Students can Download Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
Karnataka State Syllabus Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.1
Question 1.
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Answer:
If measurement of each side of each cube be l cm, then
(l)3 \(=\sqrt[3]{64}\)
∴ l = 4 cm.
Length of cubes if both joined together,
Length, l = 4 + 4 = 8 cm.
breadth, b = 4 cm.
height, h = 4 cm.
Surface area of the cuboid,
= 2lb + 2lh + 2bh
= 2(8 × 4) + 2(8 × 4) + 2(4 × 4)
= 2 × 32 + 2 × 32 + 2 × 16 = 64 + 64 + 32
= 160 sq.cm.
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.
Answer:
Diameter of hallow hemisphere = d = 14cm
∴ r = \(\frac{d}{2}=\frac{14}{2}\) = 7cm
Radius of hallow cylinder = 7 cm
Height of cylinder = 13 – 7 = 6cm
Inner surface area of the vessel = Inner surface area of cylinder + Inner surface area of hemisphere.
= 2πrh + 2πr2
= 22 × \(\frac{22}{7}\) × 7 × 6 + 2 × \(\frac{22}{7}\) × (7)2
= 44 × 6 + 44 × 7 = 264 + 308
= 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.
Answer:
Radius of cone = 3.5 cm
Radius of hemisphere = 3.5 cm
Height of the cone = 15.5 – 3.5 = 12 cm
Slant height of the cone =
Total surface area of the toy = curved surface area of hemisphere + curved surface area of cone.
= 2πr2 + πrl.
= πr (2r + l)
= \(\frac {22 }{7}\) × 3.5 (2 × 3.5 + 12.5)
= 11 × (7 + 12.5)
= 11 × 19.5
= 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.
Answer:
Greatest diameter of the hemisphere = side of the cubical block = 7 cm
∴ r = \(\frac{d}{2}=\frac{7}{2}\) = 3.5cm
Surface area of solid = TSA of cuboid – Area of the base of hemisphere + CSA of hemisphere.
= 6a2 – πr2 + 2πr2 = 6(7)2 + πr2
= 6 × 49 + \(\frac {22 }{7}\)(3.5)2
= 294 + \(\frac {22 }{7}\) × 3.5 × 3.5 = 294 + 38.5
= 332.5 cm2
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.
Answer:
Edge of cube = l
Diameter of hemisphere = l
radius = \(\frac{l}{2}\)
∴ Surface area of the remaining solid after cutting hemispherical depression
= TSA of cube – area of base of hemisphere + CSA of hemisphere.
Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 15.10). The length of the entire capsule is 14 ‘ mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer:
Cylinder:
radius = \(\frac{5}{2}\) mm and height of cylinder
= 14 – 5 = 9 mm
Hemisphere:
radius = r = \(\frac{5}{2}\)mm
Surface area of the capsule = CSA of cylinder + Surface area of two hemispheres
= 2πrh + 2 × 2πr2 = 2πr(h + 2r)
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\)(9 + 2 × \(\frac{5}{2}\))
= \(\frac{110}{7}\) × 14
= 220 mm2
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.)
Answer:
Cylinder:
diameter = 4m, radius = \(\frac{4}{2}\) = 2m
Height of cylinder = h = 2.1 m
Cone:
(ii) Radius of conical tent,
r = 2 m.
length, l = 2.8 m.
∴ Curved Surface area of cylinder = πrl
\(=\frac{22}{7} \times 2 \times 2.8\)
= 17.6 m2.
∴ Total area of the tent = 26.4 + 17.6
= 44m2
Cost of 1 aq.m. of canvas is Rs. 500.
Cost of 44 sq. m. canvas … ??
500 × 44
= Rs. 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.
Answer:
radius = \(\frac{1.4}{2}\) = 0.7 cm,
Height of cylinder = 2.4 cm and
Height of cone = 2.4 cm
Slant height = l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{5.76+0.49}=\sqrt{6.25}\) = 2.5cm
Total surface area of remaining solid = CSA of cylinder + CSA of cone + Area of upper circular base of cylinder.
= 2πrh + πrl + πr2
= πr(2h + l + r)
= \(\frac{22}{7}\) × 0.7(2 × 2.4 + 2.5 + 0.7)
= 2.2(4.8 + 2.5 + 0.7)
= 2.2 × 8.0 = 2.2 × 8
= 17.6cm2 ≈ 18cm2 (to the nearest cm2)
Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig.15.11. 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:
radius = r = 3.5 cm and
height = h = 10 cm
Total surface area of the article = CSA of cylinder + 2 × CSA of hemisphere
= 2πrh + 2 × 2πr2
= 2πr(h + 2r)
= 2 × \(\frac{22}{7}\) × 3.5 (10 + 2 × 3.5)
= 2 × 11.0 × (10 + 7)
=22 × 17
= 374 cm2.