## Geometry questions on circle triangle quadrilateral for SSC CGL with answers

Solve 10 geometry questions on circle triangle quadrilateral for SSC CGL Set 38 in 15 minutes. Verify your solutions from answers.

Answers and link to the solution set are at the end.

### 10 Questions on circle triangle quadrilateral for SSC CGL set 38 - 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 on circle triangle quadrilateral for SSC CGL Set 38

**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.

Learn to solve the questions quickly and elegantly from the paired solutions at,

**SSC CGL level Solution Set 38, Geometry 6.**

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