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Test: Geometry (March 25) - CAT MCQ


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10 Questions MCQ Test - Test: Geometry (March 25)

Test: Geometry (March 25) for CAT 2024 is part of CAT preparation. The Test: Geometry (March 25) questions and answers have been prepared according to the CAT exam syllabus.The Test: Geometry (March 25) MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Geometry (March 25) below.
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Test: Geometry (March 25) - 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.

Detailed Solution for Test: Geometry (March 25) - Question 1
  • 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
Test: Geometry (March 25) - Question 2

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

Detailed Solution for Test: Geometry (March 25) - Question 2

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Test: Geometry (March 25) - Question 3

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

Detailed Solution for Test: Geometry (March 25) - Question 3
  • 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°
Test: Geometry (March 25) - Question 4

Find the value of x in the given figure.

Detailed Solution for Test: Geometry (March 25) - Question 4

By the rule of tangents, we get:

⇒ 122 = (x + 7)x
⇒ 144 = x2 + 7x
⇒ x2 + 7x – 144 = 0
⇒ x2 +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

Test: Geometry (March 25) - Question 5

Find the value of x in the given figure.

Detailed Solution for Test: Geometry (March 25) - Question 5

By the rule of chords, cutting externally, we get:

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

Test: Geometry (March 25) - Question 6

AB is the diameter of the circle and ∠PAB=40∘
what is the value of ∠PCA?

376998

Detailed Solution for Test: Geometry (March 25) - Question 6
  • In △PAB

    ⇒  ∠PAB=40o         [ Given ]

    ⇒  ∠BPA=90o      [ angle inscribed in a semi-circle ]

    ⇒  ∠PAB+∠PBA+∠BPA=180o

    ∴   40o+∠PBA+90o=180o

    ∴   ∠PBA=180o−130o

    ∴   ∠PBA=50o

    ⇒  ∠PBA=∠PCA=50o     [ angles inscribed in a same arc PA ] 

    ∴   ∠PCA=50o

Test: Geometry (March 25) - 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

Detailed Solution for Test: Geometry (March 25) - Question 7
  • 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 

Test: Geometry (March 25) - Question 8

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

Detailed Solution for Test: Geometry (March 25) - Question 8


∠AOC of minor sector = 140°
∠AOC of major sector=360° - 140° = 220°
Theorem: The angle subtended at the centre is twice the angle formed at the circumference of the circle.
Hence,


∴ The measure of ∠x = 110°

Test: Geometry (March 25) - Question 9

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

Detailed Solution for Test: Geometry (March 25) - Question 9

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° 

Test: Geometry (March 25) - Question 10

In the given figure, AD is the bisector of ∠BAC, AB = 6 cm, AC = 5 cm and BD = 3 cm. Find DC. 

Detailed Solution for Test: Geometry (March 25) - Question 10

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