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Test Level 1: Geometry - 2 - CAT MCQ


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20 Questions MCQ Test - Test Level 1: Geometry - 2

Test Level 1: Geometry - 2 for CAT 2024 is part of CAT preparation. The Test Level 1: Geometry - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 1: Geometry - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 1: Geometry - 2 below.
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Test Level 1: Geometry - 2 - Question 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.

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

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 ⇒ 2 x 50 m = 100 m
i.e. height of tower = 100 m

Test Level 1: Geometry - 2 - Question 2

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

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

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

Then as per question

{Let the longest side of ΔDEF = x}

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Test Level 1: Geometry - 2 - Question 3

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

Detailed Solution for Test Level 1: Geometry - 2 - Question 3

Ratio of corresponding sides 

Test Level 1: Geometry - 2 - Question 4

If O is the centre of circle, find ∠x

Detailed Solution for Test Level 1: Geometry - 2 - Question 4

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

Test Level 1: Geometry - 2 - Question 5

Find the value of x in the given figure.

Detailed Solution for Test Level 1: Geometry - 2 - Question 5

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

Test Level 1: Geometry - 2 - Question 6

In the given figure, find ∠ADB.

Detailed Solution for Test Level 1: Geometry - 2 - Question 6

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°

Test Level 1: Geometry - 2 - Question 7

In the following figure, O is the centre of the circle and ∠ABO = 30°, find ∠ACB.

Detailed Solution for Test Level 1: Geometry - 2 - Question 7

OB = OA = radius of the circle
∠AOB = 180 – (30 + 30)
{Sum of angles of a triangle = 180°)
⇒ 120°
Then


because the angle subtended
by a chord at the centre is twice of what it
can subtend at the circumference. Again, ABCD is a cyclic quadrilateral;
So ∠ACB = 180° – 60° = 120° (because opposite angles of cyclic quadrilateral are supplementary).

Test Level 1: Geometry - 2 - Question 8

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 Level 1: Geometry - 2 - Question 8

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 Level 1: Geometry - 2 - Question 9

In the figure, DABC is similar to DEDC.

If we have AB = 4 cm,
ED = 3 cm, CE = 4.2 and
CD = 4.8 cm, find the value of CA and CB

Detailed Solution for Test Level 1: Geometry - 2 - Question 9

Test Level 1: Geometry - 2 - Question 10

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

Detailed Solution for Test Level 1: Geometry - 2 - Question 10

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

Test Level 1: Geometry - 2 - Question 11

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.

Detailed Solution for Test Level 1: Geometry - 2 - Question 11

BC = ED = 6 m
So AB = AC – BC = 11 – 6 = 5 m
CD = BE = 12 m
Then by Pythagoras theorem:
AE2 = AB2 + BE2 ⇒ AE = 13 m

Test Level 1: Geometry - 2 - Question 12

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

Detailed Solution for Test Level 1: Geometry - 2 - Question 12

In the DOBC, OB = 12 cm, OC = radius = 13 cm.
Then using Pythagoras theorem;
BC2 = OC2 – OB2 = 25; BC = 5 cm
Length of the chord = 2 x BC = 2 x 5 = 10 cm

Test Level 1: Geometry - 2 - Question 13

Find the value of ∠x in the given figure.

Detailed Solution for Test Level 1: Geometry - 2 - Question 13

∠AOM = 2∠ABM and
∠AON = 2∠ACN
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.
∠AON = 60° and ∠ AOM = 40°
∠X = ∠AON + ∠AOM
(∵ vertically opposite angles).
∠ X = 100°
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 180o. 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.

Test Level 1: Geometry - 2 - Question 14

Find the value of x in the given figure.

Detailed Solution for Test Level 1: Geometry - 2 - Question 14

By the rule of tangents, we know:
62 = (5 + x)5 ⇒ 36 = 25 + 5x ⇒ 11 = 5x ⇒ x = 2.2 cm

Test Level 1: Geometry - 2 - Question 15

Find the value of x in the given figure.

Detailed Solution for Test Level 1: Geometry - 2 - Question 15

By the rule of chords, cutting externally, we get
⇒ (9 + 6)6 = (5 + x)5 ⇒ 90 = 25 + 5x ⇒ 5x = 65

⇒ x = 13 cm

Test Level 1: Geometry - 2 - Question 16

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

Detailed Solution for Test Level 1: Geometry - 2 - Question 16

∠APB = 90° (angle in a semicircle = 90°)
∠PBA = 180 – (90 + 25) = 65°
∠TPA = ∠PBA (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 as the Alternate Segment Theorem.

Test Level 1: Geometry - 2 - Question 17

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.

Detailed Solution for Test Level 1: Geometry - 2 - Question 17

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°.

Test Level 1: Geometry - 2 - Question 18

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

Detailed Solution for Test Level 1: Geometry - 2 - Question 18

(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).

Test Level 1: Geometry - 2 - Question 19

In the following figure, find the value of x

Detailed Solution for Test Level 1: Geometry - 2 - Question 19

∠BAC = 30° (∵ angles subtended by an arc anywhere on the circumference in the same segment are equal).

In DBAC; ∠x = 180° – (110° + 30°) = 40°
(∵ sum of angles of a triangle = 180°)

Test Level 1: Geometry - 2 - Question 20

Find the perimeter of the given figure.

Detailed Solution for Test Level 1: Geometry - 2 - Question 20

Perimeter of the figure = 10 + 10 + 6 + 6π = 26 + 6π

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