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1. A vertical stick 20 m long casts a shadow 10 m long
on the ground. At the same time, a tower casts the shadow 50 m long on the ground. Find the height
of the tower.

Correct Answers :

[A]

Explanation :

When the length of stick = 20 m, then length of shadow = 10 m i.e. in this case length = 2 X shadow. With the same angle of inclination of the sun, the length of tower that casts a shadow of 50 m fi 2 X 50 m = 100 m i.e. height of tower = 100 m

2. In the figure, △ABC is similar to *△EDC*.

If we have*AB* = 4 cm,

*ED* = 3 cm, *CE* = 4.2 and *CD* = 4.8 cm, find the value of *CA* and *CB*

If we have

Correct Answers :

[D]

Explanation :

3. The area of similar triangles, *ABC* and *DEF* are 144 cm2
and 81 cm2 respectively. If the longest side of larger *△ABC* be 36 cm, then the longest side of smaller
*△DEF* is

Correct Answers :

[C]

Explanation :

For similar triangles ⇉ (Ratio of sides)2 = Ratio of
areas

4. Two isosceles *△*s have equal angles and their areas
are in the ratio 16 : 25. Find the ratio of their corresponding heights.

Correct Answers :

[A]

Explanation :

(Ratio of corresponding sides)2 = Ratio of area of
similar triangles ∴Ratio of corresponding sides in this question

5. The areas of two similar Ds are respectively 9 cm2
and 16 cm2. Find the ratio of their corresponding sides.

Correct Answers :

[A]

Explanation :

Ratio of corresponding sides = = ¾.

6. Two poles of height 6 m and 11 m stand vertically
upright on a plane ground. If the distance between their foot is 12 m, find the distance between their
tops.

Correct Answers :

[C]

Explanation :

So *BC* = *ED* = 6 m

*AB* = *AC* –* BC* = 11 – 6 = 5 m

*CD* = *BE* = 12 m

Then by Pythagoras theorem: *AE ^{2}* =

7. The radius of a circle is 9 cm and length of one of
its chords is 14 cm. Find the distance of the chord
from the centre.

Correct Answers :

[A]

Explanation :

In the △*OBC*; *BC* = 7 cm and *OC* = 9 cm, then using Pythagoras theorem.

*OB*^{2} = *OC*^{2} – *BC*^{2}

*OB* = √32 = 5.66 cm (approx)

8. Find the length of a chord that is at a distance of 12
cm from the centre of a circle of radius 13 cm.

Correct Answers :

[D]

Explanation :

In the △OBC, OB = 12 cm, OC = radius = 13 cm.

Then using Pythagoras theorem;

*BC*^{2} = *OC*^{2} – *OB*^{2} = 25; *BC* = 5 cm
Length of the chord = 2 x BC = 2 x 5 = 10 cm.

9. If *O* is the centre of circle, find ∠x

Correct Answers :

[A]

Explanation :

∠x = 35°; because angles subtended by an arc, anywhere on the circumference are equal.

10. Find the value of ∠x in the given figure.

Correct Answers :

[D]

Explanation :

because angle subtended by an arc at the centre of the circle is twice the angle subtended by it on the
circumference on the same segment.

Alternately, you could also solve this using the following process:

In the given figure, join the points* BD *and *CD*. Then, in the cyclic quadrilateral *ABDC*, the sum of angles
x/2 and y would be 180°. Hence, y = 180 – x/2. Also, the sum of the angles *OBD* + *OCD* = 180 – 20 - 30
= 130°. Therefore, x + y = 230 (as the sum of the angles of the quadrilateral *OBDC* is 360). Solving,
the two equations, we get x = 100.

11. Find the value of *x* in the figure, if it is given that
*AC* and *BD* are diameters of the circle.

Correct Answers :

[D]

Explanation :

The triangle BOC is an isosceles triangle with sides
OB and OC both being equal as they are the radii of the circle. Hence, the angle *OBC* = angle *OCB* =
30°. Hence, the third angle of the triangle *BOC* viz: Angle *BOC* would be equal to 120°. Also, *BOC* =
*AOD* = 120°. Hence, in the isosceles triangle *DOA*, Angle *ODA* = Angle *DAO* = x = 30°.

12. Find the value of x in the given figure.

Correct Answers :

[A]

Explanation :

By the rule of tangents, we know:
6^{2} = (5 + x)5 ⇉ 36 = 25 + 5x ⇉ 11 = 5x ⇉ x = 2.2 cm

13. Find the value of x in the given figure.

Correct Answers :

[B]

Explanation :

By the rule of tangents, we get
12^{2} = (x + 7)x ⇉ 144 = x^{2} + 7x

⇉ x^{2} + 7x – 144 = 0 ⇉ x^{2} + 16x – 9x – 144 = 0

⇉ x(x + 16) – 9(x + 16) ⇉ x = 9 or –16 –16 can’t be the length, hence this value is discarded,
thus, x = 9

14. Find the value of x in the given figure.

Correct Answers :

[A]

Explanation :

By the rule of chords, cutting externally, we get

⇉ (9 + 6)6 = (5 + x)5 ⇉ 90 = 25 + 5x ⇉ 5x = 65
⇉ x = 13 cm

15. *ABC* is a right angled triangle with *BC* = 6 cm and
*AB* = 8 cm. *A* circle with centre *O* and radius *x* has
been inscribed in △*ABC*. What is the value of *x*?

Correct Answers :

[B]

Explanation :

Use the formula: Inradius = Area/ Semi perimeter =
24/12 = 2 cm

16. In the given figure AB is the diameter of the circle and
∠*PAB* = 25°. Find ∠*TPA*.

Correct Answers :

[B]

Explanation :

Note: This is also called as the Alternate Segment Theorem.

17. In the given figure, find ∠*ADB*.

Correct Answers :

[A]

Explanation :

*ADBC* is a cyclic quadrilateral as all its four vertices
are on the circumference of the circle. Also, the opposite angles of the cyclic quadrilateral are
supplementary.

Therefore, ∠ADB = 180 – 48° = 132°

18. In the given figure, two straight lines *PQ* and *RS*
intersect each other at *O*. If ∠*SOT* = 75°, find the
value of *a, b* and *c*.

Correct Answers :

[A]

Explanation :

From the given figure we have:

4b + 2c = 180 (1)

a + b = 105 (2)

4b = a (3)

Solving these equations, we get that b = 21°; a = 84°; c = 48°.

19. In the following figure *A, B, C* and *D* are the concyclic
points. Find the value of *x*.

Correct Answers :

[B]

Explanation :

20. In the following figure, it is given
that *o* is the centre of the circle and ∠*AOC* = 140°. Find ∠*ABC*.

Correct Answers :

[A]

Explanation :

∠ADC = 140/2= 70° (because the angle subtended by an arc on the circumference is half of what it subtends at the centre). ABCD one cyclic quadrilateral So ∠ABC = 180° – 70° = 110° (because opposite angles of a cyclic quadrilateral are supplementary).

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