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 6 of 11 pages

Question 51. A ladder leans against a vertical wall. The top of the ladder is 8m above the ground. When the bottom of the ladder is moved 2m farther away from the wall, the top of the ladder rests against the foot of the wall. What is the length of the ladder ?
(CAT 2001)

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Answer: Option D. -> 17m
:
D
Option (d)

A Ladder Leans Against A Vertical Wall. The Top Of The Ladde...

Length of the ladder = x+2
Thus,

8^{2} + x^{2} = (x + 2)^{2}

Solving for x, x=15
Length of ladder = 17

Question 52. Consider the five points comprising the vertices of a square and the intersection point of its diagonals. How many triangles can be formed using these points? (CAT 1993)

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Answer: Option C. -> 8
:
C

Consider The Five Points Comprising The Vertices Of A Square...

The number of points will be $^5C_3$- 2 = 8
(2 because of the 2 diagonals)

Question 53. Answer the questions based on the following information:
A rectangle PRSU is divided into two smaller rectangles PQTU, and QRST by the line TQ.PQ = 10cm. QR = 5 cm and Rs = 10cm. Points A, B, F are within rectangle PQTU, and points C, D, E are within the rectangle QRST. The closest pair of points among the pairs (A,C), (A,D), (A,E) (F,C), (F,E), (B,C), (B,D), (B,E) are 10√3 cm apart.
Which of the following statements is necessarily true?

The closest pair of points among the six given points cannot be (F, C).
Distance between A and B is greater than that between F and C.
The closest pair of points among the six given points is (C, D), (D, E) or (C, E).
None of the above

Discuss Question

Answer: Option A. -> The closest pair of points among the six given points cannot be (F, C).
:
A
The diagonal length of a rectancle PUSR =

5 \sqrt{13}

= 18 (approx)
Among given eight pairs the shortest distance =

10 \sqrt{3}

So, the six points A, B, F, C, D and E are near corner of rectangle PUSR.
So, (F,C) cannot be the shortest distance.

Question 54. If the length of diagonals DF, AG and CE of the cube shown in the adjoining figure are equal to the three sides of a triangle, then the radius of the circle circumscribing that triangle will be :

Equal to the side of the cube
√3 times the side of the cube
1√3 times the side of the cube
impossible to find from the given information

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Answer: Option A. -> Equal to the side of the cube
:
A
The side length of a cube = AD = a
The diagonal length of a cube = AG =

a \sqrt{3}

DF = AG = CE =

a \sqrt{3}

The triangle formed was an equilateral triangle.
The circumradius of an equilateral triangle =

\frac{s \sqrt{3}}{3}

Therefore, the circumradius of that triangle =

\frac{a \sqrt{3} \sqrt{3}}{3}

= Side of a cube

Question 55.

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Answer: Option C. -> 93
:
C

Question 56.

The value of k, for which
(cos x + sin x) $^{2}$ + k sin x cos x - 1 = 0 is an identity, is

-1
-2
0
1

Discuss Question

Answer: Option B. -> -2
:
B

(b) Given, (cos x + sin x) $^{2}$ + k sin x cos x - 1 = 0, $\forall$ x
   $\Rightarrow$ cos $^{2}$ x +sin $^{2}$ x + 2cos x sin x + k sin x cos x - 1 = 0, $\forall$ x
      $\Rightarrow$ (k + 2)cos x sin x = 0, $\forall$ x
   $\Rightarrow$ k = -2

Question 57.

If $t a n (A - B) = 1, s e c (A + B) = \frac{2}{\sqrt{3}},$ then the smallest positive value of B is,

$\frac{25}{24} π$
$\frac{19}{24} π$
$\frac{13}{24} π$
$\frac{11}{24} π$

Discuss Question

Answer: Option B. ->

\frac{19}{24} π

:
B

tan(A-B)=1

$A - B = (2 n + 1) \frac{π}{4}$

n=0 since question is the smallest value possible

$A - B = \frac{π}{4}$

$A + B = 2 π - \frac{π}{6} = 11 \frac{π}{6}$

Subtract both the equations

$2 B = 11 \frac{π}{6} - \frac{π}{4}$

$2 B = (22 - 3) \frac{π}{12}$

$B = 19 \frac{π}{24}$

Question 58.

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

$\frac{1}{(a + b)^{3}}$
$\frac{a^{3} b^{3}}{(a + b)^{3}}$
$\frac{a^{2} b^{2}}{(a + b)^{2}}$
None of these

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Answer: Option A. ->

\frac{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 59.

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Answer: Option B. -> 6
:
B

Question 60.

For a positive integer n, let
$f_{n} (θ)$ = $(tan \frac{θ}{2})$ (1 + sec $θ$ )(1 + sec 2 $θ$ )(1 + sec 4 $θ$ ) .........
(1 + sec 2 $^{n}$ $θ$ ). Then

$f_{2} (\frac{π}{16})$ = 1
$f_{3} (\frac{π}{32})$ = 1
$f_{4} (\frac{π}{64})$ = 1
All of above

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Answer: Option D. -> All of above
:
D

(a,b,c) $f_{n} (θ)$ = $\frac{sin (θ / 2)}{cos (θ / 2)}$ $[\frac{2 {cos}^{2} θ / 2}{cos θ} . \frac{2 {cos}^{2} θ}{cos 2 θ} . \frac{2 {cos}^{2} 2 θ}{cos 4 θ}]$
   Combine first two factors, $f_{n} (θ)$ = $\frac{s i n θ}{c o s θ}$ $[\frac{2 {cos}^{2} θ}{cos 2 θ} . \frac{2 {cos}^{2} 2 θ}{cos 4 θ}]$
   Again combine first two factors,
   $f_{n} (θ)$ = tan 2 $θ$ $[\frac{2 {cos}^{2} 2 θ}{c o s 4 θ} . . . . . . .]$ = tan (2 $^{n}$ $θ$ )
$∴$ $f_{2} (\frac{π}{16})$ = tan $\frac{4 π}{16}$ = tan $(\frac{π}{4})$ = 1
   $f_{3} (\frac{π}{32})$ = tan $\frac{8 π}{32}$ = tan $(\frac{π}{4})$ = 1
   $f_{4} (\frac{π}{64})$ = tan $\frac{16 π}{64}$ = tan $(\frac{π}{4})$ = 1
   $f_{5} (\frac{π}{128})$ = tan $\frac{32 π}{128}$ = tan $(\frac{π}{4})$ = 1