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NCERT Solutions for Class 10 Maths Exercise 12.1 (NEW SESSION)
Solve the followings Questions.
Unless stated otherwise, use π =22/7
1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Answer:
Let l be length of each cube
Volume of each cube = l3 = 64 cm3
⇒ l3 = 64 cm3
⇒ l = 4 cm
If, we join 2 cubes each having side equal to 4 cm, we get a cuboid
Length of cuboid = L = 8 cm
Height of cuboid = H = 4 cm
Breadth of cuboid = B = 4 cm
We know that Surface Area of Cuboid = 2(L.B + B.H + H.L)
= 2((8 x 4) + (4 x 4) + (4 x 8))
= 2(32 + 16 + 32)
= 2(80)
= 160 cm2
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.
Answer:
Height of the cylindrical portion 13 − 7 = 6 cm.
Area of a Curved surface of cylindrical portion is,
= 2πrh
= 2 × 22/7 × 7 × 6
= 264 cm2
Area of a curved of hemispherical portion is
=2πr2
=2 × 22/7 × 7 × 7
= 308 cm2
∴ total surface area is = 308 + 264 = 572 cm2.
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.
Answer:
Total Surface area of toy = Surface area of hemisphere without base + Surface area of cone without base
Total height = 15.5 = r + h
h = 15.5 − 3.5 = 12 cm
A = 2πr2 + πr√(r2 + (15.5 − r)2)
A = 2π × 3.52 + π × 3.5√(3.52 + 122)
A = 214.5 cm2
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.
Answer:
(i) A cubical block of side 7 cm is surrounded by a hemisphere.
∴ Diameter of hemisphere is = 7 cm.
Radius of hemisphere, r = 3.5 cm
∴ Surface area of hemisphere = Curved surface area – Area of base of hemisphere
= 2πr2 − πr2
= 2 × 22/7 × (7/2)2 − 22/7 × (7/2)2
= 38.5
(ii) The curved surface area of square,
= 6 × I2
= 6 × (7)2
= 6 × 49
= 294 sq.cm.
∴ Total surface area of cubical block = Curved surface area of cubical block + Surface area of hemisphere
= 294 + 38.5
= 332.5 sq.cm.
5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter I of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Answer:
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.
Answer:
Surface area of capsule = 2πrh + 2 × 2πr2
Daimeter of capsule = 5mm
Radius of cylinder = Radius of hemisphere = 5/2 = 2.5mm
Height (h) =14 − 2.5 − 2.5 = 9mm
TSA = 2πr (h + 2r)
= 2 × 22/7 × 2.5 (9 + 5)
= 220 mm2
7. A tent is in the shape of a cylinder surmounted buy 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 Rs. 500 per m2. (Note that the base of the tent will not be covered with canvas.)
Answer:
Radius of the cylindrical base = 2m
Height of the cylindrical is = 2.1m
Slant Height of conical top is = 2.8
Curved Surface Area of cylindrical portion is,
= 2πrh
= 2π × 2 × 2.1
= 8.4πm2
Curved Surface Area of conical portion is,
= πrl
= π × 2 × 2.8
= 5.6πm2
Total Curved Surface Area
= 8.4π + 5.6π
= 14 × 22/7
= 44m2
Therefore,
Cost of canvas = Rate\times Surface Area
= 500 × 44
= 22000
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.
Answer:
Radius = 0.7 cm and height = 2.4 cm
Total surface area of structure = Curved surface area of cylinder + Area of top of cylinder + Curved surface area of cone
Curved surface area of cylinder = 2πrh
= 2π × 0.7 × 2.4
= 3.36 π cm2
Area of top:
= πr2
= π × 0.72
= 0.49π cm2
Slant height of cone can be calculated as follows:
l = √(h2+r2)
= √(2.42 + 0.72)
= √(5.76 + 0.49)
= √6.25 = 2.5 cm
Curved surface area of cone:
= πrl
= π × 0.7 × 2.5
= 1.75π cm2
Hence, remaining surface area of structure
= 3.36π + 0.49π + 1.75π
= 5.6π = 17.6 cm2
= 18 cm2 (approx)
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.
Answer:
Total surface area of the article = Lateral surface area of cylinder + 2 × Curved surface area of hemisphere
We know that the lateral surface are of a cylinder = 2πrh.
And the curved surface area of a hemisphere = 2πr2.
∴ TSA of article = 2πrh + 2 × 2πr2
⇒ 2 × 22/7 × 7/2 × 10 + 2 × 2 × 22/7 × (7/2)2
⇒ 22 × 10 + 22 × 7
⇒ 22 (10 + 7)
⇒ 22 × 17
∴ TSA = 374 cm2
Hence, the total surface area of the article is 374 cm2.
| CHAPTER NAME | OLD NCERT | NEW NCERT | |
| Real Numbers | EXERCISE 1.1 | ||
| EXERCISE 1.2 | 1.1 | CLICK HERE | |
| EXERCISE 1.3 | 1.2 | CLICK HERE | |
| EXERCISE 1.4 | |||
| Polynomials | EXERCISE 2.1 | 2.1 | CLICK HERE |
| EXERCISE 2.2 | 2.2 | CLICK HERE | |
| EXERCISE 2.3 | |||
| EXERCISE 2.4 | |||
| Pair of Linear Equations in Two Variables | EXERCISE 3.1 | ||
| EXERCISE 3.2 | 3.1 | CLICK HERE | |
| EXERCISE 3.3 | 3.2 | CLICK HERE | |
| EXERCISE 3.4 | 3.3 | CLICK HERE | |
| EXERCISE 3.5 | |||
| EXERCISE 3.6 | |||
| EXERCISE 3.7 | |||
| Quadratic Equations | EXERCISE 4.1 | 4.1 | CLICK HERE |
| EXERCISE 4.2 | 4.2 | CLICK HERE | |
| EXERCISE 4.3 | |||
| EXERCISE 4.4 | 4.3 | CLICK HERE | |
| Arithmetic Progressions | EXERCISE 5.1 | 5.1 | CLICK HERE |
| EXERCISE 5.2 | 5.2 | CLICK HERE | |
| EXERCISE 5.3 | 5.3 | CLICK HERE | |
| EXERCISE 5.4 | 5.4 (Optional) | CLICK HERE | |
| Triangles | EXERCISE 6.1 | 6.1 | CLICK HERE |
| EXERCISE 6.2 | 6.2 | CLICK HERE | |
| EXERCISE 6.3 | 6.3 | CLICK HERE | |
| EXERCISE 6.4 | |||
| EXERCISE 6.5 | |||
| EXERCISE 6.6 | |||
| Coordinate Geometry | EXERCISE 7.1 | 7.1 | CLICK HERE |
| EXERCISE 7.2 | 7.2 | CLICK HERE | |
| EXERCISE 7.3 | |||
| EXERCISE 7.4 | |||
| Introduction to Trigonometry | EXERCISE 8.1 | 8.1 | CLICK HERE |
| EXERCISE 8.2 | 8.2 | CLICK HERE | |
| EXERCISE 8.3 | |||
| EXERCISE 8.4 | 8.3 | CLICK HERE | |
| Some Applications of Trigonometry | EXERCISE 9.1 | 9.1 | CLICK HERE |
| Circles | EXERCISE 10.1 | 10.1 | CLICK HERE |
| EXERCISE 10.2 | 10.2 | CLICK HERE | |
| Construction | |||
| Areas Related to Circles | EXERCISE 12.1 | ||
| EXERCISE 12.2 | 11.1 | CLICK HERE | |
| EXERCISE 12.3 | |||
| Surface Areas and Volumes | EXERCISE 13.1 | 12.1 | CLICK HERE |
| EXERCISE 13.2 | 12.2 | CLICK HERE | |
| EXERCISE 13.3 | |||
| EXERCISE 13.4 | |||
| EXERCISE 13.5 | |||
| Statistics | EXERCISE 14.1 | 13.1 | CLICK HERE |
| EXERCISE 14.2 | 13.2 | CLICK HERE | |
| EXERCISE 14.3 | 13.3 | CLICK HERE | |
| EXERCISE 14.4 | |||
| Probability | EXERCISE 15.1 | 14.1 | CLICK HERE |
| EXERCISE 15.2 |