Geometry Set I(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

GEOMETRY SET I MCQs

Total Questions : 110 | Page 2 of 11 pages

Question 11. If (sec A + tan A)(sec B + tan B)(sec C + tan C) = (sec A - tan A)(sec B - tan B)(sec C - tan C), then each side is equal to

1
-1
0
None of these

Discuss Question

Answer: Option A. -> 1
:
A
(a,b) If L=M, then L

^{2}

= LM or ML = M

^{2}

Both LM = ML = 1 as sec

^{2}

A - tan

^{2}

A = 1

∴

^{2}

= M

^{2}

= 1.

Question 12. If

\frac{s i n^{4} A}{a} + \frac{c o s^{4} A}{b} = \frac{1}{a + b},

then the value of

\frac{s i n^{8} A}{a^{3}} + \frac{c o s^{8} A}{b^{3}}

is equal to

1(a+b)3
a3b3(a+b)3
a2b2(a+b)2
None of these

Discuss Question

Answer: Option A. -> 1(a+b)3
:
A
(a) It is given that

\frac{s i n^{4} A}{a} + \frac{c o s^{4} A}{b} = \frac{1}{a + b}

\Rightarrow \frac{(1 - c o s 2 A)^{2}}{4 a} + \frac{(1 + c o s 2 A)^{2}}{4 b} = \frac{1}{a + b}

\Rightarrow b (a + b) (1 - 2 c o s 2 A + c o s^{2} 2 A) + a (a + b) (1 + 2 c o s 2 A + c o s^{2} 2 A) = 4 a b

\Rightarrow {b (a + b) + a (a + b)} c o s^{2} 2 A + 2 (a + b) (a - b) c o s 2 A

Question 13. In the given figure, EADF is a rectangle and ABC is a triangle whose vertices lie on the sides of EADF. AE = 22, BE =16, CF = 16, CF = 16 and BF = 2. Find the length of the line joining the
mid-points of the sides AB and BC.
(CAT 1997)

4√4
5
3.5
none of these

Discuss Question

Answer: Option B. -> 5
:
B
Option (b)
EF = AD = 8
(EADF is a rectangle)
CD = (22 – 16) = 6

So in the right angled triangle ADC, AD = 8 and CD = 6.
Therefore AC = 10
Therefore length of the line joining the mid – points of
AB and BC =

\frac{1}{2}

(10) = 5
(length of the line joining the mid – point of two sides of a triangle is half the 3\(^{rd}\)

Question 14. Answer the questions on the basis of the information given below :In the adjoining figure, I and II are circles with centers 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 PO is 28 cm.

What is the ratio of the length of PQ to that of QO? (2004)

Discuss Question

Answer: Option B. -> 1:3
:
B
From the given diagram,

\frac{O Q}{S Q} = \frac{O P}{R P}

\frac{O Q}{O P} = \frac{S Q}{R P} = \frac{3}{4}

Therefore,

\frac{P Q}{O Q} = \frac{1}{3}

Question 15. ABCD is a rhombus with the diagonals AC and BD intersecting at the origin on the x y plane. The equation of the straight line AD is x + y =1. What is the equation of BC? (CAT 2000)

x + y = -1
x – y = -1
x + y = 1
None of these

Discuss Question

Answer: Option A. -> x + y = -1
:
A

ABCD Is A Rhombus With The Diagonals AC And BD Intersecting ...

Slope of line AD= -1
Hence, eqn of BC which passes through (0,-1) and (-1,0) and is parallel to x+y=1 is
(y-0)=-1(x+1)
Y=-x-1

\Rightarrow

x+y= -1. Option (a)

Question 16. In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If

∠

ATC = 30

^{\circ}

and

∠

ACT = 50

^{\circ}

, then the

∠

BOA is :
(CAT 2003)

In The Figure Given Below (not Drawn To Scale), A, B And C...

100∘
150∘
80∘
not possible to determine

Discuss Question

Answer: Option A. -> 100∘
:
A

∠

CAT= 100

^{\circ}

Therefore,

∠

BAC= 80

^{\circ}

∠

OCT= 90

^{\circ}

Therefore,

∠

BCT is > 90

^{\circ}

Going from answer options, answer can only be option (a)

Question 17. In triangle DEF shown below, points A, B and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC.

∠

D = 40

^{\circ}

then what is

∠

ACB in degrees?

In Triangle DEF Shown Below, Points A, B And C Are Taken On ...

(CAT 2001)

140∘
70∘
100∘
None of these

Discuss Question

Answer: Option C. -> 100∘
:
C

Option (c)
The angle in question is p
p+180-2x+180-2y=180

^{\circ}

thus p-2(x+y)= 180

^{\circ}

……………..(1)
also, BDX = x+y (exterior angle= sum of 2 interior angles not adjacent to it)
x+y= 140

^{\circ}

………(2)
using the 2 statements, p= 100

^{\circ}

Question 18. AB^ BC, BD^ AC and CE bisects

∠

∠

A = 30

^{\circ}

. Then what is CED?
(CAT 1995)

AB^ BC, BD^ AC And CE Bisects ∠C,∠ A = 30∘. Then What ...

30∘
60∘
45∘
65∘

Discuss Question

Answer: Option B. -> 60∘
:
B

∠

ACB= 60

∠

DCE=

∠

ECB= 30
Therefore,

∠

CED= 60. Option (b)

Question 19. Two opposite sides of a square are given by the lines 5x + 12y + 18 = 0 and 5x + 12y – 21 = 0. Find the length of diagonal of the square.

3√2
6√2
2√3
4√3

Discuss Question

Answer: Option A. -> 3√2
:
A

Two Opposite Sides Of A Square Are Given By The Lines 5x +...

Given lines are 5x + 12y + 18 = 0……………….(i) and 5x + 12y – 21 = 0………………….(ii)
Length of the side = distance between lines (i) and (ii) is given by
D =

∣ ∣ ∣ \frac{C_{2} - C_{1}}{\sqrt{a^{2} + b^{2}}} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{- 21 - 18}{\sqrt{25 + 144}} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{- 39}{13} ∣ ∣ ∣ = 3

Length of the diagonal =

3 \sqrt{2};

Option(a).

Question 20. Find the area of quadrilateral formed by joining the points (- 4, 2), (- 1, 0), (4, 1) and (2, 5):

20
40
36
21 .5

Discuss Question

Answer: Option D. -> 21 .5
:
D

1^{s t}

method: - The simplest way to find this is to put a rectangle around the given quadrilateral and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 40, but after subtracting the areas of the 4 right triangles, whose areas are