Exercise 12.1 Practice
Surface Areas of Combination of Solids
Q1: Joining Cubes
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Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Step 1: Volume of cube = $a^3 = 125 \Rightarrow a = 5$ cm.
Step 2: When joined end to end, dimensions of cuboid are:
Length $L = 5 + 5 = 10$ cm
Breadth $B = 5$ cm, Height $H = 5$ cm.
Length $L = 5 + 5 = 10$ cm
Breadth $B = 5$ cm, Height $H = 5$ cm.
Step 3: TSA = $2(LB + BH + HL)$
$= 2(10\times5 + 5\times5 + 5\times10)$
$= 2(50 + 25 + 50) = 2(125) = 250$.
$= 2(10\times5 + 5\times5 + 5\times10)$
$= 2(50 + 25 + 50) = 2(125) = 250$.
Surface Area is 250 cm².
Q2: Vessel Shape
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A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 28 cm and the total height of the vessel is 26 cm. Find the inner surface area of the vessel.
Step 1: Radius $r = 28/2 = 14$ cm.
Total height = 26 cm. Height of cylinder $h = 26 - 14 = 12$ cm.
Total height = 26 cm. Height of cylinder $h = 26 - 14 = 12$ cm.
Step 2: Inner Surface Area = CSA of Cylinder + CSA of Hemisphere.
$= 2\pi rh + 2\pi r^2 = 2\pi r(h + r)$.
$= 2\pi rh + 2\pi r^2 = 2\pi r(h + r)$.
Step 3: Calculate:
$= 2 \times \frac{22}{7} \times 14 \times (12 + 14)$
$= 88 \times 26 = 2288$.
$= 2 \times \frac{22}{7} \times 14 \times (12 + 14)$
$= 88 \times 26 = 2288$.
Inner Surface Area is 2288 cm².
Q3: Toy Surface Area
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A toy is in the form of a cone of radius 7 cm mounted on a hemisphere of same radius. The total height of the toy is 31 cm. Find the total surface area of the toy.
Step 1: Radius $r = 7$ cm. Total H = 31 cm.
Height of cone $h = 31 - 7 = 24$ cm.
Height of cone $h = 31 - 7 = 24$ cm.
Step 2: Slant height $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ cm.
Step 3: TSA = CSA of Cone + CSA of Hemisphere.
$= \pi rl + 2\pi r^2 = \pi r(l + 2r)$.
$= \pi rl + 2\pi r^2 = \pi r(l + 2r)$.
$= \frac{22}{7} \times 7 \times (25 + 14) = 22 \times 39 = 858$.
Total Surface Area is 858 cm².
Q4: Block & Hemisphere
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A cubical block of side 14 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Step 1: Greatest diameter = Side of cube = 14 cm.
Radius $r = 7$ cm.
Radius $r = 7$ cm.
Step 2: SA = TSA of Cube - Area of Base of Hemisphere + CSA of Hemisphere.
$= 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$.
$= 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$.
$= 6(14)^2 + \frac{22}{7}(7)^2 = 6(196) + 154$
$= 1176 + 154 = 1330$.
$= 1176 + 154 = 1330$.
Max Diameter: 14 cm; Surface Area: 1330 cm².
Q5: Depression in Cube
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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.
Edge of cube = $l$. Radius of hemisphere $r = l/2$.
Formula: SA = TSA Cube - Area of Circle + CSA Hemisphere.
$= 6l^2 - \pi(l/2)^2 + 2\pi(l/2)^2$
$= 6l^2 + \pi(l/2)^2 = 6l^2 + \frac{\pi l^2}{4}$.
$= 6l^2 - \pi(l/2)^2 + 2\pi(l/2)^2$
$= 6l^2 + \pi(l/2)^2 = 6l^2 + \frac{\pi l^2}{4}$.
Taking $l^2/4$ common:
Surface Area = $\frac{l^2}{4}(24 + \pi)$ units².
Q6: Medicine Capsule
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A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 20 mm and the diameter of the capsule is 7 mm. Find its surface area.
Diameter = 7 mm $\Rightarrow r = 3.5$ mm.
Total Length = 20 mm. Length of cylindrical part $h = 20 - 3.5 - 3.5 = 13$ mm.
Total Length = 20 mm. Length of cylindrical part $h = 20 - 3.5 - 3.5 = 13$ mm.
Formula: CSA Cylinder + 2(CSA Hemisphere)
$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$.
$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$.
$= 2 \times \frac{22}{7} \times 3.5 \times (13 + 7)$
$= 22 \times 20 = 440$.
$= 22 \times 20 = 440$.
Surface Area is 440 mm².
Q7: Tent on Canvas
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A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 4 m and 6 m respectively, and the slant height of the top is 5 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas at the rate of Rs 600 per m².
Cylinder: $h=4, r=3$. Cone: $l=5, r=3$.
Canvas Area = CSA Cylinder + CSA Cone
$= 2\pi rh + \pi rl = \pi r(2h + l)$.
$= 2\pi rh + \pi rl = \pi r(2h + l)$.
$= \frac{22}{7} \times 3 \times (2(4) + 5) = \frac{66}{7} \times 13 = \frac{858}{7} \approx 122.57$ m².
Cost = Area $\times$ Rate = $\frac{858}{7} \times 600 = \text{Rs } 73542.8$ (approx).
(Using exact fraction: $122.57 \times 600 = 73542$).
(Using exact fraction: $122.57 \times 600 = 73542$).
Area: 122.57 m²; Cost: Rs 73,543.
Q8: Hollowed Cylinder
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From a solid cylinder whose height is 12 cm and diameter 10 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.
$h = 12$ cm, $r = 5$ cm.
Slant height $l = \sqrt{12^2 + 5^2} = \sqrt{144+25} = 13$ cm.
TSA = CSA Cylinder + Area of Base + CSA Cone (Cavity)
$= 2\pi rh + \pi r^2 + \pi rl = \pi r(2h + r + l)$.
$= 2\pi rh + \pi r^2 + \pi rl = \pi r(2h + r + l)$.
$= \frac{22}{7} \times 5 \times (24 + 5 + 13) = \frac{110}{7} \times 42 = 110 \times 6 = 660$.
Total Surface Area is 660 cm².
Q9: Wooden Article
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A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 20 cm, and its base is of radius 7 cm, find the total surface area of the article.
$h = 20$ cm, $r = 7$ cm.
TSA = CSA Cylinder + 2(CSA Hemisphere)
$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$.
$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$.
$= 2 \times \frac{22}{7} \times 7 \times (20 + 14) = 44 \times 34 = 1496$.
Total Surface Area is 1496 cm².