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- Question 1 of 20
##### 1. Question

The sum of the interior angles of a polygon is 1620°. The number of sides of the polygon are

##### Hint

The sum of the interior angles of a polygon of n sides is given by the expression (2n – 4) π/2

⇒

⇒ 2n = 22

⇒ n = 11

Thus the no. of sides of the polygon are 11.

- Question 2 of 20
##### 2. Question

A cyclic parallelogram having unequal adjacent sides is necessarily a :

##### Hint

It is a rectangle.

(In a cyclic parallelogram each angle is equal to 90°. So, it is definitely either a square or a rectangle. Since the given cyclic parallelogram has unequal adjacent sides, it is a square.)

- Question 3 of 20
##### 3. Question

If the angles of a triangle are in the ratio 5 : 3 : 2, then the triangle could be :

##### Hint

Let the angles of the triangle be 5x, 3x and 2x.

Now, 5x + 3x + 2x = 180°

⇒ 10x = 180

⇒ x = 18

⇒ Angles are 36, 54 and 90°

Given ∆ is right angled.

- Question 4 of 20
##### 4. Question

If one of the diagonals of a rhombus is equal to its side, then the diagonals of the rhombus are in the ratio:

##### Hint

Let the diagonals of the rhombus be x and y and the its sides be x

Now,

⇒

⇒ 3x² = y²

⇒

⇒ y : x

- Question 5 of 20
##### 5. Question

In the given figure, ∠ ABC and ∠ DEF are two angles such that BA ⊥ ED and EF ⊥ BC, then find value of ∠ ABC + ∠ DEF.

##### Hint

Since the sum of all the angle of a quadrilateral is 360°

We have ∠ ABC + ∠ BQE + ∠ DEF + ∠ EPB = 360°

∴ ∠ ABC + ∠ DEF = 180°

[since BPE = EQB = 90° ]

- Question 6 of 20
##### 6. Question

In the given figure given below, E is the mid-point of AB and F is the midpoint of AD. if the area of FAEC is 13, what is the area of ABCD?

##### Hint

As F is the mid-point of AD, CF is the median of the triangle ACD to the side AD.

Hence area of the triangle FCD = area of the triangle ACF.

Similarly area of triangle BCE = area of triangle ACE.

∴ Area of ABCD = Area of (CDF + CFA + ACE + BCE)

= 2 Area (CFA + ACE) = 2 × 13 = 26 sq. units.

- Question 7 of 20
##### 7. Question

Give that segment AB and CD are parallel, if lines ℓ, m and n intersect at point O. Find the ratio of θ to ∠ODS

##### Hint

Let the line m cut AB and CD at point P and Q respectively

∠ DOQ = x (exterior angle)

Hence, Y + 2x (corresponding angle)

∴ y = x …(1)

Also . ∠ DOQ = x (vertically opposite angles)

In ∆ OCD, sum of the angles = 180⁰

∴ y + 2y + 2x + x =180°

⇒ 3x + 3y = 180°

⇒ x + y = 60 …(2)

From (1) and (2)

x = y = 30 = 2y = 60

∴ ∠ ODS = 180 – 60 = 120°

∴ θ = 180 – 3x = 180 – 3(30) = 180 – 90 = 90°.

∴ The required ratio = 90 : 120 = 3 : 4.

- Question 8 of 20
##### 8. Question

AB ⊥ BC and BD ⊥ AC. CE bisects the angle C. ∠A = 30°. Then, what is ∠CED?

##### Hint

In ∆ ABC, = ∠C = 180 – 90 – 30 = 60°

Again in ∆ DEC, = ∠ CED = 180 – 90 – 30 = 60°

- Question 9 of 20
##### 9. Question

In triangle ABC, angle B is a right angle. If (AC) is 6 cm, and D is the mid – point of side AC. The length of BD is

##### Hint

In a right angled ∆, the length of the median is ½ the length of the hypotenuse .

Hence BD = ½ AC = 3 cm

- Question 10 of 20
##### 10. Question

ABCD is a square of area 4, which is divided into four non overlapping triangles as shown in the fig. Then the sum of the perimeters of the triangles is

##### Hint

ABCD is square a² = 4 ⇒ a = 2

perimeters of four triangles

= AB + BC + CD + DA + 2(AC + BD)

- Question 11 of 20
##### 11. Question

The sides of a quadrilateral are extended to make the angles as shown below :

What is the value of x?##### Hint

Sum of all the interior angles of a polygon taken in order is 360°.

i.e., x + 90 + 115 + 75 = 360

⇒ x = 360° – 280° = 80°

⇒ x = 80°

- Question 12 of 20
##### 12. Question

Instead of walking along two adjacent sides of a rectangular field, a boy took a short cut along the diagonal and saved a distance equal to half the longer side. Then the ratio of the shorter side to the longer side is

##### Hint

According to question,

- Question 13 of 20
##### 13. Question

If two parallel lines are cut by two distinct transversals, then the quadrilateral formed by these four lines will always be a :

##### Hint

The quadrilateral obtained will always be a trapeziam as it has two lines which are always parallel to each other.

- Question 14 of 20
##### 14. Question

In the following figure, find ∠ADC.

##### Hint

∠ACB = ∠BAC (Angles opposite equal sides are equal)

Similarly, ∠ADC = ∠CAD

∴ ∠ACB = ∠BAC

⇒ ∠ADC = ∠CAD

- Question 15 of 20
##### 15. Question

In a triangle ABC, the internal bisector of the angle A meets BC at D. If AB = 4, AC = 3 and ∠A = 60°, then the length of AD is

##### Hint

Using the theorem of angle of bisector,

In ∆ABD, by sine rule,

…(1)

In ∆ABC, by sine rule;

⇒ [putting the value of sin B from (1)]

- Question 16 of 20
##### 16. Question

In the figure AG = 9, AB = 12, AH = 6, Find HC.

##### Hint

m ∠ AHG = 180 – 108 = 72⁰

∴ ∠ AHG = ∠ ABC …..(same angle with different names)

∴ ∆ AHG – ∆ABC …..(AA test for similarity)

AH/AB = AG/AG;

⇒ 6/12 = 9/AC

∴ AC = = 18

∴ HC = AC – AH = 18 – 6 = 12

- Question 17 of 20
##### 17. Question

In ∆ABC, DE || BC and AD/DB = 3/5. If AC = 5.6 cm, find AE.

##### Hint

In ∆ABC, DE || BC

By applying basic Proportionality theorem,

But

∴

⇒

⇒

⇒

⇒ 8AE = 3 × 5.6

⇒ AE = 3 × 5.6/8

∴ AE = 2.1 cm.

- Question 18 of 20
##### 18. Question

In the given fig. AB || QR, find the length of PB.

##### Hint

∆PAB ~ ∆PQR

∴ PB = 2cm

- Question 19 of 20
##### 19. Question

In a triangle ABC, the lengths of the sides AB, AC and BC are 3, 5 and 6 cm, respectively. If a point D on BC is drawn such that the line AD bisects the angle A internally, then what is the length of BD?

##### Hint

As AD bisects ∠BAC, we have

⇒

⇒

⇒

⇒

⇒

= 2.25 cm

- Question 20 of 20
##### 20. Question

The number of tangents that can be drawn to two non-intersecting circles is :

##### Hint

Four tangents can be drawn to two non-intersecting circles in the following manner :