Surface Areas and Volumes: 9 Math


mathematics Class Nine

NCERT Exercise 13.7 q5-9: 9th math

Surface Areas And Volumes Class nine Math NCERT Exercise 13.7  Question (5) A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Solution

Given, Diameter of the conical pit = 3.5 m

Therefore, radius of the given conical pit = 3.5/2 = 1.75 m

And, Depth of the conical pit = 12 m

Thus, Capacity of the given conical pit in kiloliters = ?

Here, Capacity of the given conical pit = Volume of the given conical pit

We know that Volume of a cone = 1/3 ℿ r2 h

Therefore, volume of the given conical pit

= `1/3xx22/7` × (1.75 m)2 × 12 m

= `1/3xx22/7` × 1.75 m × 1.75 m × 12 m

= 22 × 0.25 m × 1.75 m × 4 m

= 22 × 1.75 m3

= 38.5 m3

Thus, volume of the given pit = 38.5 m3

Now, since 1 m3 = 1 kilolitre

Therefore, 38.5 m3 = 38.5 kilolitre

Thus, volume of the given pit = 38.5 kilolitre Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.7  Question (6) The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find

(i) Height of the cone

(ii) Slant height of the cone

(iii) Curved surface area of the cone

Solution

Given, Volume of the cone = 9856 cm3

And, diameter of the base of the cone = 28 cm

Therefore, radius of the cone = 28/2 = 14 cm

9 math surface areas and volumes ncert exercise 13.7 answer of question 69 math surface areas and volumes ncert exercise 13.7 answer of question 6

(i) Height of the cone

We know that Volume of a cone = 1/3 ℿ r2 h

Therefore, volume of the given cone

= 9856 cm3 = `1/3xx22/7` × 14 cm × 14 cm × h

⇒ 9856 cm3 × 3 = 22 × 2 cm × 14 cm × h

⇒ 9856 cm3 × 3 = 22 × 28 cm2 × h

`=> h = (9856 cm^3xx3)/(22xx28cn^2)`

⇒ h = 16 cm × 3

⇒ h = 48 cm

Thus, height of the given cone = 48 cm

(ii) Slant height of the cone

Now, we know that,

(Slant height of a cone)2 = (radius)2 + (height)2

⇒ [Slant height (`l`)]2 = (14 cm)2 + (48 cm)2

⇒ `l` = 196 cm2 + 2304 cm2

⇒ `l`2 = 2500 cm2

`=>l = sqrt(2500cm^2)`

⇒ `l` = 50 cm

Thus, slant height of the given cone = 50 cm Answer

(iii) Curved surface area of the cone

We know that, Curved surface area of a cone `=\ pi\ r\ l`

Here, radius of the cone = 14 cm

(∵ As per question diameter = 28 cm)

And, slant height of the cone = 50 cm

[As calculated in section (ii) of this question]

Thus, curved surface area of the given cone

= `22/7` × 14 cm × 50 cm

= 22 × 2 cm × 50 cm

= 22 × 100 cm2

= 2200 cm2

Thus, curved surface area of the given cone = 2200 cm2 Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.7  Question (7) A right triangle ABC with sides 5 cm, 12 cm, and 13 cm is revolved about the side 12 cm, find the volume of the solid so obtained.

Solution

Given,

The sides of a right angle triangle = 5 cm, 12 cm and 13 cm

Now, we know that, in a right angle triangle, the hypotenuse is the biggest side.

Thus, here, hypotenuse = 13 cm, and base = 12 cm and height = 5 cm

Now, as per question, the given right angle triangle is revolved about the side 12 cm

Thus, by revolving the given right angle triangle, a cone is formed.

Thus, Radius of the base of the cone (r) = 5 cm

And, height of the cone (h) = 12 cm

And, slant height of the cone (`l`) = 13 cm

Thus, volume of the solid so obtained, i.e. of the cone = ?

We know that Volume of a cone = 1/3 ℿ r2 h

Thus, volume of the cone formed by revolving the right angled triangle ABC

= `1/3` × ℿ × (5 cm)2 × 12 cm

= ℿ × 5 cm × 5 cm × 4 cm

= ℿ 100 cm3 - - - - (i)

= `22/7` × 100 cm3

= 314.285 cm3

Thus, volume of the cone formed = ℿ 100 cm3 or 314.285 cm3 Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.7  Question (8) If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volume of the two solids obtained in Question 7 and 8.

Solution

Given, As in question (7) the sides of a right angled triangle ABC are 5 cm, 12 cm and 13 cm.

By revolving the given triangle about the side 5 cm, a solid cone is formed.

Then the volume of the solid cone = ?

And, the ratio of the volume of the solid obtained in question 7 and in this question = ?

Here, since given right angled triangle revolves about the side 5 cm

Therefore, the height of the cone so formed = 5 cm

And, radius of cone = 12 cm

Now, we know that, Volume of a cone `=1/3\ pi\ r^2\ h`

Thus, volume of the given cone = `1/3` ℿ × (12 cm)2 × 5 cm

= `1/3` ℿ 12 cm × 12 cm × 5 cm

= ℿ 4 cm × 12 cm × 5 cm

= ℿ 240 cm3 - - - - (i)

`=22/7xx240` cm3

= 754.285 cm3

Thus, volume of the solid cone formed by revolving the given right angled triangle = ℿ 240 cm3 or 754.285 cm3

Now, volume of the solid cone formed in question (7) = ℿ 100 cm3

Thus, ratio of volume of cone formed in question (7) and volume of cone formed in question (8)

`=(pi\ 100 cm^3)/(pi\ 240\ cm^3)`

= 100/240 = 5/12

= 5: 12

Thus, volume of the solid cone formed by revolving the given right angled triangle = ℿ 240 cm3 or 754.285 cm3 and ratio of cone formed in question 7 and cone formed in this question = 5:12 Answer

i.e. volume = ℿ 240 cm3 or 754.285 cm3 and ratio = 5:12 Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.7  Question (9) 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.

Solution

Given, Diameter of the conical heap = 10.5 m

Therefore, radius of the conical heap = 10.5/2 m = 5.25 m

And, Height of the conical heap heap = 3 m

Therefore, volume of the conical heap = ?

And, area of canvas to cover the heap of wheat from rain = ?

Now, we know that volume of a cone = 1/3 ℿ r2 h

Thus, volume of the given conical heap of wheat

= `1/3xx22/7` × 5.25 m × 5.25 m × 3 m

= 22 × 0.75 m × 5.25 m × 1 m

= 22 × 3.7375 m3

= 86.625 m3

Thus, volume of the conical heap of wheat = 86.625 m3

Now, area of canvas required = curved surface area of cone

And, we know that, Curved Surface Area of a Cone `=pi\ r\ l`

Here, we have radius (r) of the conical heap = 5.25 m

And, height (h) of the conical heap = 3 m

Here, we need slant height of the conical heap in order to calculate its curved surface area.

Now, we know that,

[Slant height of a cone (`l`)]2 =(radius)2 + (height)2

⇒ (slant height)2 = (3 m)2 + (5.25 m)2

⇒ (slant height)2 = 9 m 2 + 27.5625 m2

⇒ (slant height)2 = 36.5625 m2

⇒ Slant height `=sqrt(36.5625 m^2)`

⇒ Slant height (`l`) = 6.05 m

Now since, curved surface area of a cone = ℿ r `l`

Therefore, curved surface area of the given conical heap of wheat

= `22/7` × 5.25 m × 6.05 m

= 22 × 0.75 m × 6.05 m

= 22 × 4.5375 m2

= 99.825 m2

Thus, volume of the conical heap of wheat = 86.625 m3 and area of canvas required to cover that heap = 99.825 m2 Answer

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