SSC CGL Tier II level Question Set 4, Geometry 1 | SureSolv

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SSC CGL Tier II level Question Set 4, Geometry 1

4th SSC CGL Tier II level Question Set, 1st on Geometry

SSC CGL Tier2 level question set 4 geometry1 top

The ten SSC CGL Tier II level questions in this set are specially selected for variety and depth with no two questions of same type.

The questions are on,

  • triangles,
  • circles,
  • medians,
  • parallelogram,
  • interior bisectors and
  • inscribed circle.

For best results,

  1. this question set should be used as a mini-mock test and
  2. after timed completion and self-scoring from answers,
  3. the difficulties faced should be cleared up from the corresponding solution set.

Link of the detailed solutions is at the end.

Now set the timer and take the test. By the way don't bother much if you can't solve the questions in 12 minutes scheduled time. That's an ideal. But with intelligent preparation and practice, you can surely reach this level of competence.

4th question set - 10 problems for SSC CGL exam: 1st on topic Geometry - answering time 12 mins

Problem 1.

If P and Q are two points on sides AB and AD of a parallelogram ABCD respectively, and areas of $\triangle CPD=A_1$ and that of $\triangle BQC=A_2$, then,

  1. $2A_1=A_2$
  2. $A_1=A_2$
  3. $A_1=A_2$
  4. $2A_1=3A_2$

Problem 2.

If three non-colllinear points A, B and C lie on the periphery of a circle so that $AB=AC=BC=3$ cm, then the radius of the circle (in cm) is,

  1. $\sqrt{3}$
  2. $\displaystyle\frac{1}{\sqrt{3}}$
  3. $\displaystyle\frac{\sqrt{3}}{2}$
  4. $\displaystyle\frac{2}{\sqrt{3}}$

Problem 3.

Two similar triangles have areas 96 cm$^2$ and 150 cm$^2$. If the largest side of the larger triangle is 20 cm, the largest side of the smaller triangle (in cm) is,

  1. 20
  2. 15
  3. 16
  4. 18

Problem 4.

The interior bisectors of $\angle B$ and $\angle C$ of $\triangle ABC$ meet at point P. If $\angle A=70^0$, then the value of $\angle BPC$ is,

  1. $55^0$
  2. $125^0$
  3. $150^0$
  4. $135^0$

Problem 5.

In $\triangle ABC$ the medians BE and CF intersect at point G. If GD=3 cm, then the length of AD is,

  1. 12 cm
  2. 4.5 cm
  3. 6 cm
  4. 9 cm

Problem 6.

D is the mid-point of side AB of a right angled $\triangle ABC$ with right angle at B. If AD subtends an angle $\alpha$ at C and BC is $n$ times of AB, then $\tan \alpha$ is,

  1. $\displaystyle\frac{n}{2n^2+1}$
  2. $\displaystyle\frac{n}{n^2+1}$
  3. $\displaystyle\frac{n}{n^2-1}$
  4. $\displaystyle\frac{n^2-1}{n^2+1}$

Problem 7.

If A, B and C are three points lying on the same plane and $AB=5$ cm and $BC=10$ cm, then the possible length of AC (in cm) is,

  1. 15
  2. 5
  3. 3
  4. 6

Problem 8.

In the following figure, a square ABCD is formed with its vertices as the mid-points of a larger square PQRS. A circle is inscribed in square ABCD and the $\triangle EFG$ is an equilateral triangle inscribed in the circle. If length of side of square PQRS is $a$, the area of the $\triangle EFG$ is,


  1. $\displaystyle\frac{\sqrt{3}a^2}{16}$
  2. $\displaystyle\frac{3\sqrt{3}a^2}{32}$
  3. $\displaystyle\frac{5\sqrt{3}a^2}{32}$
  4. $\displaystyle\frac{5\sqrt{3}a^2}{64}$

Problem 9.

The radii of two non-intersecting circles are $r_1$ and $r_2$ with $r_1$ being the larger one and the smaller circle inscribed within the larger circle. If the least distance between their circumference be $S$, the distance between their centres is,

  1. $r_1-r_2+S$
  2. $r_1-r_2$
  3. $r_1+r_2-S$
  4. $r_1-r_2-S$

Problem 10.

If $\triangle ABC$, $AD$, $BE$ and $CF$ are the altitudes and $AD$ and $BE$ intersect at $G$ with $BE+EG=BG$, then,

  1. $FC+CG=FG$
  2. $CF+FG=CG$
  3. $CG+GF=CF$
  4. $CF-FG=CG$

Answers to the problems

Problem 1. Answer: b: $A_1=A_2$.

Problem 2. Answer: a: $\sqrt{3}$.

Problem 3. Answer: c: 16.

Problem 4. Answer: b: $125^0$.

Problem 5. Answer: d: 9 cm.

Problem 6. Answer: a: $\displaystyle\frac{n}{2n^2+1}$.

Problem 7. Answer: d: 6.

Problem 8. Answer: b: $\displaystyle\frac{3\sqrt{3}a^2}{32}$.

Problem 9. Answer: d: $r_1-r_2-S$.

Problem 10. Answer: a: $FC+CG=FG$.

Detailed Solution to this question set

SSC CGL Tier II level Solution Set 4, Geometry 1

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