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here we are providing mensuration solutions for class 10.

Exercise 1.

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

Solution:

Given volume of cube=side³ =64 cm³

Then side = ∛(64) =4 cm.

According to problem two cubes joined end to end and formed a cuboid.

Then length of resulting cuboid =4cm+4cm=8cm.

Breadth=b=4 cm and height=h =4 cm.

⇒Surface area Surface area of resulting cuboid =2(lb+bh+lh)

=2[(8 ×4)+(4 ×4)+(4 ×8)]

=2(32+16+32)

=2 × 80=160 cm²

∴ Surface area of resulting cuboid=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.

Solution:

Solution :

Given diameter of hemisphere= d=14 cm.

Then radius of Hemisphere=radius of cylinder =ᵈ/₂=r=7 cm.

∵ Given that hemisphere is mounted on cylinder.

Height of cylinder =h=Total height of the vessel – hemisphere height(radius)=

h=13-7=6 cm.

Inner surface area of the vessel=2πrh+2πr²

=2πr(h+r)

=2 ×²²/₇ ×7(6+7)

=44 ×13= 572 cm²

∴ Inner surface area of the vessel= 572 cm².

3 A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere same of radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution :

Given height of the toy =15.5 cm.

Toy is in the form of cone mounted on hemisphere.

∴ Radius of cone=Radius of hemisphere=r=3.5 cm.

Height of cone =h=Height of the toy -height(radius) of hemisphere

=15.5-3.5=12cm.

Slant height of the cone=l=

$l=\sqrt{r\xb2+h\xb2}=\sqrt{(3.5)\xb2+\left(12\right)\xb2}=\sqrt{12.25+144}$

$=\sqrt{156.25}=12.5$

The total surface area of the toy=CSA of the cone +CSA of hemisphere

=πrl+2πr²=πr(l+2r)

=²²/₇ ×3.5 ×[12.5+2(3.5)]

=11 ×19.5=214.5 cm².

∴ Total surface area of the toy=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.

Solution:

Given side of cubical block=7 cm.

The maximum diameter of hemisphere=Side of cubical block=a=7cm.

Radius of Hemisphere=r=⁷/₂=3.5 cm.

The surface area of the solid=Surface area of cubical block+CSA of hemisphere- Area of base of hemisphere.

=6a²+2πr²-πr²

=6a²+πr²

$=6\left(7\right)\xb2+\frac{22}{7}\times (3.5)\xb2$

=294+38.5

=332.5 cm²

∴ The surface area of the solid =332.5 cm².

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 :

Given that diameter of Hemisphere=edge of cube =l.

Radius of Hemisphere =r =½(l).

The surface area of the remaining solid =TSA of cubical block+CSA of hemisphere-Area of base of Hemisphere.

=6 l²+2πr²-πr²

=6 l²+πr²

$=6l\xb2+\pi {\left(\frac{l}{2}\right)}^{2}=\left(6+\frac{\pi}{4}\right)l\xb2$

Surface area of the remaining solid$=\left(6+\frac{\pi}{4}\right)l\xb2$

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 5mm. find the surface area.

Solution :

Given diameter of cylinder of capsule=5mm.

From figure

Radius of capsule (cylindrical part)=Radius of Hemisphere=r=⁵/₂ =2.5 mm.

Given Length of entire capsule=14mm.

Length of cylindrical part =h=14-(2.5+2.5)=9mm.

Surface area of capsule=

CSA of cylindrical part+ 2 ×CSA of Hemispherical part.

=2πrh+2(2πr²)

=2πr(h+2r)

=2π ×2.5 [9+2(2.5)]

=5π(14)=70π

$=70\times \frac{22}{7}=220mm\xb2$

∴ The surface area of capsule=220 mm².

7 A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical partt are 2.1 m and 4m 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)

Solution :

Given height of cylinder=h=2.1m, diameter of cylinder=4m.

Then radius of the cylindrical part =Radius of conical part=r=⁴/₂ =2m

∵ Cylinder is surmounted by conical top.

Given that slant height of conical part=l=2.8m.

Area of Canvas used =CSA of conical part +CSA of cylindrical part.

=πrl+2πrh=πr(l+2h)

=π (2)[2.8+2(2.1)]

=2π ×7.

$=2\times \frac{22}{7}\times 7=44m\xb2.$

The cost of 1m² Canvas= Rs 500.

∴The cost of 44m² Canvas=44 ×500= Rs 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.

Solution;

Given that height of cylinder= Height of cone =h=2.4 cm.

Diameter of cylinder=Diameter of cone =1.4 cm.

Then radius of cylinder= Radius of the cone=r =0.7 m.

Slant height of cylindrical part =l=

$=\sqrt{r\xb2+h\xb2}=\sqrt{(0.7)\xb2+(2.4)\xb2}=\sqrt{0.49+5.76}$

$=\sqrt{6.25}=2.5cm$

l=2.5cm

The total surface area of remaining solid=

CSA of cylinder+CSA of conical part+Area of base of cylinder.

=2πrh+πrl+πr²

=πr(2h+l+r)

=π ×0.7[2(2.4)+2.5+0.7]

=π×0.7(8)

=5.6 ×π

$=5.6\times \frac{22}{7}=0.8\times 22=17.6cm\xb2$

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 :

Given that height of cylinder=h=10m.

Radius of cylindrical part=Radius of Hemispherical part=r=3.5 cm.

The total surface area of article =

CSA of cylindrical part+CSA of two Hemispherical part.

=2πrh+2(2πr²)

=2πr(h+2r)

=2π ×3.5 [10+2(3.5)]

=7π ×17

=119π

$=119\times \frac{22}{7}=17\times 22=374cm\xb2.$

∴ The total surface area of the article=374 cm².