SSC CGL 2025 Tier-II · Paper-I, 19 January 2026 · Mathematical Abilities · Question 11 of 150

We transcribed this text from the pictures of a candidate's response sheet for this paper; the answer is the key SSC marked with a green tick on that sheet. Solution last reviewed 11 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. This paper reaches us as a candidate's response sheet saved as pictures, so we transcribed the text from those pictures, and a second reader checked every question against them; where the layout carries meaning, we show the picture too. Each answer is the key SSC marked with a green tick on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. UnlockIAS wrote the solutions.

In a right-angled triangle ABC, the angle at B is 90° . If AB = 6 cm and BC = 8 cm, what is the distance between the centroid G and the orthocenter H?

(a)5 cm
(b)10⁄3 cm
(c)8 cm
(d)20⁄3 cm

Answer

Why

Correct — B. Locate each centre, then measure between them.

Orthocentre: the legs AB and BC are both altitudes, so they meet at H = B
Hypotenuse by Pythagoras: AC = √(6² + 8²) = √100 = 10 cm
Median from B to the midpoint M of AC: M is equidistant from A, B and C, so BM = AC ÷ 2 = 5 cm

Centroid: G lies two-thirds of the way from B to M, so BG = 2⁄3 × 5 = 10⁄3 cm
Distance: GH = GB = 10⁄3 cm → option (b)

Why the others are wrong

(a)
5 cm — 5 cm is the whole median BM, which is also the distance from H (at B) to the circumcentre M. The centroid sits only two-thirds of the way along it, at 10⁄3 cm.
(c)
8 cm — 8 cm is the leg BC, not a distance between centres. G lies on the median BM, which is only 5 cm long, so GH cannot exceed 5 cm.
(d)
20⁄3 cm — 20⁄3 cm is two-thirds of the hypotenuse (2⁄3 × 10). The 2 : 1 split belongs to the median BM = 5 cm, and 20⁄3 cm is longer than BM itself.

Concept

A triangle has several centres, each fixed by a different set of lines. The centroid G is where the medians meet, and it cuts each median 2 : 1 from the vertex.

The orthocentre H is where the altitudes meet. The circumcentre O is the point equidistant from all three vertices.

In a right-angled triangle, H is the right-angle vertex and O is the midpoint of the hypotenuse, so the median to the hypotenuse runs from H to O.

Check with the Euler line: H, G and O lie on one line with HG = 2 × GO. Here H = B and O = M, so HO = 5 cm and HG = 2⁄3 × 5 = 10⁄3 cm.

Key facts

The centroid divides each median in the ratio 2 : 1, the longer part next to the vertex.
In a right-angled triangle the orthocentre is the right-angle vertex.
In a right-angled triangle the circumcentre is the midpoint of the hypotenuse, so the median to the hypotenuse is half the hypotenuse.
On the Euler line the centroid divides HO so that HG = 2 × GO.

Study next

Inradius of a right-angled triangle: r = (a + b − c) ⁄ 2, with c the hypotenuseMedian lengths from Apollonius' theorem: AB² + BC² = 2(BM² + AM²)Where the centres coincide (equilateral triangle) and where H falls outside (obtuse triangle)

Common traps

Placing the orthocentre inside the triangle out of habit: with a right angle at B it sits at B.
Taking two-thirds of the hypotenuse instead of the median, which gives 20⁄3 cm.

Here the right angle fixes the orthocentre for free.

Triangle centres also appear at 19 Jan 2026, 11:00 AM, Maths Q.4 (the circumcentre of an equilateral triangle coincides with its centroid), and the centroid's 2 : 1 split of a median decides 11 Sep 2024, 16:00, Quant Q.8.

Related PYQs

No directly related past PYQ was found.

Topic: Triangle and its various kinds of centres — all 9 SSC CGL questions on it, every solved shift.