## 15th SSC CGL Tier II level Question Set, 4th on Geometry

This is the 15th question set of 10 practice problem exercise for SSC CGL Tier II exam and the 4th on topic Geometry.

A few of the problems look simple but deceptively so. In the same way, a few of the problems seem to be quite difficult, but can be solved quickly by identifying key geometric patterns and applying corresponding methods that are based on core concepts. In a few cases, awareness of algebraic patterns and methods becomes important for quick solution.

**Answers** to the questions, links of **detailed solution set**, **tutorials** and **other question and solution sets on Geometry** are given at the end. Preferably you should go through the **first three tutorials** to refresh your knowledge before going through this mini-mock test.

### 15th SSC CGL Tier II level question set - 10 problems for SSC CGL exam: 4th on topic Geometry - answering time 15 mins

**Problem 1. **

If G is the centroid of an equilateral triangle ABC with side length 10 cm, then the length of AG is,

- $5\sqrt{3}$ cm
- $10\sqrt{3}$ cm
- $\displaystyle\frac{5\sqrt{3}}{3}$ cm
- $\displaystyle\frac{10\sqrt{3}}{3}$ cm

**Problem 2.**

If the ratio of the angles of a quadrilateral is $2:7:2:7$, then the shape

- is a rhombus
- is a parallelogram
- is a trapezium
- cannot be identified

**Problem 3.**

If $D$, $E$ and $F$ are the mid-points of sides $BC$, $CA$ and $AB$ respectively of $\triangle ABC$, then the ratio of the area of the parallelogram $DEFB$ and the area of the trapezium $CAFD$ is,

- $2:3$
- $3:4$
- $1:3$
- $1:2$

**Problem 4. **

If a chord of a circle is equal to its radius, the angle subtended by the chord at the minor arc of the circle is,

- $60^0$
- $150^0$
- $120^0$
- $75^0$

**Problem 5. **

In a cyclic quadrilateral ABCD, AB and DC when extended meet at point P. If PA= 8 cm, PB = 6 cm and PC = 4 cm, the length of PD in cm is,

- 10 cm
- 12 cm
- 6 cm
- 8 cm

**Problem 6.**

In $\triangle ABC$, $DE||BC$ and $AD:DB=5:4$. Then the ratio, $DE:BC$ is,

- $4:5$
- $9:5$
- $5:9$
- $4:9$

** Problem 7.**

In a $\triangle ABC$, $BC=12$ cm. A line CD is drawn to intersect AB at D internally. If $DB=9$ cm, $CD=6$ cm and $\angle BCD=\angle BAC$, ratio of the perimeter of $\triangle ADC$ to that of $\triangle BDC$ is,

- $\displaystyle\frac{5}{9}$
- $\displaystyle\frac{6}{9}$
- $\displaystyle\frac{7}{9}$
- $\displaystyle\frac{8}{9}$

** Problem 8.**

In the figure below, chord $CD$ is parallel to diameter $AB$ of a circle with centre at $O$.

If $\angle CEB=65^0$, then $\angle BCD$ is,

- $25^0$
- $35^0$
- $55^0$
- $45^0$

**Problem 9.**

In $\triangle ABC$, the line $BB_1$ through $B$ intersects side $AC$ at $B_1$. A line through $A$ parallel to $BB_1$ meets $CB$ extended at $A_1$ and another line through $C$ parallel to $BB_1$ meets $AB$ extended at $C_1$. Then which of the following is true?

- $\displaystyle\frac{1}{AA_1}-\displaystyle\frac{1}{CC_1}=\displaystyle\frac{2}{BB_1}$
- $\displaystyle\frac{1}{BB_1}-\displaystyle\frac{1}{AA_1}=\displaystyle\frac{2}{CC_1}$
- $\displaystyle\frac{1}{CC_1}+\displaystyle\frac{1}{AA_1}=\displaystyle\frac{1}{BB_1}$
- $\displaystyle\frac{1}{CC_1}-\displaystyle\frac{1}{AA_1}=\displaystyle\frac{1}{BB_1}$

** Problem 10.**

In an isosceles right $\triangle ABC$, $\angle C=90^0$. If $D$ is any point on $AB$ then $AB^2+BD^2$ is equal to,

- $2CD^2$
- $4CD^2$
- $3CD^2$
- $CD^2$

**Note:** The detailed explanatory solutions are available in the corresponding solution set,

**SSC CGL Tier II level Solution Set 15 on Geometry 4.**

Answers to the questions are given below.

### Answers to the questions

**Problem 1.** Answer: Option d: $\displaystyle\frac{10\sqrt{3}}{3}$ cm.

**Problem 2.** Answer: Option d: cannot be identified.

**Problem 3.** Answer: Option a: $2:3$.

**Problem 4.** Answer: Option b: $150^0$.

**Problem 5.** Answer: Option b: 12 cm.

**Problem 6.** Answer: Option c: $5:9$.

**Problem 7.** Answer: Option c: $\displaystyle\frac{7}{9}$.

**Problem 8.** Answer: Option a: $25^0$.

**Problem 9.** Answer: Option c: $\displaystyle\frac{1}{CC_1}+\displaystyle\frac{1}{AA_1}=\displaystyle\frac{1}{BB_1}$.

**Problem 10.** Answer: Option a: $2CD^2$.

**Related resources that should be useful for you**

**You may refer to:**

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

**Concept tutorials for SSC CGL and other competitive exams on Geometry**

**Basic and rich Geometry concepts part 7, Laws of sines and cosines**

**Basic and rich Geometry concepts part 6, proof of triangle area from medians**

**Basic and rich Geometry concepts part 5, proof of median relations**

**Basic and rich Geometry concepts part 4, proof of arc angle subtending concept**

**Geometry, basic and rich concepts part 3, Circles**

**Geometry, basic concepts part 2, Quadrilaterals polygons and squares**

**Geometry, basic concepts part 1, points lines and triangles**

**How to solve difficult Geometry problems quickly in a few steps**

**How to solve intriguing SSC CGL level Geometry problem in a few steps 4**

**How to solve difficult SSC CGL Geometry problems in a few steps 3**

**How to solve difficult SSC CGL Geometry problems in a few steps 2**

**How to solve difficult SSC CGL Geometry problems in a few steps 1**

**SSC CGL Tier II level question sets and solution sets on Geometry**

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