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 10 of 11 pages

Question 91.

Determine the radius of the circle, two of whose tangents are the lines 2x+3y-9 = 0 and 4x+6y+19 =0

$\frac{13}{4 \sqrt{37}}$
$\frac{35}{4 \sqrt{13}}$
$\frac{11}{4 \sqrt{13}}$
$\frac{37}{4 \sqrt{13}}$

Discuss Question

Answer: Option D. ->

\frac{37}{4 \sqrt{13}}

:
D

The given tangents are

2x+3y-9 = 0

4x+6y+19 =0

Or 2x + 3y + $\frac{19}{2}$ = 0

Which are parallel, than distance between parallel tangents must be diameter of the circle than diameter.
$\frac{∣ ∣ ∣ \frac{19}{2} - (- 9) ∣ ∣ ∣}{\sqrt{2^{2} + 3^{2}}} = \frac{37}{2 \sqrt{13}}$

Radius = $\frac{1}{2}$ diameter = $\frac{37}{4 \sqrt{13}}$

Question 92.

Find the equation of circle whose radius is 5 and which touches the circle $x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ at the point (5, 5).

$x^{2} + y^{2} - 18 x - 16 y + 120 = 0$
$x^{2} + y^{2} + 6 x - 3 y - 54 = 0;$
$x^{2} + y^{2} - 18 x - 17 y + 50 = 0$
$x^{2} + y^{2} - 18 x - 8 y + 60 = 0$

Discuss Question

Answer: Option A. ->

x^{2} + y^{2} - 18 x - 16 y + 120 = 0

:
A

The given circle is $x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ with centre (1, 2) and radius = $\sqrt{1 + 4 + 20} = 5$ .

And required circle has radius 5 hence circles touch each other externally.

Since point of contact is P(5, 5).

P is the mid point of $C_{1}$ and $C_{2}$ , let co – ordinate of centre $C_{2}$ is (h, k) then

$\frac{1 + h}{2} = 5; h = 9$

$\frac{2 + k}{2} = 5$ ; k = 8

And, Equation of required circle is $(x - 9)^{2} + (y - 8)^{2} = 5^{2}$

$x^{2} + y^{2} - 18 x - 16 y + 120 = 0;$ Option(a)

Question 93.

What is the equation of a line which has an x-intercept of 4 units and passes through the point (-2,3)?

x+4y-4=0
x+2y-4=0
x+6y-16=0
none of these

Discuss Question

Answer: Option B. -> x+2y-4=0
:
B

option (b)

To have an x-intercept 4, the value at y=0 should give x=4.

Put y=0 in all 3 answer options.

Option a and b give x=4

Question 94.

Find the equation of the straight line passing through the point (2, 1) and through the point of intersection of the lines x + 2y = 3 and 2x – 3y= 4.

5x + 3y – 13 = 0
4x – 7y – 1= 0
2x – 7y – 20=0
x – 7y + 13 =0

Discuss Question

Answer: Option A. -> 5x + 3y – 13 = 0
:
A

$1^{s t}$ method: - equation of any straight line passing through the intersection of the lines x + 2y = 3 and 2x – 3y= 4 is

$λ$ (x + 2y – 3) + (2x – 3y – 4) = 0

Since it passes through the point (2, 1)

$λ$ (2 + 2 – 3) + (4 – 3 – 4) = 0

$λ$ - 3 = 0

$λ$ = 3

Now substituting this value of $λ$ in (i), we get

3(x + 2y – 3) + (2x – 3y – 4) = 0

5x + 3y – 13 = 0

$2^{n d}$ method: - The straight line passing through the point (2, 1), put x = 2 and y = 1only option (a) and (b) will satisfy. Now, intersection point of the lines x + 2y = 3 and 2x – 3y= 4 is $(\frac{17}{7}, \frac{2}{7}),$ Put x = $\frac{17}{7}$ and y = $\frac{2}{7}$ only option(a) will satisfy. Option(a).

Question 95.

Find the co-ordinates of the foot of the perpendicular from (2, 4) upon the line x + y = 4

(1, 3)
(3, - 1)
(2, 1)
(4, 0)

Discuss Question

Answer: Option A. -> (1, 3)
:
A

From the given figure, we can easily determine that the co-ordinates of the foot of the perpendicular from (2, 4) upon the line x + y = 4 are (1, 3)

Question 96.

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
$\frac{1}{\sqrt{3}}$ times the side of the cube
impossible to find from the given information

Discuss Question

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 97.

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 98.

Answer the questions on the basis of the information given below:
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 PO is 28 cm.

What is the radius of the circle II?

2 cm
3 cm
4 cm
5 cm

Discuss Question

Answer: Option B. -> 3 cm
:
B

$\frac{P Q}{O Q} = \frac{1}{3}$
$P Q = \frac{1}{4} (28) = 7$
$\frac{M P}{M Q} = \frac{4}{3} (M i s p o i n t o f i n t e r s e c t i o n o f t w o c i r c l e s)$
$M Q = 3 c m$

Question 99.

8√3 cm
10√3 cm
12√3cm
14√3 cm

Discuss Question

Answer: Option C. -> 12√3cm
:
C

O S = \sqrt{21^{2} - 3^{2}} = 12 \sqrt{3}

Question 100.

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)