### SSC CHSL- Practice Aptitude Questions (Geometry) With Solutions Set-42

SSC CHSL- Practice Aptitude Questions (Geometry) With Solutions Set-42:

Dear Readers, Important Aptitude Questions for SSC CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

Directions (1-10): Find The value for the Given below Geometry Questions

1). A vertical pole and a vertical tower are standing on the same level ground. Height of the pole is 10 metres. From the top of the pole the angle of depression of the foot of the tower are 60o and 30o respectively. The height of the tower is
a)    20 m
b)    30 m
c)    40 m
d)    50 m
2). Area of a regular hexagon with side ‘a’ is
a)    [(3√3)/4] a2 sq. unit
b)    [12/(2√3)] a2 sq. unit
c)    [9/(2√3)] a2 sq. unit
d)    [6/√2] a2 sq. unit
3). If the sum of the dimensions of rectangular parallel piped is 24 cm and the length of the diagonal is 15 cm, then the total surface are of its is
a)    420 cm2
b)    275 cm2
c)    351 cm2
d)    378 cm2
4). A horse takes 2 ½ seconds to complete a round around a circular field. If the speed of the horse was 66 m/sec, then the radius of the field is,  { given π=22/7 }
a)    25.62 m
b)    26.52 m
c)    25.26 m
d)    26.25 m
5). A flask in the shape of a right circular cone of height 24 cm is filled with water. The water is poured in a right circular cylindrical flask whose radius 1/3 rd is of the radius of the base of the circular cone. Then the height of the water in the cylindrical flask is
a)    32 cm
b)    24 cm
c)    48 cm
d)    72 cm
6). If the three medians of a triangle are same, then the triangle is
a)    Equilateral
b)    Isosceles
c)    Right-angled
d)    Obtuse-angled
7). The external fencing of a circular path around a circular plot of land is 33 m more than its interior fencing. The width of the path around the plot is
a)    5.52 m
b)    5.25 m
c)    2.55 m
d)    2.25 m
8). The Perimeters of two similar triangles ∆ABC and ∆PQR are 36 cm and 24 cm respectively. If PQ = 10 cm, The AB is
a)    15 cm
b)    12 cm
c)    14 cm
d)    26 cm
9). If the sides of a right-angled triangle are three consecutive integers, then the length of the smallest side is
a)    3 units
b)    2 units
c)    4 units
d)    5 units
10). Two circles intersect each other at the point A and B, A straight line parallel to AB intersects the circles at C, D, E and F. If CD = 4.5 cm, then the measure of EF is
a)    1.50 cm
b)    2.25 cm
c)    4.50 cm
d)    9.00 cm

1).c)  2).c)  3).c)  4).d)  5).d)  6).a)  7).b)  8).a)  9).a)  10).c)

Solutions:

1).

In ∆ECB
tan 60o = BE/CE à √3 = BE/CE
BE = √3 CE …………… (i)
In ∆ACD,
tan 30 = CD/DA
∆ 1/√3 = 10/DA
DA = 10√3
But  CE = DA = 10√3
(i)  à  BE = √3 CE
= √3 ×10√3 = 30m
Height of the tower = AE + BE = 10 + 30 = 40m

2).
Area of an equaliteral triangle  = √3 a2  / 4

Area of regular hexagon
= 6 ×[ √3 a2 / 4 ]
= [ 3√3 / 2 ]a2
= [3√3 / 2] a2 × √3/√3
= 9 a2 / 2√3 sq. unit

3).
Let the dimensions of the rectangular parallelepiped be x, y and z
Then x + y + z = 24
Length of the diagonal
√(x2+y2+z2) = 15
x2 +y2 + z2 = 225
Now , (x+y+z)2 = x2+y2+z2+2(xy+yz+zx)
(24)2 = 225+2(xy+yz+zx)
2(xy+yz+zx) = 576-225 =351
Total surface area = 2(xy+yz+zx) = 351 cm2

4). Perimeter of the circle = Time taken × Speed
= 66 × 2 ½
= 66 × 5/2
= 33 × 5 = 165 m
2πr      = 165
2 × 22/7  × r   = 165
r = (165 × 7)  / (2 × 22)
r = 26.25 m

5).Let the radius of the cone be R and height be H
Let the radius and height of the cylinder be r and h
Then  , Volume of the cylindrical flask = Volume of the cone
π r2h = 1/3 π R2H
(R/3)2h = 1/3 × R2 24
h = (R2×24×3×3) / (R2×3) = 72 cm

6). If three medians of a triangular are the same, then the triangle is equilateral
7).
Let the radius of the external circle and the internal circle be R and r respectively
Then                           2πR-2πr = 33
R-r      = 33/2π
= (33×7) / (2×22)
= 21/4
= 5.25 m

8).
∆ ABC and ∆ PQR are similar
Perimeter of ∆ ABC / Perimeter of ∆ PQR = AB / PQ
36 / 24 = AB / 10
AB = 36×10 / 24
= 15 cm

9). Let the sides of the right angle triangle be x, (x+1) and (x+2)
By Pythagoras theorem
(x+2)2 = (x+1)2 + x2
x2 + 4x + 4 = x2 + 2x + 1 + x2
x2-2x-3 =0
(x-3)(x+1) = 0
x = 3 or x = -1
Side cannot be negative.
X = 3 (Smallest side)