3 Marks Question

Question 13 Marks

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing is Rs. 20 per square metre. Find the

inside surface area of the dome.
Volume of the air inside the dome.

Answer

Inside surface area of the dome $={4989.60\over20}$ = 249.48 m²
Let the radius of the hemisphere be r m.
Inside surface area = 249.48 m²
$\Rightarrow$ 2$\pi $r² = 249.48
$\Rightarrow$ 2 × $22\over7$× r² = 249.48
$\Rightarrow$ r² = ${249.48\times7}\over2\times22$
$\Rightarrow$ r² = 39.69
$\Rightarrow$ r = $\sqrt{39.69}$
$\Rightarrow$ r = 6.3 m
$\therefore$ Volume of the air inside the dome = ${2\over3}\pi r^3$
$={2\over3}\times{22\over7}\times(6.3)^3$ = 523.9 m³

View full question & answer→

Question 23 Marks

The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction is the volume of the moon of the volume of the earth?

Answer

Let diameter of earth be $x$
$\therefore$ Radius of earth $\left( r \right)=\frac{x}{2}$
Now, Volume of earth = $\frac{4}{3}\pi {{r}^{3}}$ [$\because$ Earth is considered to be a sphere]
=$\frac{4}{3}\times \pi \times \frac{x}{2}\times \frac{x}{2}\times \frac{x}{2}$
=$\frac{1}{8}\times \frac{4}{3}\pi {{x}^{3}}$ ………..(i)
According to question, Diameter of moon = $\frac{1}{4}\times$ Diameter of earth= $\frac{1}{4}\times x=\frac{x}{4}$
$\therefore$ Radius of moon (R) = $\frac{x}{8}$
Now, Volume of Moon = $\frac{4}{3}\pi {{\text{R}}^{3}}$ [$\because$ Moon is considered to be a sphere]
=$\frac{4}{3}\times \pi \times \frac{x}{8}\times \frac{x}{8}\times \frac{x}{8}$
=$\frac{1}{512}\times \frac{4}{3}\pi {{x}^{3}}$
=$\frac{1}{64}\times \left[ \frac{1}{8}\times \frac{4}{3}\pi {{x}^{3}} \right]$
=$\frac{1}{64}\times$ Volume of Earth [From eq. (i)]
$\therefore$ Volume of moon is $\frac{1}{64}th$ the volume of earth.

View full question & answer→

Question 33 Marks

A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Answer

For heap of wheat
Diameter = 10.5 m
∴ Radius (r) = $10.5\over2$ cm = 5.25 m
Height (h) = 3 m
∴ Volume = ${1\over3}\pi r^2h$
$={1\over3}\times{22\over7}\times(5.25)^2\times3$
= 86.625 m³
Slant height, $l=\sqrt{r^2+h^2}$
$= \sqrt{(5.25)^2+(3)^2}=\sqrt{27.5625+9}$
$=\sqrt{36.5625}$ = 6.05 m
∴ Curved surface area =$\pi$rl
= $22\over7$ × 5.25 × 6.05 = 99.825 m²
∴ The area of the canvas required is 99.825 m²

View full question & answer→

Question 43 Marks

A right triangle ABC with its sides 5 cm, 12 cm, and 13 cm is revolved about the side 12 cm. Find the volume of the solid so formed. If the triangle ABC is revolved about side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained.

Answer

Let ABC be a right triangle with AB = 12 cm, BC = 5 cm and AC = 13 cm.
When this triangle is revolved about AB, it forms a right circular cone of radius = BC = 5 cm and height AB = 12 cm.

$\therefore$ V₁ = Volume of the solid formed = Volume of the cone of radius 5 cm and height 12 cm
$= \frac { 1 } { 3 } \times \pi$ $\times$ 5 $\times$ 5 $\times$ 12 cm³ = 100 $\pi$ cm³
When triangle ABC is revolved about BC, it forms a right circular cone of radius AB = 12 cm and height BC = 5 cm.
$\therefore$ V₂ = Volume of the solid formed
$\Rightarrow$ V₂ = Volume of the cone of radius 12 cm and height 5 cm
$\Rightarrow$ V₂ $= \frac { 1 } { 3 } \times \pi$ $\times$ 12 $\times$ 12 $\times$ 5 cm³ = 240 $\pi$ cm³
$\therefore$ Required ratio = V_{1 :}V₂ = 100$\pi$ : 240 $\pi$ = 5 : 12

View full question & answer→

Question 53 Marks

The diameter of the moon is approximately one fourth the diameter of the earth. Find the ratio of their surface areas.

Answer

Let diameter of Earth = x
$\therefore$ Radius of Earth $\left( r \right)=\frac{x}{2}$
$\therefore$ Surface area of Earth = $4\pi {{r}^{2}}$= $4\pi \times \frac{x}{2}\times \frac{x}{2}=\pi {{x}^{2}}$

Now, Diameter of Moon = $\frac{1}{4}th$ of diameter of Earth = $\frac{x}{4}$
$\therefore$ Radius of Moon $\left( r \right)$=$\frac{x}{8}$
Surface area of Moon = $4\pi {{r}^{2}}$= $4\pi \times \frac{x}{8}\times \frac{x}{8}=\frac{\pi {{x}^{2}}}{16}$
Now, Ratio = $\frac{\text{Surface area of Moon}}{\text{Surface area of Earth}}$=$\frac{\frac{\pi {{x}^{2}}}{16}}{\pi {{x}^{2}}}$=$\frac{\pi {{x}^{2}}}{16}\times \frac{1}{\pi {{x}^{2}}}$=$\frac{1}{16}$
$\therefore $ Required ratio = 1: 16

View full question & answer→

Question 63 Marks

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m², what will be the cost of painting all these cones? (Use $\pi$ = 3.14 and take $\sqrt {1.04}$ = 1.02)

Answer

Diameter of cone = 40 cm
$\Rightarrow$ Radius of cone (r) = $\frac{40}{2}$
= 20 cm
= $\frac{20}{100}$m
= 0.2 m
Height of cone (h) = 1 m
Slant height of cone $\left( l \right)=\sqrt{{{r}^{2}}+{{h}^{2}}}$
=$\sqrt{{{\left( 0.2 \right)}^{2}}+{{\left( 1 \right)}^{2}}}$
= $\sqrt{1.04}$ m
Curved surface area of cone = $\pi rl$
= $3.14\times 0.2\times \sqrt{1.04}$
= 0.64056 m²
$\because$ Cost of painting 1m² of a cone = Rs.12
$\therefore$ Cost of painting 0.64056m² of a cone = 12 $\times$ 0.64056 = Rs. 7.68672
$\therefore$ Cost of painting of 50 such cones = 50 $\times$ 7.68672 = Rs. 384.34 (approx.)

View full question & answer→

Question 73 Marks

A Joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Answer

Radius of cap (r) = 7 cm, Height of cap (h) = 24 cm
Slant height of the cone (l) = $\sqrt{{{r}^{2}}+{{h}^{2}}}$ = $\sqrt{{{\left( 7 \right)}^{2}}+{{\left( 24 \right)}^{2}}}$
= $\sqrt{49+576}$
= $\sqrt{625}$
= 25 cm
Area of sheet required to make a cap = CSA of cone = $\pi rl$ $\pi$rl
= $\frac{22}{7}\times 7\times 25$
$=\text{ }550\text{ }c{{m}^{2}}$=550 cm²
$\therefore$ Area of sheet required to make 10 caps =10 $\times$ 550 = 5500 cm²

View full question & answer→

Question 83 Marks

A conical tent is 10 m high and the radius of its base is 24 m. Find

Slant height of the tent.
Cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs. 70.

Answer

h = 10 m , r = 24 m
$l = { \sqrt{r^2+h^2}}={\sqrt{(24)^2+(10)^2}}$
$= {\sqrt{576+100}}={\sqrt{676}}$
= 26 m
$\therefore$ the slant height of the tent is 26 m.
Curved surface area of the tent =$\pi rl$
$={22\over7}\times24\times26\ m^2$
$\therefore$ Cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs. 70.
$={22\over7}\times24\times26\times70=Rs.137280$
∴ The cost of the canvas is Rs. 137280.

View full question & answer→

Question 93 Marks

Monica has a piece of Canvas whose area is 551 m². She uses it to have a conical tent made with a base radius of 7m. Assuming that all the stitching margins and wastage incurred while cutting, amounts to approximately 1 m². Find the volume of the tent that can be made with it.

Answer

Since the area of the canvas = 551 m² and area of the canvas lost in wastage is 1 m², therefore the area of canvas available for making the tent is (551 - 1) m² = 550 m²
Now, the surface area of the tent = 550 m² and the required base radius of the conical tent = 7 m
Therefore, curved surface area of tent = 550 m²
That is, $\pi rl$ = 550
or, $\frac { 22 } { 7 }$ $\times$ 7 $\times$ l = 550
l = $\frac { 550 } { 22 }$m = 25 m
Now, l² = r² + h²
Therefore, h = $\sqrt { l ^ { 2 } - r ^ { 2 } } = \sqrt { 25 ^ { 2 } - 7 ^ { 2 } } m = \sqrt { 625 - 49 m } = \sqrt { 576 m }$
= 24 m
So, the volume of the conical tent = $\frac { 1 } { 3 }$ $\pi$r²h = $\frac { 1 } { 3 } \times \frac { 22 } { 7 }$ $\times$ 7 $\times$ 7 $\times$ 24m³ = 1232 m³

View full question & answer→

Question 103 Marks

A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ₹ 5 per 100 cm².

Answer

Since only the rounded surface of the dome is to be painted, we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2$\pi r$
2 $\times \frac { 22 } { 7 } r$ = 17.6 m
So, the radius of the dome = 17.6 $\times \frac { 7 } { 2 \times 22 }$ m = 2.8 m
The curved surface area of the dome = 2$\pi r$²
= 2 $\times$ $\frac { 22 } { 7 }$ $\times$ 2.8 $\times$ 2.8 m²
= 49.28 m²
Now, the cost of painting 100 cm² is Rs. 5.
So, the cost of painting 1 m² = Rs. 500
Therefore, the cost of painting the whole dome
= Rs. 500 $\times$ 49.28
= Rs. 24640

View full question & answer→

Question 113 Marks

A corn cob (see Fig.), shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length as 20 cm. If each 1 cm² of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob?

Answer

Since the grains of corn are found on the curved surface of the corn cob.
So, Total number of grains on the corn cob = Curved surface area of the corn cob $\times$ Number of grains of corn on 1 cm²
Now, we will first find the curved surface area of the corn-cob.
We have, r = 2.1 and h = 20
Let l be the slant height of the conical corn cob. Then,
$l = \sqrt { r ^ { 2 } + h ^ { 2 } } = \sqrt { ( 2.1 ) ^ { 2 } + ( 20 ) ^ { 2 } } = \sqrt { 4.41 + 400 } = \sqrt { 404.41 }$ = 20.11
$\therefore$ Curved surface area of the corn cub = $\pi r l$
= $\frac { 22 } { 7 } \times$ 2.1 $\times$ 20.11 cm²
= 132.726 cm² = 132.73 cm²
Hence, Total number of grains on the corn cob = 132.73 $\times$ 4 = 530.92
So, there would be approximately 531 grains of corn on the cob.

View full question & answer→

Maths 3 Marks Question

Test yourself on this topic