CBSE Class 10

These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3

Unless stated otherwise, take π = \(\frac { 22 }{ 7 }\)

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Given: radius of metallic sphere = 4.2 cm

∴ Volume = \(\frac { 4 }{ 3 }\)π(4.2)³ …. (i)
∵ Sphere is melted and recast into a cylinder of radius 6 cm and height h.
∴ Volume of the cylinder =πr²h = π(6)² x h … (ii)
According to question,
Volume of the cylinder = Volume of the sphere
310.464 cm³ = \(\frac{22 \times 36}{7}\) h
310.464 cm³ = 113.142 cm³h
h = \(\frac{310.464}{113.142}\) cm³
h = 2.74 cm.
Height of cylinder = 2.74 cm.

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Radius of sphere (r₁) = 6 cm.

So the volume = \(\frac { 4 }{ 3 }\)π x (r₁)³
= \(\frac { 4 }{ 3 }\)π (6)³ cm.
Radius of sphere (r₂) = 8 cm.

So the volume = \(\frac { 4 }{ 3 }\)π x (r₂)²
= \(\frac { 4 }{ 3 }\)π (6)² cm.
Radius of sphere (r₃ ) = 10 cm.

So the volume = \(\frac { 4 }{ 3 }\)π x (r₃)²
= \(\frac { 4 }{ 3 }\)π (10)² cm.
Now, spheres are melted and form a single
Sphere of radii R. Volume of single sphere = \(\frac { 4 }{ 3 }\)πR³

Radius of single sphere = 12 cm.

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Given: diameter of the well = 7 m Radius = \(\frac { 7 }{ 2 }\)m
and depth of the well = 20 m
Volume of the earth taken out from the well = πr²

= \(\frac { 22 }{ 7 }\) x \(\frac { 7 }{ 2 }\) x \(\frac { 7 }{ 2 }\) x 20
= 770 cm³.
To form a platform of 22m , 14 m ,and h height.
∴ Volume = l x b x h
Volume = 22 x 14 x h
Volume of earth = Volume of platform
770m³ = 22 x 14 x h
So, h = \(\frac{770 m^{3}}{308 m^{3}}\)
h= 2.5 m
Height of the platform = 2.5 m

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 3 m
Inner radius of well = \(\frac { 3 }{ 2 }\)m

Volume of the earth dug out = (πr²h) m³
Width of circular ring = 4 m
Outer radius of well = \(\frac { 3 }{ 2 }\) + 4 = \(\frac { 11 }{ 2 }\) m
Volume of the earth = \(\frac { 22 }{ 1 }\) x \(\frac { 3 }{ 2 }\) x \(\frac { 3 }{ 2 }\) x 14 m
= 99m³
Area of shaded region = π(R² – r²)

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
We have
Radius of the cylinder = 6 cm
Height of the cylinder = 15 cm
∴ Volume of the cylinder = πr²h
= π x 6² x 15 cm.
= 540π cm³
Radius of the ice cream cone = 3 cm
Height of the ice cream cone = 12 cm

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?
Solution:
Given: diameter of each coin = 1.75 cm ⇒ radius = \(\frac { 1.75 }{ 2 }\)cm
and thickness of each coin = 2 mm
Let the number of coins be n. So volume of coins of
= \(\frac { 22 }{ 7 }\) x (0.875)2 x (0.2) cm x n
= \(\left(\frac{4.4}{7} \times 0.765\right) n\)
= \(\frac{(3.3687) n}{7}\)
= (0.48125) n
Coins melted to form a cuboid of dimensions 5.5 cm, 10 cm, 3.5 cm,
= 5.5 x 10 cm x 3.5 cm
= 192.5 cm³.
∴ 0.48125 n = 192.5
∴ n = \(\frac{192.5}{0.48125}\) = 400
No. of coins = 400

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Given: radius of the cylindrical bucket = 18 cm
and height = 32 cm

Height of conical heap = 24 cm. Let the radius of conical heap be r¹ and slant height, l then the volume of conical heap

Volume of cylinder = Volume of conical heap
3258.14 cm³ = 25.1428 r² cm².

Slant height of conical heap = 12\(\sqrt{13}\) cm.

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Given: width of canal = 6m, depth = 1.5 m
Rate of flowing water – 10 km/h
Volume of the water flowing in 30 minutes = \(\frac { 6×1.5×30×10 }{ 60 }\)km³
= \(\frac { 6×1.5×10×1000×30 }{ 10×60 }\)km³ = 45000 m³
We require water for standing up to height = 8 cm = \(\frac { 8 }{ 100 }\) m
Let the required area he A
∴ Volume of water required = A(\(\frac { 8 }{ 100 }\))m³
According to question. 45000 = \(\frac { A×8 }{ 100 }\)
⇒ \(\frac { 45000×100 }{ 8 }\) = A ⇒ A = 562500 m²
Area will it irrigate in 30 minutes = 562500 m² or 56.25 hectares.

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his Held, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Suppose the tank is filled in x hours. Since water is flowing at the rate of 3 km/hr. Therefore, length of the water of the water column in x hours – 3x km = 3000x meters. Clearly, the water column forms a cylinder of radius r = \(\frac { 20 }{ 2 }\) cm = 10 cm = \(\frac { 1 }{ 10 }\) m and h = height (length) = 300x meters.
∴ Volume of the water that flows in the tank in x hours

Since the tank is filled in x hours
∴ Volume of the water that flows in the tank in x hours = Volume of the tank

These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2

Unless stated otherwise, take π = \(\frac { 22 }{ 7 }\)

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of n.
Solution:
Radius of cone = 1 cm. and radius of hemisphere is also = 1 cm.
Volume of solid = Volume of cone + Volume of hemisphere

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
Volume of air contained in the model = Total volume of the solid
Diameter of base of each cone = 3 cm
∴ Radius of base of each cone = \(\frac { 3 }{ 2 }\)
Height of each cone = 2 cm

Volume of the air inside the model = Volume of air inside = Volume of cone + volume of cylinder + volume of other cone.

Volume of the model = 66 cm³

Question 3.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).

Solution:
Volume of one piece of gulab jamun = Volume of the cylindrical portion + Volume of the two hemispherical ends 1 2 8
Radius of each hemispherical portion = \(\frac { 2.8 }{ 2 }\) = 1.4 cm

Radius of gulab jamun = r = \(\frac { 2.8 }{ 2 }\) = 1.4 cm and
height = 5 cm so, height of cylinder (h) = 5 – (2.8) = 2.2.
Volume 45 gulab jamuns = 25.05 x 45 = 1127.279
30% of its volume = \(\frac{1127.279 \times 30}{100}\)
= 338.18 = 338 cm³.

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm.
Find the volume of wood in the entire stand (see figure).

Solution:
Pen stand is made in the shape of a cuboid whose length, breadth and height are respectively = 15 cm, 10 cm, 3.5 cm.
So, the volume of pen stand = length x breadth x height
= 15 cm x 10 cm x 3.5 cm = 525 cm²
Each hole, is in the shape of a cone, so the volume of cone
= \(\frac { 1 }{ 3 }\)r²h
Radius of hole = 0 0.5 cm, and height = 1.4 cm.
∴ Volume of 4 (holes) cone = 4 x \(\frac { 1 }{ 3 }\)r²h
= \(\frac { 4 }{ 3 }\) x \(\frac { 22 }{ 7 }\) 0.5 x 0.5 x 1.4 cm³
= \(\frac{4 \times 4.4 \times 0.25}{3}\) = \(\frac { 4 }{ 3 }\) = 1.466 cm³
Volume of wood in the entire stand = Volume of cuboid – Volume 4 holes.
= (525 – 1.466) cm³.
= 523.533 cm³.

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:

When lead shots are dropped into the vessel, then
Volume of water flows out = Volume of leads shots

Number of lead shots dropped in the vessel = 100.

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14)

Solution:
Given: radius of 1^st cylinder = 12 cm
and height of 1^st cylinder = 220 cm
∴ Volume of 1^st cylinder = πr²h
= π(12)² (220) cm³
= 144 x 220π cm³
= 144 x 220 x 3.14 cm³
= 99475.2 cm³ … (i)
Given: radius of 2^nd cylinder = 8 cm
and height of 2^nd cylinder = 60 cm
∴ Volume of 2^nd cylinder = πr²h
= π(8)² (60) cm³ = 64 x 60π cm³
= 64 x 60 x 3.14 cm³
= 12057.6 cm³ … (ii)
Total volume of solid = Volume of 1^st cylinder + Volume of 2^nd cylinder
= 99475.2 cm³ + 12057.6 cm³ = 111532.8 cm³
Given: mass of 1 cm³ of iron = 8 g
∴ Mass of 111532.8 cm³ of iron = 111532.8 x 8 g
= 892262.4 g = 892.262 kg

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Height of right circular cone = 120 cm.
and radius of hemisphere and cone = 60 cm.
Radius of cylinder = 60 cm, and height = 180 cm.

Volume of water, left in the cylinder = Volume of cylinder – (Volume of cone + volume hemisphere)

So the left in the cylinder = 1,31 m³. (approx).

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter, the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Volume of water the glass vessel can hold = 345 cm³ (Measured by the child)
Radius of the cylindrical part = \(\frac { 2 }{ 2 }\) = 1 cm

Height of the cylindrical part = 8 cm
∴ Volume of the cylindrical part = πr²h
= 3.14 x (1)² x 8 cm³
Diameter of the spherical part Radius = 8.5 cm
∴ Radius = \(\frac { 8.5 }{ 2 }\) cm
= \(\frac{7713.41}{24}\)
= 321.392 cm³.
Total volume of the glass vessel = Volume of the cylindrical part + Volume of the spherical part
= 25.12 cm³ + 321.39 cm³ = 346.51 cm³
Volume measured by child is 345 cm³, which is not correct. Correct volume is 346.51 cm³.

These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
Diameters of two circular ends are 4 cm and 2 cm
∴ Radius are 2 cm and 1 cm.
Volume of the furstum of the cone

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Let r₁ and r₁ be the radii of the circular bases of the frustum, l be the slant height and h be its height.
We have, l = 4 cm, 2πr₁ = 18 and 2πr₂ = 6
⇒ l = 4 cm, r₁ = — and r₂ = —
Curved surface area = l
= π\(\left(\frac{9}{\pi}+\frac{3}{\pi}\right)\) x 4 cm²
= π\(\left(\frac{9+3}{\pi}\right)\) x 4 cm²
= (12 x 4) cm²
= 48 cm².

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:
Radius of open side (r₁) = 10 cm
Radius of upper base (r₂) = 4 cm
Slant height (l) = 15 cm
Area of material used for making = Curved surface area of frustum of cone + area of closed side

Area of material used for cap = 710\(\frac { 2 }{ 3 }\) cm².

Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm². (Take π = 3.14)
Solution:
Radius of the lower end (r₁) = 8 cm
Radius of the upper end (r₂) = 20 cm
Height of the frustum (h) = 16 cm

Cost of milk at the rate of ₹ 20 per litre = ₹(20 x 10.45) = ₹ 209
Now, Total surface area of the frustum = π(r₁) + r₂) l + πr₂² [∵ Top is open]
= [3.14 (20 + 8) x 20 + 3.14 x 8²] cm²
= 3.14 x (560 + 64) cm²
= 3.14 x 624 cm²
= 1959.36 cm²
Cost of metal used = ₹\(\left(\frac{1959.36 \times 8}{100}\right)\) = ₹ 156.75

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\frac { 1 }{ 16 }\) cm, find the length of the wire.
Solution:
Height of right circular cone = 20 cm
In ∆ABP,

Diameter of wire = \(\frac { 1 }{ 16 }\) cm
Radius of wire = \(\frac { 1 }{ 32 }\) cm
Volume of wire = πr²h (h = 1)
= π(\(\frac { 1 }{ 32 }\))² x l … (2)
From (1) and (2)
π(\(\frac { 1 }{ 32 }\))² x l = \(\frac { 7000π }{ 9 }\)
l = \(\frac{7000 \times 32 \times 32}{9}\)
= 79644.44 cm = 7964.4 m
∴ Length of the wire = 7964.4 m

These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1

Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:
Volume of one cube = 64 cm³
Let edge of one cube = a
Volume of the cube = (edge)³
a³ = 64 ⇒ a = 4 cm
Similarly, edge of the another cube = 4 cm.
Now, both cubes are joined together and a cuboid is formed as shown in the figure.
Now, length of the cuboid (l) = 8 cm
breadth of the cuboid (b) = 4 cm
height of the cuboid (h) = 4 cm
Surface area of the cuboid so formed = 2 (lb + bh + hl)
= 2(8 x 4 + 4 x 4 + 4 x 8)
= 2(32 + 16 + 32) = 160 cm²

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution:
Given: diameter of the hemisphere = 14 cm
Radius = \(\frac { 14 }{ 2 }\) = 7 cm
Curved surface area of the hemisphere = 2πr² = 2 x \(\frac { 22 }{ 7 }\) x 7 x 7 cm²
= 14 x 22 cm² = 308 cm²
Here, total height of the vessel = 13 cm
Height of the cylinder = Total height – Height of the hemisphere = 13 cm – 7 cm = 6 cm
and radius of the cylinder = radius of the hemisphere = 7 cm
Inner surface area of the cylinder = 2πrh = 2 x \(\frac { 22 }{ 7 }\) x 7 x 6
= 2 x 22 x 6 = 264 cm²
Inner surface area of the vessel = Inner surface area of the cylinder + curved surface area of the hemisphere
= 264 cm² + 308 cm² = 572 cm².

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius of the hemispheree, r = 3.5 cm
Radius of the base of the cone, r = 5 cm
Height of the cone, h = (15.5 – 3.5) cm = 12 cm

∴ total surface area of the toy = (curved surface area of the hemisphere + (curved surface area of the cone)

Hence, the total surface area of the toy is 214.5 cm².

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:
Greatest diameter of hemisphere is = 7 cm (side of cube)
Radius of hemisphere \(\frac { 7 }{ 2 }\) = cm.
Surface area of solid

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Diameter of hemisphere = l
So radius of hemisphere = \(\frac { l }{ 2 }\)

Surface area of cube = 6 (edge)² – 6l².
Surface area of remaining solid = Total surface area of cube-back area of hemisphere + Curved surface area of hemisphere.

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:
Radius of the cylinder = radius of the
hemispherical ends = r = \(\frac { 5 }{ 2 }\) mm
Height of the cylinder = h = 14 – 2 x \(\frac { 5 }{ 2 }\) = 9 mm
∴ Total surface area = curved surface area of the cylinder + surface area of two hemispherical ends
= (2πrh + 2 x πr²) mm²
= 2πr (h + 2r) cm²
= 2 x \(\frac { 22 }{ 7 }\) x \(\frac { 5 }{ 2 }\)(9 + 2 x \(\frac { 5 }{ 2 }\))
= \(\frac { 110 }{ 7 }\)(9 + 5)
⇒ = \(\frac { 110 }{ 7 }\) x 14
∴ Surface area = 220mm²

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution:
Area of the canvas used = Curved surface area of the cylinder + (Curved surface area of the cone)

Cost of the canvas of the tent at the rate ₹ 500 per m².
= ₹ (44 x 500) = ₹ 22,000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
Solution:
Height of solid cylinder is 2.4 cm
Diameter of cylinder is 1.4 cm
∴ Radius = \(\frac { 1.4 }{ 2 }\) cm = 0.7 cm

Total surface area of remaining solid
= 2πrh + πrl² + nrl … (i)
where πrl curved surface area of cone

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:
Radius of the cylinder = Radius of the hemispherical ends = r = 3.5 cm
Height of the cylinder = 10 cm.
Total surface area of the article
= Curved surface area of the cylinder + Surface area of the two hemispherical ends.
= (2πrh + 2 x 2πr²) cm².
= 2πr (h + 2r) cm².
= 2 x \(\frac { 22 }{ 7 }\) x 3.5 (10 + 7)
= 22 (14) = 374 cm².

These NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.2

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct a pair of tangents to the circle and measure their lengths.

Solution:
Given: A circle of radius 6 cm and a point P is located outside the circle at a distance of 10 cm from the centre O of the circle, i.e.. OP = 10 cm.

Required : To draw two tangents to the circle from P and to measure their lengths also.

Steps of Construction :
(i) Draw a circle of radius 6 cm arid let O be the centre of it
(ii) Produce O to X and from OX cut OP = 10 cm.
(iii) Draw a perpendicular bisector ST of OP to intersect OP at M.
(iv) Taking M as centre and MO and MP as radius draw another circle so as to intersect the previously drawn circle at Q and R.
(v) Join PQ and PR.
Hence. PQ and PR and the required tangents.
(vi) Then measure the tangents FQ and PR. You will find that length of tangents PQ and PR will be 8 cm each.

Calculation:

Question 2.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation:
Solution:
Steps of Construction:

(i) Draw a circle of radius OP = 6 cm.
(ii) Draw a consentric circle of radius ON = 4 cm
(iii) Draw the ⊥ bisector of OP which and it at M.
(iv) Taking M as the centre and radius equal to NQ.
Draw a circle which intersect the inner circle at P and Q.

Justification:
In dotted circle PO is the diameter and ∠PRQ = 90° (Angle in semi-circle)
∴ OQ ⊥ PQ
Thus, PQ will be the tangent because we know that angle between radius (OQ) and tangent (PQ) is 90°.

Question 3.
Draw a circle with radius 3 cm. Take ; two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre, Draw tangent to the circle from these two points P and Q.
Solution:
Steps of Construction:

(i) Draw a circle with radius 3 cm and with centre O.
(ii) Taking two points P and Q ouside the circle such that OP = 7 cm and OQ = 7 cm and bisect OP and OQuestion Let the mid-points of OP be M and OQ be N.
(iii) Taking M and N as centres draw two circles as MO and NO are the radius respectively. The two circles cut the main circles at the points R, S, T and U.
(iv) Join P to R and S and Join Q to T and U, Then PR, PS and NT, NU are the required tangents of the circle,

Justification:
In left circle of is the diameter
∴ ∠PRO = 90° (angle in a semi-circle)
∴ OR x PR
Thus, PR is tangent (∵ Angle between radius and tangent is 90°)
Similarly PS, QT and QU are tangents.

Question 4.
Draw a pair of tangents to a circle of ‘ radius 5 cm which are inclined to each other at an angle of 60°.
Solution:
Steps in Construction:
(i) Draw a circle of radius 5 cm.
(ii) Draw a diameter QM.
(iii) At Q draw a perpendicular QA.

(iv) At O make the angle of 60° so that ∠QOR = 120°
(v) At R draw a perpendicular RB.
(vi) Both the perpendicular intersect at P.
(vii) Thus, PQ and PR are the required tangents which are inclined at 60°.
Justification:
In PQOR
∠P + ∠Q + ∠O + ∠R = 360°
or ∠P + 90° + 120° + 90° = 360°
or ∠P + 300° = 360°
or ∠P = 360° – 300° = 60°

Question 5.
Draw a line segment AB of length 8 cm. Taking A as centre draw a circle of radius 4 cm and taking B as centre draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:
Steps of Construction:

(i) Draw a line segment AB = 8 cm and bisect it. Let M be the mid-point of AB.
(ii) Taking A as centre draw a circle with j radius 4 cm and taking B as centre draw another circle with radius 3 cm and taking M as centre draw’ circle with MA as radius it cuts both circles at the points P, Q and R. S.
(iii) Join A to R and S. Join B to P and Q.
Therefore AR, AS and BP, BQ are the required tangents.

Justification : In middle class circle AB is the diameter
∠APB = 90° and ∠APB = 90° (Angie in a semi-circle)
∴ AP ⊥ BP and BR ⊥ AR (Angle between radius and tangent is 90°)
Thus, BP and AR are tangents similarly AS and BQ are tangents.

Question 6.
Let ABC be a right triangle in which AB = 6 cm BC = 8 cm and ∠B = 90°, BD is the perpendicular from B on AC, The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Steps of Construction:
(i) Construct a AABC in which BC = 8 cm, AB = 6 cm and ∠B = 90°
(ii) From the point B draw a perpendicular BP which cut the AC at D.

(iii) Now draw the per perpendicular bisector of CD and BC which intersect at M.
∴ BM ⊥ AB (Angie between radius and tangent is 90°)
Thus, AB is tangent.
AQ is also tangent which is equal to AB because tangents drawn from the outer point are equal.

Question 7.
Draw a circle with the help of bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:
Steps of Construction;
(i) Draw’ a circle of any radius.
(ii) Take a point P outside the circle.
(iii) Take PM = PA.

(iv) Now take BA as diameter and draw its perpendicular bisector which cut B at M.
(v) Taking M as centre draw the semi-circle.
(vi) From P draw a perpendicular which met the semi-circle at D.
(vii) Now taking PD as radius draw two arc which cut the circle at Q and R.
(viii) join P to Q and P to R.
(ix) Thus we get PQ and PR as the required tangents.
(x) Taking M as the centre draw a circle which passes through the B, C and D.
(xi) Taking AB = 6 cm in compass draw an arc which cut the circle at Q.
(xii) AB and AQ are the required tangents.

justification : Since ∠BDC = 90° (Angle in semi-circle)
∴ BM is-the radius.
and ∠B = 90°

These NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.3

Question 1.
Find the area of the shaded region in the given figure, if PQ = 24cm, PR = 7cm and O is the centre of the circle.
Solution:

Question 2.
Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 400.
Solution:
∠AOC = 40° (given)
Radius of the sector AOC = 14 cm

Question 3.
Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution:
ABCD is a square
Given: side of the square = 14 cm
∴ Area of the square = (side)² = (14)² = 196 cm²
Radius of the semicircle APD = \(\frac { 1 }{ 2 }\)(side of square) = \(\frac { 1 }{ 2 }\) x 14 = 7 cm
Area of the semicircle APD = \(\frac { 1 }{ 2 }\) πr² = \(\frac { 1 }{ 2 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 = 11 × 7 = 77cm²
Similarly, area of the semicircle BPC = 77 cm²
Total area of both the semicircles = 77 + 77 = 154 cm²
Area of the shaded region = Area of square – area of both semicircles
= 196 – 154 = 42 cm²

Question 4.
Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
Solution:
Area of the equilateral triangle OAB

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.

Solution:
In the figure, area of the shaded region = area of the square ABCD with side 4 cm – area of the circle of radius 1 cm, centrally placed – area of the four quarter circles of radii 1 cm each, placed | at each corner of the square ABCD.

∴ Area of the remaining portion of the square = \(\frac { 68 }{ 7 }\) cm²

Question 6.
In a circular table cover of the radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).

Solution:

Question 7.
In the figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Solution:
Edge of the square ABCD = 14 cm
Area of square ABCD = (14)² = 196 cm²
Here radius of each circle is 7 cm.

Area of shaded region = Area of square – Area of 4 sectors
196 – 154 = 42cm².

Question 8.
The given figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:
(i) the distance around the track along its inner edge.
(ii) the area of the track.

Solution:
(i) ABCD and EFGH are two rectangles and corner has two semicircles. The distance around the track along its inner edge

= 2 x length of rectangle + 2 x circumference of semicirlce

(ii) Area of the track = 2 [Area of rectangles] + 2 [Area of outer semicircle] – [Area of inner semicircles]

Question 9.
In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:
Since AB ⊥ CD,
therefore, ∠COB = ∠COA
∠DOA = ∠DOB = 90°
In the figure, area of the shaded region, area of the small circle of diameter (OD = OA = 7cm) + (area of the segment BMC with central angle BOC = 90c and radius (OB = OC = 7cm) + area of the segment ANC with central angle AOC = 90° and radius (OA = OC) = 7cm

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region. (Use π = 3.14 and \(\sqrt{3}\) = 1.73205).
Solution:
∆ABC is an equilateral triangle. About the angular points A, B and C as centres three circles half the length of the side of the triangle are described.
Area of equilateral ∆ABC(given) = 17320.5 cm²

Since ABC is an equilateral triangle, therefore ∠A = ∠B = ∠C = 60°
= 17320.5 cm²
There are 3 equal sectors in the figure of central angles 60° and radii 100 cm [∵ \(\frac { 200 }{ 2 }\) = 10 cm]
Then, area of the shaded region = area of ∆ABC – 3 (area of one sector of central angle 60° and radius 100 cm)
= 17320.5 cm² – [3 x \(\frac { 60° }{ 360° }\) x π x (100)²]
= 17320.5 cm² – (\(\frac { 3 }{ 6 }\) x 3.14 x 1000) cm²
= 17320.5 cm² – (5000 x 3.14) cm²
= (17320.5 – 15700) cm² = 1620.5 cm²
∴ area of the shaded region is 1620.5 cm²

Question 11.
On a square handkerchief, nine circular designs each of the radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.

Solution:
ABCD is a square
Edge of the square ABCD = 3 x diameter of circle
= 3 x 4 = 42 cm
Area of square ABCD = (42)²
= 1764 cm²
Radius of one circle = 7 cm
Area of one circle = πr²
= \(\frac { 22 }{ 7 }\) x (7)²
= 154 cm²
Area of nine circles = 9 x Area of one circle
= 9 x 154 = 1386 cm²
Area of remaining portion = Area of square – Area of circle
= 1764 – 1386 = 378 cm²

Question 12.
In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region.
Solution:
(i) Let r be the radius of the circle and r = 35 cm \(\frac { 35 }{ 10 }\) = \(\frac { 7 }{ 2 }\) cm
Radius of the circle = Radius of quadrant of circle
= r = \(\frac { 7 }{ 2 }\) cm
Area of the quadrant of the circle = \(\frac { 1 }{ 4 }\)πr²

Since it is quadrant of circle, therfore the central angle of it θ = ∠AOB = 90°

(ii) Area of the shaded portion = Area of the sector for area of the quadrant of circle) – area of ∆AOD

Question 13.
In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)
Solution:
OABC is a square and OA is 20 cm. Join OB. Now we have a triangle QAB.
By Pythagoras theorem
(OB)² = (OA)² + (AB)²
(OB)² = (20)² + (20)²
= 400 + 400

Area of square
OABC = (20)² = 400 cm².

Area of shaded region = Area of sector OPBQ – Area of square OABC
= 628 – 400
= 228 cm²

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠AOB = 30°, find the area of the shaded region.
Solution:

Question 15.
In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
Let us find BC By Pythagoras theorem BC² = AB² + AC²
BC² = 14² + 14²
BC = 14\(\sqrt{2}\) cm
Required Area = Area of semicircle BRC – [Area of quadrant – area of ∆ABC)
Required Area = Area BCQB – (Area BACQB – Area of ∆ABC)

Question 16.
Calculate the area of the designed region in the figure common between the two quadrants of the circles of the radius 8 cm each.

Solution:

Area of square = (8)² = 64 cm²
Area of shaded region = Area of both sectors – Area of square
= \(\frac { 704 }{ 7 }\) – 64
= \(\frac { 704-448 }{ 7 }\) = \(\frac { 256 }{ 7 }\) cm²