## GEOMETRY SET I MCQs

Total Questions : 110 | Page 1 of 11 pages
Question 1. Find the slope of the line joining the points (7, 5) and (9, 7):
1.    1
2.    2
3.    3
4.    4
:
A
Option(a)
Slope of the line = y2y1x2x1=7597=1
Here (x1,y1) = (7, 5) and(x2,y2) = (9, 7)
Question 2. Find the area of pentagon formed by joining the points (4, 4), (2, 2), (6, 2), (2, - 3) and (6, -3):
1.    24
2.    20
3.    16
4.    12
:
A

Area of pentagon = Area of triangle ABE + Area of rectangle BCDE
12×4×2+5×4=24
Question 3. The lines 2x + 2y + 1 = 0 and 8x – 2y + 9 = 0 intersect in the
:
B
2x + 2y + 1 = 0 …………….(i)
8x – 2y + 9 = 0 …………………………(ii)
After solving equation (i) and (ii)
x = - 1 and y = 12
(1,12); which is in 2nd quadrant.
Question 4. Determine the radius of the circle, two of whose tangents are the lines 2x+3y-9 = 0  and 4x+6y+19 =0
1.    134√37
2.     354√13
3.     114√13
4.     374√13
:
D
The given tangents are
2x+3y-9 = 0
4x+6y+19 =0
Or 2x + 3y + 192 = 0
Which are parallel, than distance between parallel tangents must be diameter of the circle than diameter.
192(9)22+32=37213

Radius = 12 diameter = 37413
Question 5. Three points A, B and C are collinear such that AB = 2BC. If the coordinates of the points A and B are (1, 7) and (6, -3) respectively, then the coordinates of the point C can be
1.    (72;−2)
2.     (172;−8)
3.    (−172;−8)
4.    (172;−2)
:
B
Given, AB = 2BC
AB : BC = 2 : 1
Let C BE (h, k), then
(2h±12±1;2k±72±1) = (6, - 3) [B may divide AC internally or externally]
2h + 1 = 18 and 2k + 7 = - 9
(or) 2h - 1 = 6 , 2k - 7 = - 3
2h = 17 and 2k = - 16 (or) 2h = 7, 2k = 4
h = 172 and k = - 8 (or) h = 72; k = 2
C = (172;8) or (172;2)
Option(b)
Question 6. Let α,β be such that π < (α - β) < 3π. If sin α + sin β = -2165 and cos α + cos β = -2765, then the value of \
cos αβ2
1.    −665
2.    3√130
3.    665
4.    -3√130
:
D
(d)sin α + sin β = -2165 , cosα + cos β = -2765
Now (sin α + sin β)2 + (cosα + cos β)2 = (2165)2 +(2765)2
2 + 2 sin α sin β + 2 cos α cos β = 441652 + 729652
2 + 2[cos (α - β)] = 1170(65)2 2.2 cos2(α+β2)=1170(65)2
cos (αβ2) = 3130130 = 3130
Therefore cos (αβ2) = 3130 , { π2 < αβ2 < 3π2 }
Question 7. Let A0A1A2A3A4A5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A0A1, A0A2 and A0A4 is
1.    34
2.    3√3
3.    3
4.    3√32
:
C
(c) Each triangle is an equilateral triangle

Hence A0A1 = 1
A0A20 =A0A21 +A1A20 - 2A0A1A1A2 cos 120
= 1 + 1 - 2.1.1(-12) = 3
A0A2 = 3 =A0A4
A0A1 ×A0A2 ×A0A4 = 1.3.3
Question 8. The maximum value of cosα1.cosα2........cosαn, under the restrictions 0α1,α2,.........αn π2 and
cotα1.cotα2........cotαn = 1 is
1.    12n/2
2.    12n
3.    12n
4.    1
:
A
(a) Here (cot α1).(cot α2)....(cot αn)= 1
cosα1.cosα2........cosαn = sinα1.sinα2........sinαn
Now, (cosα1.cosα2........cosαn)2
= (cosα1.cosα2........cosαn) (cosα1.cosα2........cosαn)
= (cosα1.cosα2........cosαn) (sinα1.sinα2........sinαn)
= 12n sin 2α1.sin 2α2........sin 2αn
But each of sin 2αi 1
(cosα1.cosα2........cosαn)212n
But each of cos αi, is positive.
cosα1.cosα2........cosαn 12n = 12n/2.
Question 9. If α, β, γ, δ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive
quantity k, then the value of 4 sin α2 + 3 sin β2+ 2 sin γ2 + sin δ2 is equal to
1.    2√1−k
2.    12√1+k.
3.    2√1+k.
4.    None of these
:
C
(c) Given α < β < γ < δ and sin α = sin β = sin γ = sin δ = k. Alsoα, β,γ,δ are smallest positive angles satisfying above two conditions.
We can take β = π - α, γ = 2π + α, δ = 3π - α
Given expression
= 4 sin α2 + 3 sin(π2α2) + 2 sin(π+α2) +3 sin(3π2α2)
=4 sin α2 + 3cos α2 - 2sin α2 -cos α2 = 2 (sinα2+cosα2)
= 2 (sin12α+cos12α)2 = 2 1+sinα = 21+k.
Question 10. If asin2θ+bsinθ cosθ+c cos2θ12(a+c)12k, then k2 is equal to
1.     b2+(a−c)2
2.    a2+(b−c)2
3.    c2+(a−b)2
4.    None of these