NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7
NCERT Solutions for Class 9 Maths Chapter 13 Surface
Areas and Volumes Ex 13.7 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 13 Surface
Areas and Volumes Ex 13.7.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7
Ex 13.7 Class 9 Maths Question 1.
Find the volume of the right
circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
Solution:
(i) Here, radius of the cone (r) = 6
cm
Height of the cone (h) = 7 cm
Ex 13.7 Class 9 Maths Question 2.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm
Solution:
(i) Here, radius (r) = 7 cm and slant height (l) = 25
cm
Thus, the required capacity of the conical vessel is 1.232 litres.
(ii) Here, height (h) = 12 cm and slant height (l) =
13 cm
∴ Capacity of the conical vessel
Thus, the required capacity of the conical vessel is 11/35 litres.
Ex 13.7 Class 9 Maths Question 3.
The height of a cone is 15 cm. If its volume is 1570
cm³, find the radius of the base. (Use Ï€ = 3.14)
Solution:
Here, height of the cone (h) = 15 cm
Volume of the cone = 1570 cm3
Let the radius of the base be ‘r’ cm.
Thus, the required radius of the base is 10 cm.
Ex 13.7 Class 9 Maths Question 4.
If the volume of a right circular cone of height 9
cm is 48 cm³, find the diameter of its base.
Solution:
Volume of the cone = 48 cm3
Height of the cone (h) = 9 cm
Let the base radius be r cm.
Diameter = 2 × radius
∴ Diameter of the base of the cone = (2 × 4) cm = 8 cm
Ex
13.7 Class 9 Maths Question 5.
A conical pit of top diameter 3.5 m is 12 m deep.
What is its capacity in kilolitres?
Solution:
Here, diameter of the conical pit = 3.5 m
Radius of the conical pit (r) = 3.5/2 m
= 35/20 m,
Depth (h) = 12 m
⇒ Volume (capacity) = 1/3 × Ï€r2h 
Thus, the capacity of the conical pit is 38.5 kl.
Ex
13.7 Class 9 Maths Question 6.
The volume of a right circular cone is 9856 cm3.
If the diameter of the base is 28 cm, find:
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
Solution:
Volume of the cone = 9856 cm3
Diameter of the base = 28 cm
Radius of the base (r) = 28/2 = 14 cm
(i) Let the height of the cone be h cm.
Thus, the required height is 48 cm.
(ii) Let the slant height be l cm.
⇒ l2 = r2 + h2
⇒ l2 = 142 + 482 =
196 + 2304 = 2500
∴ l = 50
Thus, the required slant height is 50 cm.
(iii)
The curved surface area of a cone = πrl
∴ Curved surface area = 22/7 × 14 × 50 cm2
= 2200 cm2
Thus, the curved surface area of the cone is 2200 cm2.
Ex 13.7 Class 9 Maths Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13
cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution:
Sides of the right triangle ABC are 5 cm, 12 cm and
13 cm.
The right-angled triangle is revolved about the side 12 cm.
Thus,
we have radius of the base of the cone so formed (r) = 5 cm
Height (h) = 12 cm
∴ Volume of the cone so obtained = 1/3 × Ï€r2h
= 1/3 × Ï€ × (5)2 ×
12 cm3
= 100 π cm3
Thus, the required volume of the cone is 100Ï€ cm3.
Ex 13.7 Class 9 Maths Question 8.
If the triangle ABC in Question 7 above is revolved
around the side 5 cm, then find the volume of the solid so obtained. Find also
the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
Since the right triangle is revolved around the side
5 cm.
∴ Height of the cone so obtained (h) = 5 cm
Radius of the cone (r) = 12 cm
Thus, the required ratio is 5 : 12.
Ex 13.7 Class 9 Maths Question 9.
A heap of wheat is in the form of a cone whose
diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be
covered by canvas to protect it from rain. Find the area of the canvas
required.
Solution:
Here, the heap of wheat is in the shape of a cone
with base diameter = 10.5 m
Thus, the required volume = 86.625 m3
Now,
the area of the canvas to cover the heap must be equal to the curved surface
area of the conical heap.
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