## 38th SSC CGL level Question Set, 6th on topic Geometry

This is the 38th question set of 10 practice problem exercise for SSC CGL exam and 6th on topic Geometry. You need to take this test first before referring to the corresponding solution set. As expected some of the problem pictorial representation seemed to be complex, but once you represent a geometric figure properly even in a quick sketch, rest should not take much time.

### Method for taking the test and get the best results from the test set:

**Before start,**go through the**tutorials on****Geometry basic concepts part 1 on points lines and triangles ,****Geometry basic concepts part 2 on Quadilaterals Squares Rectangles,****Geometry basic and rich concepts part 3 on Circles,**or any other short but good material to refresh your concepts if you so require.**Basic and rich Geometry concepts part 4 on proof of arc angle subtending concept****Answer the questions**in an undisturbed environment with no interruption, full concentration and alarm set at 15 minutes.**When the time limit of 15 minutes is over,**mark up to which you have answered,**but go on to complete the set.****At the end,**refer to the answers given at the end to mark your score at 15 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time.**Identify and analyze**the problems that**you couldn't do**to learn how to solve those problems.**Identify and analyze**the problems that**you solved incorrectly**. Identify the reasons behind the errors. If it is because of**your shortcoming in topic knowledge**improve it by referring to**only that part of concept**from the best source you get hold of. You might google it. If it is because of**your method of answering,**analyze and improve those aspects specifically.**Identify and analyze**the**problems that posed difficulties for you and delayed you**. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques.**Give a gap**before you take a 10 problem practice test again.

Important:bothandpractice testsmust be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance.mock tests

Now set the stopwatch alarm and start taking this test. It is not difficult.

### 38th question set- 10 problems for SSC CGL exam: 6th on Geometry - answering time 15 mins

#### Problem 1.

If I be the incentre of $\triangle ABC$, $\angle ABC=60^0$ and $\angle ACB=50^0$, then $\angle BIC$ is,

- $70^0$
- $65^0$
- $125^0$
- $55^0$

#### Problem 2.

If ratio of number of sides of two polygons is 5 : 6, and ratio of their internal angles is 24 : 25, the number of sides of the two polygons are,

- 20, 24
- 5, 6
- 15, 18
- 10, 12

#### Problem 3.

If the inradius of an equilateral triangle is 3 cm, its side length is,

- $3\sqrt{3}$ cm
- $6$ cm
- $9\sqrt{3}$ cm
- $6\sqrt{3}$ cm

#### Problem 4.

ABC is an isosceles triangle such that AB=AC and AD is the median to the base BC with $\angle ABC=35^0$. Then $\angle BAD$ is,

- $35^0$
- $55^0$
- $110^0$
- $70^0$

#### Problem 5.

Two circles touch each other internally. Their radii are 2 cm and 3 cm. The biggest chord of the larger circle which is outside the inner circle is of length,

- $2\sqrt{2}$ cm
- $4\sqrt{2}$ cm
- $2\sqrt{3}$ cm
- $3\sqrt{2}$ cm

#### Problem 6.

In a right $\triangle ABC$, point D lies on side BC. If AB be the hypotenuse, then,

- $AB^2 + CD^2=BC^2 + AD^2$
- $AB^2=AD^2+BD^2$
- $AD^2 + AC^2=2AD^2$
- $CD^2+BD^2=2AD^2$

#### Problem 7.

If each angle of a triangle is less than the sum of the other two angles, then the triangle is,

- Obtuse angled
- Right angled
- Acute angled
- Not a triangle

#### Problem 8.

The side AB of a parallelogram ABCD is extended to E such that BE=AB and DE intersects BC at Q. Then the point Q divides BC in the ratio,

- 1 : 2
- 2 : 1
- 1 : 1
- 2 : 3

#### Problem 9.

A cyclic quadrilateral ABCD is such that AB=BC, AD=DC, AC perpendicular to BD and $\angle CAD=\theta$. Then the $\angle ABC=$,

- $\theta$
- $\displaystyle\frac{\theta}{2}$
- $3\theta$
- $2\theta$

#### Problem 10.

ABCD is a rhombus whose side AB = 4 cm and $\angle ABC = 120^0$. Then the length of the diagonal BD is,

- 4 cm
- 2 cm
- 1 cm
- 3 cm

### Answers to the questions

**Problem 1.** Option c: $125^0$

**Problem 2.** Option d: 10, 12.

**Problem 3.** Option d: $6\sqrt{3}$ cm

**Problem 4.** Option b: $55^0$

**Problem 5.** Option b: $4\sqrt{2}$ cm

**Problem 6.** Option a: $AB^2 + CD^2=BC^2 + AD^2$.

**Problem 7.** Option c: Acute angled.

**Problem 8.** Option c: 1:1.

**Problem 9.** Option d: $2\theta$.

**Problem 10.** Option a: 4 cm.

For detailed explanation of the solutions clarifying the concepts used for elegant solutions, you should refer to the corresponding **SSC CGL level Solution Set 38, Geometry 6.**

**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**

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