Class X_Delhi_Math_Set-3
SECTION - C
Question Numbers 15 to 24 carry three marks each.
15.
A vessel is in the form of hemispherical bowl surmounted by a hollow cylinder of same
diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the
22 

vessel is 13 cm. Find the total surface area of the vessel.  Use π = 
7

Ans.
Let the radius and height of cylinder be r cm and h cm respectively.
Diameter of the hemispherical bowl = 14 cm
 Radius of the hemispherical bowl = Radius of the cylinder = r =
14
cm = 7 cm
2
Total height of the vessel = 13 cm
 Height of the cylinder, h = Total height of the vessel − Radius of the hemispherical bowl
= 13 cm – 7 cm = 6 cm
Total surface area of the vessel = 2 (curved surface area of the cylinder + curved surface
area of the hemisphere) (Twice because the vessel is hollow)
22
 2  2rh  2r 2   4r  h  r   4   7   6  7  cm 2
7
2
 1144 cm
Thus, the total surface area of the vessel is 1144 cm2.
Class X_Delhi_Math_Set-3
16.
A wooden toy was made by scooping out a hemisphere of same radius from each end of a
solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find
22 

the volume of wood in the toy.  Use π = 
7

Ans.
Height of the cylinder, h = 10 cm
Radius of the cylinder = Radius of each hemisphere = r = 3.5 cm
Volume of wood in the toy = Volume of the cylinder – 2 × Volume of each hemisphere
2
  r2 h  2   r3
3
22
4 22
2
3
   3.5 cm  10 cm     3.5 cm 
7
3 7
539
 385 cm3 
cm3
3
616

cm3
3
 205.33 cm3 (Approx)
Thus, the volume of the wood in the toy is approximately 205.33 cm3.
17.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the
22 

length of the arc (ii) area of the sector formed by the arc.  Use π = 
7

Ans.
It is given that, radius = 21 cm.
The arc subtends an angle of 60.
(i) Length (l) of the arc is given by:

l
 2πr , where r = 21 cm and  = 60
360
Class X_Delhi_Math_Set-3
60
22
 2   21 cm
360
7
 22 cm

(ii) Area, A of the sector formed by the arc is given by

A
 πr 2 , where r = 21 cm and  = 60
360
60 22

  21 21 cm 2
360 7
 231 cm 2
18.
In Fig.5, AB and CD are two diameters of a circle with centre O, which are perpendicular
to each other. OB is the diameter of the smaller circle. If OA = 7 cm, find the area of the
22 

shaded region.  Use π = 
7

Ans.
AB and CD are the diameters of a circle with centre O.
 OA = OB = OC = OD = 7 cm (Radius of the circle)
Area of the shaded region
= Area of the circle with diameter OB + (Area of the semi-circle ACDA – Area of ACD)
2
1
2
7  1

         7    CD  OA 
2
2  2

22 49
1 22
1
 
cm 2    49 cm 2   14 cm  7 cm
7 4
2 7
2
77
 cm 2  77 cm 2  49 cm 2
2
 66.5 cm 2
Thus, the area of the shaded region is 66.5 cm2.
Class X_Delhi_Math_Set-3
19.
Find the ratio in which the y-axis divides the line segment joining the points (–4, – 6) and
(10, 12). Also find the coordinates of the point of division.
Ans.
Let the y-axis divide the line segment joining the points (–4, –6) and (10, 12) in the ratio
 : 1. and the point of the intersection be (0, y).
So, by section formula, we have:
 10   4  12   6  
,

   0, y 
 1 
  1
10  4

 0  10  4  0
 1
4 2
 

10 5
 24  30 
2
12   6  12  5  6  5 
6
 y



2
 1
7
 25
1


5
 5 
Thus, the y-axis divides the line segment joining the given points in the ratio 2 : 5
6

and the point of division is  0,   .
7

20.
The horizontal distance between two poles is 15 m. The angle of depression of the top of
first pole as seen from the top of second pole is 30°. If the height of the second pole is 24
m, find the height of the first pole.  Use 3 1.732
Ans.
Let AB and CD be two poles, where CD = 24 m.
It is given that angle of depression of the top of the pole AB as seen from the top of the
pole CD is 30 and horizontal distance between the two poles is 15 m.
 CAL = 30 and BD = 15 m.
Class X_Delhi_Math_Set-3
To find: Height of pole AB
Let the height of pole AB be h m.
AL = BD = 15 m and AB = LD = h
Therefore, CL = CD  LD = 24  h
Consider right ACL:
Perpendicular CL
tan CAL 

Base
AL
24  h
 tan 30 
15
1
24  h


15
3
 24  h 
15
3
 24  h  5 3
 h  24  5 3
 h  24  5  1.732
Taking 3 1.732 


 h  15.34
Therefore, height of the pole AB = h m = 15.34 m.
21.
For what values of k, the roots of the quadratic equation (k + 4) x2 + (k + 1) x + 1 = 0 are
equal.
Ans.
The given quadratic equation is (k + 4)x2 + (k + 1)x + 1 = 0.
For equal roots, its discriminant, D is 0.
 b2 – 4ac = 0 where a = k + 4, b = k + 1, c = 1
 (k + 1)2 – 4(k + 4) × 1 = 0
 k2 + 2k + 1 – 4k – 16 = 0
 k2 – 2k – 15 = 0
 k2 – 5k + 3k – 15 = 0
 k(k – 5) + 3(k – 5) = 0
 (k – 5) (k + 3) = 0
 k = 5 or k = –3
Thus, for k = 5 or k = –3, the given quadratic equation has equal roots.
Class X_Delhi_Math_Set-3
22.
The sum of first n terms of an AP is 3n2 + 4n. Find the 25th term of this AP.
Ans.
The sum of first n terms (Sn) is given as Sn  3n2  4n .
So, first term (a1)  S1  3 1  4 1  7
2
S2  a1  a2  3  2   4  2   20
a2 = 20 – a1 = 20 – 7 = 13
So, common difference (d) = a2– a1 = 13 – 7 = 6
nth term is given by, an = a + (n – 1) d
2
Thus, 25th term = a25 = 7 + (25 – 1) × 6 = 7 + 24 × 6 = 7 + 144 = 151
23.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of
radius 6 cm.
Ans.
Steps of construction:
1. Draw two concentric circles with centre O and radii 4 cm and 6 cm. Take a point P on
the outer circle and then join OP.
2. Draw the perpendicular bisector of OP. Let the bisector intersects OP at M.
3. With M as the centre and OM as the radius, draw a circle. Let it intersect the inner
circle at A and B.


4. Join PA and PB. Therefore, PA and PB are the required tangents.
24.
Show that the points (–2, 3), (8, 3) and (6, 7) are the vertices of a right triangle.
Ans.
The given points are A (–2, 3), B (8, 3) and C (6, 7).
By, distance formula, AB 
8   2  3  3
2
2
Class X_Delhi_Math_Set-3
 AB2  102  0
 AB2  100
 6  8    7  3
2
 BC2   2   42
BC 
2
2
 BC2  4  16
 BC2  20
CA 
 CA
 2  6   3  7 
2
2
  8   4 
2
2
 CA 2  64  16
 CA 2  80
It is observed that BC2 + CA2 = 20 + 80 = 100 = AB2
So, by the converse of Pythagoras Theorem, ABC is right angled at C.