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This mock test of Test: Geometry- 1 for UPSC helps you for every UPSC entrance exam.
This contains 15 Multiple Choice Questions for UPSC Test: Geometry- 1 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Geometry- 1 quiz give you a good mix of easy questions and tough questions. UPSC
students definitely take this Test: Geometry- 1 exercise for a better result in the exam. You can find other Test: Geometry- 1 extra questions,
long questions & short questions for UPSC on EduRev as well by searching above.

QUESTION: 1

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

Solution:

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

⇒ Height of tower =**100 m**

QUESTION: 2

The area of similar triangles, ABC and DEF are 144cm^{2} and 81 cm^{2} respectively. If the longest side of the larger △ABC be 36 cm, then the longest side of the smaller △DEF is:

Solution:

For similar triangles

⇒ (Ratio of sides)^{2 } = Ratio of areas

Then as per question = (36/x)^{2 }= 141/81

Let the longest side of ΔDEF = x

⇒ 36/x = 12/9

⇒ x = 27cm

QUESTION: 3

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

Solution:

- 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 i.e. Angle BOC would be equal to 120°.

**⇒ BOC = AOD = 120°** - Also, in the isosceles triangle DOA:

Angle ODA = Angle DAO =**x = 30°**

QUESTION: 4

Find the value of x in the given figure.

Solution:

__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) = 0

⇒ x = 9 or –16

–16 can’t be the length, hence this value is discarded. Thus,** x = 9**

QUESTION: 5

Find the value of x in the given figure.

Solution:

__By the rule of chords, cutting externally, we get:__

(9 + 6) * 6 = (5 + x) * 5

90 = 25 + 5x

5x = 65

**x = 13 cm**

QUESTION: 6

In the given figure ㄥPAB = 25º. AB is the diameter. Find ㄥTPA.

Solution:

**ㄥAPB = 90°**(angle in a semicircle = 90°)- ㄥPBA = 180 – (90 + 25) = 65°
**ㄥTPA = ㄥPBA**(the angle that a chord makes with the tangent, is subtended by the chord on the circumference in the alternate segment)**= 65°**

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

QUESTION: 7

In the figure, AB is parallel to CD and RD || SL || TM || AN, and BR : RS : ST : TA = 3 : 5 : 2 : 7. If it is known that CN = 1.333 BR. Find the ratio of BF : FG : GH : HI : IC

Solution:

- Since the lines,
**AB and CD****are parallel**to each other, and the lines**RD and AN are parallel**, it means that the triangles RBF and NCI are similar to each other. Since the ratio of**CN : BR = 1.333**, if we take BR as 3, we will get CN as 4. - This means that the ratio of BF : CI would also be
**3 : 4**.

Also, the ratio of BR : RS : ST : TA = BF : FG : GH : HI = 3 : 5 : 2 : 7 (given).

Hence, the correct answer is** 3 : 5 : 2 : 7 : 4**

QUESTION: 8

In the following figure, it is given that O is the centre of the circle and ㄥAOC = 140°. Find ㄥABC.

Solution:

∠AOC of minor sector = **140°**

∠AOC of major sector=360° - 140° = **220°**

__ Theorem:__ The angle subtended at the centre is

Hence,

**∴ The measure of ∠x = 110°**

QUESTION: 9

In the figure below, PQ = QS, QR = RS and angle SRQ = 100°. How many degrees is angle QPS?

Solution:

In ΔQRS, QR = RS

⇒ **ㄥRQS = ㄥRSQ** (because angles opposite to equal sides are equal).

__Thus:__

ㄥRQS + ㄥRSQ = 180° - 100° = 80°

ㄥRQS = ㄥRSQ = 40°

ㄥPQS = 180° – 40° = 140° (sum of angles on a line = 180°)

Then again, **ㄥQPS = ㄥQSP** (since angles opposite to equal sides are equal)

ㄥQPS + ㄥQSP = 180° – 140° = 40°

**ㄥQPS = ㄥQSP = 20° **

QUESTION: 10

In the given figure, AD is the bisector of ΔBAC, AB = 6 cm, AC = 5 cm and BD = 3 cm. Find DC. It is given that ∠ABD = ∠ACD.

Solution:

We know that the** internal bisector** of angle of a triangle divides the **opposite side** internally in the ratio of the sides containing the angle.

__Hence:__

AB/BD = AC/CD

6/3 = 5/CD

**CD = 2.5 cm**

QUESTION: 11

In the given figure, straight line AB and CD intersect at O. IF ∠δ =3∠v, then ∠v = ?

Solution:

COD is a straight line

∴ ∠δ + ∠v =180⁰ ⇒ 3v +v =180 ⇒ 4v = 180 ⇒ v =45⁰.

QUESTION: 12

In a triangle ABC, the incentre is at 0. If ㄥBOC = 100°, find ㄥBAC.

Solution:

In ∠BOC

⇒ x + y = 80°

⇒ 2x + 2y. = 160°

Also, 2x + 2y + 2z = 180°

⇒ 160° + 2z = 180°

⇒ ∠BAC = 2z = 20°

QUESTION: 13

A rectangular enclosure 40 m x 36 m has a horse tethered to a corner with a rope of 14 m in length. What is the ratio of the respective areas it can graze if it is outside the enclosure and if it is inside the enclosure?

Solution:

- Imagine a circle on the corner of the
**rectangle**. - 3 quarters of the circle lie outside the rectangle and 1 quarter lies inside.
- Hence,
**Required ratio = 3 : 1**

QUESTION: 14

Read the passage below and solve the questions based on it.

The area of a square is equal to the area of a rectangle. Moreover, the perimeter of the square is also equal to the perimeter of the rectangle.

Q. The length of the rectangle is equal to the:

Solution:

This is possible only when both the length and breadth of the rectangle are** equal** to the **side of the square.**

QUESTION: 15

A quadrilateral is inscribed in a circle. If an angle is inscribed in each of the four segments outside the quadrilateral, then what is the sum of these four angles?

Solution:

Required sum = θ + 180° - θ + 180° - θ + 180° - θ + θ + 8 = **540°**

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