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Test Level 2: Geometry - 1 - CAT MCQ


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10 Questions MCQ Test Level-wise Tests for CAT - Test Level 2: Geometry - 1

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

If the medians of a right-angled triangle, which are drawn from the vertices of the acute angles, measure 6 cm and 8 cm, find the length of the hypotenuse.

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

In ΔABC, AE and CD are the medians of the triangle.
So, AE = 6 and CD = 8

hence, BE = EC = 1/2BC and AD  = DB= 1/2AB

Adding (l) and (2), we get 

Test Level 2: Geometry - 1 - Question 2

If O is the centre of the circle and R is the point of contact of the tangent, then what is the measure of ∠ORQ?

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

∠PQR = 50° (Alternate Segment Theorem)

∠PQO + ∠ OQR = 50°
30° + ∠OQR = 50°
∠OQR = 20°

Since PO = OQ, (Radii of the same circle)
∠OPQ = ∠OQP = 30°
∠OQR = ∠ORQ = 20°

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Test Level 2: Geometry - 1 - Question 3

Find the measure of angle PQS, if the measure of arc PMQ is 150°.

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

As shown in the figure, join PO and QO, where O is the centre.
Arc PMQ = 150°
(∵ Given ∠POQ = 150°)
Now, join PN and QN as shown in the figure.

∠PNQ = 1/2 ∠POQ = 75° [∵ The angle subtended by an arc or a chord of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle]
Now, ∠PQS = ∠PNQ [∵ Angles in alternate segments are always equal]
∠PQS = 75°

Test Level 2: Geometry - 1 - Question 4

What is the length of PN, if AB = 30, PM = 8, DC = 16 and P is the centre of the circle?

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

PM ⊥ AB
Any perpendicular from the centre to the chord bisects the chord.

In ΔAMP
AP2 = MP2 + AM2
AP2 = 82 + 152 = 64 + 225 = 289

In ΔCNP
CP2 = NP2 + CN2
172 = NP2 + 82
172 - 82 = NP2
NP = 15

Test Level 2: Geometry - 1 - Question 5

Four tangents to a circle form a parallelogram PQRS. Find the length of side PS.

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

Opposite sides of a parallelogram are parallel and equal.
Therefore, PQ = SR and PS = QR.
Suppose the length of SC is x units and that of CR is y units.
∵ SR = PQ
∴ x + y = 8 ... (1)
SC and SD are tangents to the circle, so SC = SD = x units.
CR and RB are tangents to the circle, so CR = BR = y units.
SP = (x + 3) = RQ = (5 + y)
x - y = 2 ... (2)
Solving the equations (1) and (2), we get
x = 5 units and y = 3 units
∴ PS = 3 + x = 3 + 5 = 8 units

Test Level 2: Geometry - 1 - Question 6

If PS || QR, ∠1 = ∠2 and ∠3 = ∠4, then what is the measure of ∠STR?

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

Given; PS || RQ, ∠1 = ∠2, ∠3 = ∠4
Let ∠STR = θ
∠PSR + ∠SRQ = 180° {Sum of co-interior angles is 180°}
∠1 + ∠2 + ∠3 + ∠4 = 180°
2∠2 + 2∠3 = 180°
∠2 + ∠3 = 90° ...(1)
Now, ∠2 + ∠θ + ∠3 = 180° {Sum of the angles in a triangle is 180°}
∠θ + 90° = 180° {from (1)}
∠θ = 180° - 90°
∠θ = 90°
∴ ∠STR = ∠θ = 90°

Test Level 2: Geometry - 1 - Question 7

The sum of the interior angles of a triangle is 180°. How many sides does a polygon have if the sum of its interior angles is 2520°?

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

Sum of interior angles of a polygon = (n - 2) x 180°, where n is the number of sides of polygon.
Now, (n - 2) x 180° = 2520°
n - 2 = 14
n = 16

Test Level 2: Geometry - 1 - Question 8

In the given figure, AB || CD. Find angle BGF.

Detailed Solution for Test Level 2: Geometry - 1 - Question 8

Given, AB || CD.
70° = 30° +∠ECD
∠ECD = 40° 
∠ECD + ∠CEF = 140° + 40° = 180° 
So, EF || CD.
∠EFG = ∠FGB
∠BGF = 50°

Test Level 2: Geometry - 1 - Question 9

In the following figure, OA bisects ∠A, ∠ABO = ∠ACO and ∠BOC = 100°. Find the measure of ∠AOB.

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

In triangle AOB and AOC, two angles are correspondingly equal.
⇒ ∠AOB = ∠AOC = z
Now, ∠AOB + ∠AOC + ∠BOC = 360°
z + z + 100° = 360°
Or z = 130°
Hence, option (2) is correct.

Test Level 2: Geometry - 1 - Question 10

Six circles, each of radius r feet, are placed as shown in the figure below. If A is the area bounded by but not included in the circles (in square feet), and B is the height of the stack (in feet), then find the value of A + Br.

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

Draw a triangle running through the centres of the circles as shown above. Since the circles are of equal radii, ΔABC will be an equilateral triangle with side 4r.
Area bounded by but not included in the circles (A) = Area of the equilateral triangle – (Area of 3 semicircles + Area of 3 sectors of angle 60° each)

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