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Test Level 1: Geometry - 1 - CAT MCQ


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10 Questions MCQ Test - Test Level 1: Geometry - 1

Test Level 1: Geometry - 1 for CAT 2024 is part of CAT preparation. The Test Level 1: Geometry - 1 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 1: Geometry - 1 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 1: Geometry - 1 below.
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Test Level 1: Geometry - 1 - Question 1

In the adjoining figure, I and II are circles with centres P and Q, respectively. The two circles touch each other and have a common tangent that touches them at points R and S, respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in the ratio 4 : 3. It is also known that the length of PQ is 28 cm.

What is the ratio of the length of PQ to that of QO?

Detailed Solution for Test Level 1: Geometry - 1 - Question 1

Let the diameter of circle I and II be 4x and 3x, respectively.
ΔOSQ is similar to ΔORP by AA criteria.

Test Level 1: Geometry - 1 - Question 2

Two perpendicular lines that intersect each other at the centre of a circle of radius 1 unit divides the circle into four parts. A smaller circle is inscribed in one of those parts as shown in the figure below. What is the radius of the smaller circle?

Detailed Solution for Test Level 1: Geometry - 1 - Question 2

Let x denote the radius of the smaller circle. The figure shows the lengths marked to the right as follows.

Here,

BO2 = AB2 + AO2

BO2 = x2 + x2

BO = √2x

BO + OC = 1

Hence, this is the correct answer.

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Test Level 1: Geometry - 1 - Question 3

If the ratio of areas of two squares is 9 : 1, then the ratio of their perimeters will be

Detailed Solution for Test Level 1: Geometry - 1 - Question 3

Let area of the first square be a1, whose side is s1.
Let area of the second square be a2, whose side is s2.

Then,

So, required ratio of perimeter

Test Level 1: Geometry - 1 - Question 4

In the given figure, if PM and PN are tangents to the circle with centre Q, radius = 7 cm and PM = 7 cm, then what is the length of PQ?

Detailed Solution for Test Level 1: Geometry - 1 - Question 4

PM is a tangent and MQ is the radius.
⇒ QM perpendicular to PM
⇒ PMQ is a right-angled Δ.
PQ2 = PM2 + MQ2 (By Pythagoras Theorem)
PQ2 = (7)2 + (7)2 = 2(7)2
PQ2 = 98

PQ = 7√2 cm

Test Level 1: Geometry - 1 - Question 5

CD is parallel to EF. AD = DF, CD = 4 units and DF = 3 units. What is the measure of EF?

Detailed Solution for Test Level 1: Geometry - 1 - Question 5

 In ΔADC and ΔAFE:
∠A = ∠A {Common}
AD = DF {Given}
∠ADC = ∠AFE
{Corresponding angles between two || lines}

Hence, ΔADC is similar to ΔAFE.

We know: AF = AD + DF = 3 + 3 = 6 units

Test Level 1: Geometry - 1 - Question 6

ΔABC, ΔCDE, ΔEFG, ΔGHI, ΔIJK and ΔKLM are congruent to one another and similar to ΔANM. What is the ratio of the area of ΔANM to the area of ABC?

Detailed Solution for Test Level 1: Geometry - 1 - Question 6

Since all small triangles are congruent, AC = CE = EG = GI = IK = KM and 6AC = AM.

Now, as ΔANM ~ ΔABC,

Required Ratio = 36 : 1

Test Level 1: Geometry - 1 - Question 7

The figure given below has 2 circles with centers A and B.What is the measure of ∠APT?

Detailed Solution for Test Level 1: Geometry - 1 - Question 7

Since AT = AP, (Radii)
∴ ∠APT = ∠ATP = 40°
ΔAPT is an isosceles Δ, i.e. with AP = AT and the angles subtended by them will be the same.

Test Level 1: Geometry - 1 - Question 8

ABCD is a cyclic trapezium with segments AB and DC parallel to each other. If ∠ABC = 80°, then what is the measure of the angle subtended by major arc ABC at the centre?

Detailed Solution for Test Level 1: Geometry - 1 - Question 8

Angle subtended by the minor arc ADC at centre = 2 × (∠ABC) = 160°
Angle subtended by the complementary major arc ABC at centre = (360 - 160)° = 200°

Test Level 1: Geometry - 1 - Question 9

If C is the centre of the following circle, RS = 6 units and SC = 5 units, then what is the length of PT?

Detailed Solution for Test Level 1: Geometry - 1 - Question 9

RS = 6 units, SC = 5 units = CT

RP = 6/2 = 3 units = PS, as perpendicular from the centre to chord divides the chord into 2 equal parts.

Test Level 1: Geometry - 1 - Question 10

The chord AB is perpendicular to the diameter MN at P. The lengths MN and AB are two-digit integral numbers and the length AB is obtained by reversing the digits of the length MN. The length OP is a non-zero rational number. Find the diameter of the circle.

Detailed Solution for Test Level 1: Geometry - 1 - Question 10

Put MN = 10m + n
AB = 10n + m

As OP is rational, so 11 must divide (m + n)(m - n) and m and n are single digit numbers.
Hence, m = 6, n = 5.
Therefore, diameter MN = 10(6) + 5 = 65 units

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