part of what the theorem already guarantees.

Why the others are wrong

(a)
4 — 4 is QR ÷ 3. No triangle rule produces a third here — the midsegment is fixed at exactly half the third side, which is 6 cm.
(b)
8 — 8 is ⅔ of 12. That 2:1 ratio belongs to the centroid on a median, not to the segment joining two midpoints.
(d)
10 — 10 is longer than half of QR. Since ST = QR ÷ 2 for every triangle with QR = 12 cm, no shape of ∆PQR can stretch ST to 10.

Concept

The midpoint theorem says that in any triangle, the segment joining the midpoints of two sides is parallel to the third side and equal to half of it.

Its converse is just as useful: a line drawn through the midpoint of one side, parallel to a second side, bisects the third.

The result does not depend on the shape of the triangle. ST = 6 cm whether ∆PQR is right-angled, obtuse or equilateral, so long as QR = 12 cm.

Stating the parallelism in the question makes it look like a similarity problem needing a ratio. It is not: the two midpoints alone force both the parallelism and the halving.

Key facts

The segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
With QR = 12 cm, ST = 6 cm for any triangle PQR.
∆PST is similar to ∆PQR in the ratio 1:2, so its area is one quarter of ∆PQR.
Drawing all three midsegments cuts a triangle into four congruent triangles.

Study next

The converse of the midpoint theorem, used to prove that a point is a midpointThe medial triangle formed by joining all three midpoints, as at 24 Sep 2024, 16:00, Quant Q.2Area ratios in a triangle cut by a midsegment, as at 11 Sep 2024, 12:30, Quant Q.5

Common traps

Treating the given parallelism as new information and setting up similarity ratios.
Halving the wrong side — ST halves QR, the side the two midpoints do not touch.
Confusing the midsegment's 1:2 ratio with the centroid's 2:1 ratio on a median.

SSC asks the midpoint family in several shapes: the congruence of ∆DEF at 24 Sep 2024, 16:00, Quant Q.2, and an equilateral version at 13 Sep 2024, 12:30, Quant Q.11.

It also asks for the area left when a midsegment cuts a corner off — 11 Sep 2024, 12:30, Quant Q.5 wants quadrilateral MNRQ and is keyed 34.5 from an area of 46.

The plain length version here needs only the halving.

Related PYQs

No directly related past PYQ was found.