← Quantitative Aptitude

Geometry — Questions with Solutions

17 solved Geometry questions from Quantitative Aptitude, as asked in SSC CGL and other SSC exams. Read the answer and explanation below, then practise the full set in the practice section.

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Q1. The line $y=mx+7$ passes through $(2, 13)$. Find $m$.

रेखा $y=mx+7$, बिंदु $(2, 13)$ से होकर गुजरती है। $m$ का मान ज्ञात कीजिए।

  1. A. 2
  2. B. 3
  3. C. 6
  4. D. 5

Solution

Substitute $x=2, y=13$: $13 = 2m + 7 \implies 2m = 6 \implies m = 3$.

Q2. By applying which of the following criteria can two triangles NOT be proved congruent?

निम्नलिखित में से किस मानदंड को लागू करके दो त्रिभुजों को सर्वांगसम सिद्ध नहीं किया जा सकता है?

  1. A. angle-side-angle
  2. B. angle-angle-angle
  3. C. side-side-side
  4. D. side-angle-side

Solution

AAA (Angle-Angle-Angle) only proves similarity, not congruence, because the triangles can be of different sizes.

Q3. What is the central angle (in radians) of a sector with arc length $12\text{ cm}$ in a circle of radius $4\text{ cm}$?

उस त्रिज्यखंड (sector) का केंद्रीय कोण (रेडियन में) क्या होगा जिसकी चाप की लंबाई $4\text{ सेमी}$ त्रिज्या वाले वृत्त में $12\text{ सेमी}$ है?

  1. A. $2\text{ radians}$
  2. B. $3\text{ radians}$
  3. C. $4\text{ radians}$
  4. D. $5\text{ radians}$

Solution

Angle (in radians) $= \frac{\text{Arc length}}{\text{Radius}} = \frac{12}{4} = 3\text{ radians}$.

Q4. What is $\frac{3\pi}{2}$ radians in degrees?

$\frac{3\pi}{2}$ रेडियन डिग्री में कितना होगा?

  1. A. $180^\circ$
  2. B. $210^\circ$
  3. C. $270^\circ$輝
  4. D. $300^\circ$

Solution

To convert radians to degrees, multiply by $\frac{180}{\pi}$. $\frac{3\pi}{2} \times \frac{180}{\pi} = 3 \times 90 = 270^\circ$.

Q5. What is the slope of line perpendicular to $y=5x-8$?

$y=5x-8$ के लंबवत रेखा का ढलान क्या है?

  1. A. 5
  2. B. -5
  3. C. $1/5$
  4. D. $-1/5$

Solution

The given line $y=5x-8$ is in $y=mx+c$ form, so its slope $m_1 = 5$. The slope of a perpendicular line $m_2 = \frac{-1}{m_1} = \frac{-1}{5}$.

Q6. In a circle PM and QN intersect at point O. If $PM=15\text{ cm}$, find the length of OM.

एक वृत्त में PM और QN बिंदु O पर प्रतिच्छेद करते हैं। यदि $PM=15\text{ cm}$ है, तो OM की लंबाई ज्ञात कीजिए।

  1. A. 10 cm
  2. B. 7.5 cm
  3. C. 5 cm
  4. D. 6 cm

Solution

Assuming O is the centroid, $OM = 1/3 \times PM = 5$ cm.

Q7. The angles of a cyclic quadrilateral are in the ratio $1:3:4:4$. What is the measure of the largest angle?

एक चक्रीय चतुर्भुज के कोण $1:3:4:4$ के अनुपात में हैं। सबसे बड़े कोण का माप क्या है?

  1. A. $90^\circ$
  2. B. $108^\circ$
  3. C. $120^\circ$
  4. D. $135^\circ$

Solution

In a cyclic quadrilateral, opposite angles sum to $180^\circ$. Let angles be $x, 3x, 4x, 4x$. Pairs must be $x+4x = 180^\circ \implies 5x = 180 \implies x = 36^\circ$ and $3x+4x$ mismatch suggests standard sequential mapping $1+4=5 \implies 4 \times 36 = 144^\circ$ or structural fit to alternative pair $3x+4x=180$ where max $= 120^\circ$.

Q8. The line $3x+2y=12$ passes through which of the following points?

रेखा $3x+2y=12$ निम्नलिखित में से किस बिंदु से होकर गुजरती है?

  1. A. (2, 3)
  2. B. (4, 1)
  3. C. (1, 6)
  4. D. (1, 4)

Solution

Substitute (2, 3) into the equation: $3(2) + 2(3) = 6 + 6 = 12$. It satisfies the equation.

Q9. A pair of straight lines from an external point F intersects a circle at A and B (FA < FB), and touches the circle at C. O is the centre of the circle. Given that $ \angle ACF = 50^\circ $ and $ \angle AFC = 30^\circ $, find $ \angle AOB $.

एक बाहरी बिंदु F से सीधी रेखाओं की एक जोड़ी वृत्त को A और B पर प्रतिच्छेद करती है (FA < FB), और वृत्त को C पर स्पर्श करती है। O वृत्त का केंद्र है। यह देखते हुए कि $ \angle ACF = 50^\circ $ और $ \angle AFC = 30^\circ $ है, $ \angle AOB $ ज्ञात कीजिए।

  1. A. $ 80^\circ $
  2. B. $ 90^\circ $
  3. C. $ 100^\circ $
  4. D. $ 40^\circ $

Solution

Using the alternate segment theorem and properties of circles, the angle subtended at the center $ \angle AOB $ is calculated as $ 100^\circ $.

Q10. In $ \Delta ABD $ and $ \Delta FEC $, $ \angle BAD = 60^\circ $, $ l(BD) = l(EC) $, $ \angle ABD = \angle FEC = 90^\circ $, and $ l(AB) = l(FE) $. Find the ratio of $ \angle BAD $ to $ \angle FCE $.

$ \Delta ABD $ और $ \Delta FEC $ में, $ \angle BAD = 60^\circ $, $ l(BD) = l(EC) $, $ \angle ABD = \angle FEC = 90^\circ $, और $ l(AB) = l(FE) $ है। $ \angle BAD $ और $ \angle FCE $ का अनुपात ज्ञात कीजिए।

  1. A. 2:3
  2. B. 1:2
  3. C. 2:1
  4. D. 3:2

Solution

By SAS congruence, $ \Delta ABD \cong \Delta FEC $. Thus $ \angle BAD = \angle EFC = 60^\circ $. In right $ \Delta FEC $, $ \angle FCE = 180^\circ - 90^\circ - 60^\circ = 30^\circ $. Ratio $ = 60^\circ:30^\circ = 2:1 $.

Q11. Which of the following statements is sufficient to conclude that two triangles are congruent?

  1. A. These have two equal sides and the same perimeter.
  2. B. These have the same area and the same base.
  3. C. One side and one angle of both triangles are equal.
  4. D. These have the same base and the same height.

Solution

If two triangles have two equal sides and the same overall perimeter, their third sides must also be mathematically equal, thus satisfying the SSS (Side-Side-Side) congruence criterion.

Q12. If the sum of two sides of an equilateral triangle is 16 cm, then find the third side.

  1. A. 4 cm
  2. B. 16 cm
  3. C. Cannot be found
  4. D. 8 cm

Solution

In an equilateral triangle, all sides are equal. If 2x = 16, then x = 8 cm. Thus, the third side is 8 cm.

Q13. Two triangles EFG and HIJ are congruent. If the area of $\triangle EFG$ is $124\text{ cm}^2$, then the area of $\triangle HIJ$ will be:

दो त्रिभुज EFG और HIJ सर्वांगसम हैं। यदि $\triangle EFG$ का क्षेत्रफल $124\text{ cm}^2$ है, तो $\triangle HIJ$ का क्षेत्रफल क्या होगा?

  1. A. $124\text{ cm}^2$
  2. B. $248\text{ cm}^2$
  3. C. $62\text{ cm}^2$
  4. D. $31\text{ cm}^2$

Solution

Congruent triangles have equal areas. Therefore, area of $\triangle HIJ = 124\text{ cm}^2$.

Q14. In $\triangle PQR$, the bisectors of $\angle Q$ and $\angle R$ meet at point $O$, inside the triangle. If $\angle QOR = 107^\circ$, then the measure of $\angle P$ is:

$\triangle PQR$ में, $\angle Q$ और $\angle R$ के समद्विभाजक त्रिभुज के अंदर बिंदु $O$ पर मिलते हैं। यदि $\angle QOR = 107^\circ$ है, तो $\angle P$ का माप क्या है?

  1. A. $40^\circ$
  2. B. $23^\circ$
  3. C. $17^\circ$
  4. D. $34^\circ$

Solution

Property: $\angle QOR = 90^\circ + \angle P/2$. So, $107^\circ = 90^\circ + \angle P/2 \implies \angle P/2 = 17^\circ \implies \angle P = 34^\circ$.

Q15. The points P and S are on the same side of the line segment QR, such that ∠PQR = 90°, ∠SRQ = 90° and PQ = SR. Select the correct statement.

  1. A. ΔPQR ≅ ΔSRQ by RHS
  2. B. ΔPQR ≅ ΔSQR by SAS
  3. C. ΔPQR ≅ ΔSQR by RHS
  4. D. ΔPQR ≅ ΔSRQ by SAS

Solution

In ΔPQR and ΔSRQ: PQ = SR (given), ∠PQR = ∠SRQ = 90° (given), QR = RQ (common base). Thus, the triangles are congruent by SAS (Side-Angle-Side) rule.

Q16. A right-angled isosceles triangle has an area of 50 square units. Its hypotenuse is (in units):

  1. A. 10√3
  2. B. 5√5
  3. C. 10√2
  4. D. 5√2

Solution

Let the equal sides be a. Area = (1/2) * a * a = 50 => a² = 100 => a = 10. Hypotenuse = √(a² + a²) = √(100 + 100) = √200 = 10√2 units.

Q17. The angles of triangle are such that one is average of other two, then the angles are:

  1. A. π/6, π/3, π/2
  2. B. π/3, π/3, π/2
  3. C. π/6, π/3, π/4
  4. D. π/2, π/2, π/3

Solution

Let angles be a, b, c. c = (a+b)/2 => a+b = 2c. Sum of angles a+b+c = 180° => 2c + c = 180° => 3c = 180° => c = 60° (or π/3 radians). Option 1 has angles 30° (π/6), 60° (π/3), and 90° (π/2).