Exams > Cat > Quantitaitve Aptitude
GEOMETRY SET I MCQs
Total Questions : 110
| Page 3 of 11 pages
Answer: Option C. -> (3,1)
:
C
1st method: - In a right angled triangle
Circumradius = hypotenuse2
The coordinates of circumcentre = (3,1)
2nd method: - If O(x1, y1) is the circumcentre, then OA2=OB2.
(x1−2)2+(Y1−3)2=(X1−4)2+(Y1+1)2
−4X1−6Y1+8X1−2Y1=17−13=4
i.e., 4X1−8Y1=4 or X1−2Y1=1 ..................(I)
OB2=OC2
X1−4)2+(Y1+1)2=(X1−4)2+(Y1−3)2
2Y1+6Y1=8
8Y1=8 ....................(II)
Y1=1
From (i) X1=3
X1,Y1)=(3,1)
:
C
1st method: - In a right angled triangle
Circumradius = hypotenuse2
The coordinates of circumcentre = (3,1)
2nd method: - If O(x1, y1) is the circumcentre, then OA2=OB2.
(x1−2)2+(Y1−3)2=(X1−4)2+(Y1+1)2
−4X1−6Y1+8X1−2Y1=17−13=4
i.e., 4X1−8Y1=4 or X1−2Y1=1 ..................(I)
OB2=OC2
X1−4)2+(Y1+1)2=(X1−4)2+(Y1−3)2
2Y1+6Y1=8
8Y1=8 ....................(II)
Y1=1
From (i) X1=3
X1,Y1)=(3,1)
Answer: Option D. -> (√2,313)
:
D
Even without calculations it is clear that (√2,313) cannot lie on the liney = 5x + 3. If x is irrational, y must also be irrational. Option(d)
:
D
Even without calculations it is clear that (√2,313) cannot lie on the liney = 5x + 3. If x is irrational, y must also be irrational. Option(d)
Answer: Option A. -> 27
:
A
The simplest way to find this is to put a rectangle around the given triangle and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 126, but after subtracting the areas of the 3 right triangles, whose areas are 54, 3, and 42, we are left with an area of 27.
:
A
The simplest way to find this is to put a rectangle around the given triangle and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 126, but after subtracting the areas of the 3 right triangles, whose areas are 54, 3, and 42, we are left with an area of 27.
Answer: Option A. -> 5x + 3y – 13 = 0
:
A
1st 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
2nd 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 (177,27), Put x = 177 and y = 27 only option(a) will satisfy. Option(a).
:
A
1st 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
2nd 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 (177,27), Put x = 177 and y = 27 only option(a) will satisfy. Option(a).
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
:
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
Answer: Option A. -> x2+y2−18x−16y+120=0
:
A
The given circle is x2+y2−2x−4y−20=0 with centre (1, 2) and radius = √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 C1 and C2, let co – ordinate of centre C2 is (h, k) then
1+h2=5;h=9
2+k2=5; k = 8
And, Equation of required circle is (x−9)2+(y−8)2=52
x2+y2−18x−16y+120=0; Option(a)
:
A
The given circle is x2+y2−2x−4y−20=0 with centre (1, 2) and radius = √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 C1 and C2, let co – ordinate of centre C2 is (h, k) then
1+h2=5;h=9
2+k2=5; k = 8
And, Equation of required circle is (x−9)2+(y−8)2=52
x2+y2−18x−16y+120=0; Option(a)
Answer: Option B. -> 18
:
B
(b) We havek = sinπ18.sin5π18.sin7π18
= cos(π2−π18)cos(π2−5π18)cos(π2−7π18)
= cosπ9cos2π9cos4π9 = sin23π923sinπ9 = sin8π98sinπ9
= sin(π−π9)8sinπ9 = 18
:
B
(b) We havek = sinπ18.sin5π18.sin7π18
= cos(π2−π18)cos(π2−5π18)cos(π2−7π18)
= cosπ9cos2π9cos4π9 = sin23π923sinπ9 = sin8π98sinπ9
= sin(π−π9)8sinπ9 = 18
Answer: Option D. -> All of above
:
D
(a,b,c) fn(θ) = sin(θ/2)cos(θ/2)[2cos2θ/2cosθ.2cos2θcos2θ.2cos22θcos4θ]
Combine first two factors,fn(θ) = sinθcosθ[2cos2θcos2θ.2cos22θcos4θ]
Again combine first two factors,
fn(θ) = tan 2θ[2cos22θcos4θ.......] = tan (2nθ)
∴f2(π16) = tan 4π16 = tan (π4)= 1
f3(π32) = tan 8π32 = tan (π4) = 1
f4(π64) = tan 16π64 = tan (π4) = 1
f5(π128) = tan 32π128 = tan (π4) = 1
:
D
(a,b,c) fn(θ) = sin(θ/2)cos(θ/2)[2cos2θ/2cosθ.2cos2θcos2θ.2cos22θcos4θ]
Combine first two factors,fn(θ) = sinθcosθ[2cos2θcos2θ.2cos22θcos4θ]
Again combine first two factors,
fn(θ) = tan 2θ[2cos22θcos4θ.......] = tan (2nθ)
∴f2(π16) = tan 4π16 = tan (π4)= 1
f3(π32) = tan 8π32 = tan (π4) = 1
f4(π64) = tan 16π64 = tan (π4) = 1
f5(π128) = tan 32π128 = tan (π4) = 1
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