SSC CGL 2024 Tier-I · 26 September 2024, Shift 1 · Quantitative Aptitude · Question 18 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.

From a point A, which is at a distance of 17 cm from the centre C of a circle with radius 8 cm, the pair of tangents AB and AD to the circle are drawn. The area of the quadrilateral ABCD is _____ cm 2 .

(a)192
(b)60
(c)120
(d)360

Answer

Why

Correct — C. CB and CD are radii drawn to the points of contact, so a tangent meets the radius at 90° and triangles ABC and ADC are both right-angled.

AB² = AC² − CB²
= 17² − 8² = 289 − 64 = 225
AB = 15 cm, and AD = 15 cm because tangents from one external point are equal.

Area of ΔABC = ½ × AB × CB = ½ × 15 × 8 = 60 cm²
ABCD = ΔABC + ΔADC = 2 × 60 = 120 cm² → option (c).

Why the others are wrong

(a)
192 — 192 needs each right triangle to be 96 cm², that is a tangent of 24 cm. A leg can never be longer than the hypotenuse AC = 17 cm, so 24 cm is impossible here.
(b)
60 — 60 cm² is one triangle, ΔABC alone. The quadrilateral ABCD is made of two such triangles, ABC and ADC, so its area is double.
(d)
360 — 360 is triple the true area. ABCD is a kite with diagonals AC = 17 cm and chord of contact BD = 240⁄17 ≈ 14.1 cm, and ½ × 17 × 14.1 gives 120 cm².

Concept

Two tangents from an external point make a kite: the two radii to the points of contact are equal, the two tangent segments are equal, and the angle at each point of contact is a right angle.

That turns an area question into two identical right triangles. Get the tangent length by Pythagoras on the radius and the distance to the centre, then double the area of one triangle.

The quadrilateral is named ABCD, so B and D are the points of contact and C is the centre. Its sides are two tangents and two radii, not four chords.

Key facts

A tangent is perpendicular to the radius at the point of contact, which is what makes ΔABC and ΔADC right-angled.
The two tangents drawn from an external point to a circle are equal in length, so AB = AD = 15 cm.
Tangent length from an external point = √(d² − r²), where d is the distance from the point to the centre.
8, 15, 17 is a Pythagorean triple, and recognising it removes the arithmetic entirely.

Study next

Length of the direct and the transverse common tangent to two circlesChord of contact BD and the kite that two tangents form with the radiiArea of a kite as ½ × product of its diagonals

Common traps

Answering 60, which is one of the two right triangles rather than the whole quadrilateral.
Treating AC = 17 cm as a leg instead of the hypotenuse of the right triangle.
Using the radius in place of the tangent length when computing the triangle area.

The same circle is asked both ways: 13 Sep 2024, 12:30, Quant Q.7 gives the identical 8 cm radius and 17 cm external distance but wants the tangent length.

12 Sep 2024, 12:30, Quant Q.18 inverts it — tangent 32 cm, distance from the centre 40 cm, find the diameter.

Related PYQs

No directly related past PYQ was found.