{"id":602317,"date":"2022-05-13T11:16:28","date_gmt":"2022-05-13T11:16:28","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=602317"},"modified":"2022-05-14T06:17:19","modified_gmt":"2022-05-14T06:17:19","slug":"ml-aggarwal-solutions-for-class-8-maths-chapter-18-mensuration","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ml-aggarwal-solutions-for-class-8-maths-chapter-18-mensuration\/","title":{"rendered":"ML Aggarwal Solutions for Class 8 Maths Chapter 18- Mensuration"},"content":{"rendered":"\n

Class 8: Maths Chapter 18 solutions. Complete Class 8 Maths Chapter 18 Notes.<\/p>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 18- Mensuration<\/h2>\n\n\n\n

ML Aggarwal 8th Maths Chapter 18, Class 8 Maths Chapter 18 solutions<\/p>\n\n\n\n

Exercise 18.1<\/h4>\n\n\n\n

1. The length and breadth of a rectangular field are in the ratio 9 : 5. If the area of the field is 14580 square metres, find the cost of surrounding the field with a fence at the rate of \u20b93.25 per metre.
Solution:<\/strong><\/p>\n\n\n\n

Let the length of rectangle be 9x and its breadth be 5x<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Area = l \u00d7 b<\/p>\n\n\n\n

\u21d2 14580 = 9x \u00d7 5x<\/p>\n\n\n\n

45x2<\/sup> = 14580<\/p>\n\n\n\n

x2<\/sup> = 14580\/45 = 324<\/p>\n\n\n\n

x = \u221a324<\/p>\n\n\n\n

x = 18<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Length = 9 \u00d7 18 = 162 m and Breadth = 5 \u00d7 18 = 90 m<\/p>\n\n\n\n

Now, Perimeter = 2(l + b)<\/p>\n\n\n\n

= 2 (162 + 90) = 2(252)<\/p>\n\n\n\n

= 504 m.<\/p>\n\n\n\n

Therefore, cost for fencing the surrounding 504 m at the rate of \u20b93.25 per metre = \u20b9(504 \u00d7 3.25) = \u20b91638<\/p>\n\n\n\n


2. A rectangle is 16 m by 9 m. Find a side of the square whose area equals the area of the rectangle. By how much does the perimeter of the rectangle exceed the perimeter of the square?
Solution:<\/strong><\/p>\n\n\n\n

Area of rectangle = (16 \u00d7 9) m2<\/sup> = 144 m2<\/sup><\/p>\n\n\n\n

Given condition,<\/p>\n\n\n\n

Area of square = Area of rectangle<\/p>\n\n\n\n

\u2234 (Side)2<\/sup> = 144<\/p>\n\n\n\n

Side = \u221a144 = 12 m<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Perimeter of square = 4 \u00d7 side = 4 \u00d7 12 = 48 m<\/p>\n\n\n\n

Perimeter of rectangle = 2(l + b) = 2 (16 + 9) = 50 m<\/p>\n\n\n\n

Hence, difference in their perimeters = 50 \u2013 48 = 2 m<\/p>\n\n\n\n


3. Two adjacent sides of a parallelogram are 24 cm and 18 cm. If the distance between longer sides is 12 cm, find the distance between shorter sides.
Solution:<\/strong><\/p>\n\n\n\n

\"ML
<\/strong>
Let take 24 cm as the base of parallelogram, then its height is 12 cm.<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

Area of parallelogram = base \u00d7 height<\/p>\n\n\n\n

= 24 \u00d7 12 = 288 cm2<\/sup><\/p>\n\n\n\n

Let\u2019s consider d cm to be the distance between the shortest sides.<\/p>\n\n\n\n

\u2234 Area of parallelogram = (18 \u00d7 d) cm2<\/sup><\/p>\n\n\n\n

18 \u00d7 d = 288<\/p>\n\n\n\n

\u21d2 d = 288\/18 = 16 cm<\/p>\n\n\n\n

Therefore, the distance between the shorter sides is 16 cm.<\/p>\n\n\n\n


4. Rajesh has a square plot with the measurement as shown in the given figure. He wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of \u20b950 per m2<\/sup>.

Solution:<\/strong><\/p>\n\n\n\n

Given,
<\/strong>
Side of square plot = 24 m<\/p>\n\n\n\n

Length of house (l) = 18 m<\/p>\n\n\n\n

and breadth (b) = 12m<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of square plot = (24)2<\/sup> m2<\/sup> = (24 \u00d7 24) m2 <\/sup>= 576 m2<\/sup><\/p>\n\n\n\n

And,<\/p>\n\n\n\n

Area of house = 18 \u00d7 12 = 216 m2<\/sup><\/p>\n\n\n\n

Remaining area of the garden = 576 m2 <\/sup>\u2013 216 m2<\/sup> = 360 m2<\/sup><\/p>\n\n\n\n

The cost of developing the garden = \u20b950 per m2<\/sup><\/p>\n\n\n\n

Therefore, the total cost = \u20b950 \u00d7 360 = \u20b918000<\/p>\n\n\n\n


5. A flooring tile has a shape of a parallelogram whose base is 18 cm and the corresponding height is 6 cm. How many such tiles are required to cover a floor of area 540 m2<\/sup>? (If required you can split the tiles in whatever way you want to fill up the comers).
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"MLBase of the parallelogram-shaped flooring tile = 18 cm and its height = 6 cm<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Area of one tile = Base \u00d7 Height<\/p>\n\n\n\n

= 18 \u00d7 6<\/p>\n\n\n\n

= 108 cm2<\/sup><\/p>\n\n\n\n

We have the area of floor = 540 m2<\/sup><\/p>\n\n\n\n

Hence, number of tiles = Total area\/ Area of one tile<\/p>\n\n\n\n

= (540 x 100 x 100)\/108  [As, 1 m2<\/sup> = (100 x 100) cm2<\/sup>]<\/p>\n\n\n\n

= 50000<\/p>\n\n\n\n


6. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food piece would the ant have to take a longer round?

Solution:<\/strong><\/p>\n\n\n\n

(a) Diameter of semicircle = 2.8 cm<\/p>\n\n\n\n

\u2234 Perimeter = \u03c0r + 2r<\/p>\n\n\n\n

= 22\/7 \u00d7 2.8 + 2 \u00d7 2.8<\/p>\n\n\n\n

= 8.8 + 5.6 cm<\/p>\n\n\n\n

= 14.4 cm<\/p>\n\n\n\n

(b) Total perimeter = 1.5 + 1.5 + 2.8 + semi circumference (\u03c0r, where r = 2.8\/2 = 1.4 cm)<\/p>\n\n\n\n

= 1.5 + 1.5 + 2.8 + (22\/7 x 1.4)<\/p>\n\n\n\n

= 5.8 + 8.8<\/p>\n\n\n\n

= 14.6 cm<\/p>\n\n\n\n

(c) Total perimeter = 2 + 2 + Semi circumference (\u03c0r, where r = 2.8\/2 = 1.4 cm)<\/p>\n\n\n\n

= 4 + 8.8<\/p>\n\n\n\n

= 12.8 cm<\/p>\n\n\n\n

Hence, it is clearly seen that distance of (b) i.e. 14.6 is the longest.<\/p>\n\n\n\n


7. In the adjoining figure, the area enclosed between the concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, calculate the radius of the inner circle.

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Radius of outer circle (R) = 21 cm.<\/p>\n\n\n\n

Radius of inner circle (r) = r cm.<\/p>\n\n\n\n

Area of shaded portion = 770 cm2<\/sup><\/p>\n\n\n\n

\u21d2 \u03c0 (R2<\/sup> \u2013 r2<\/sup>) = 770<\/p>\n\n\n\n

(212<\/sup> \u2013 r2<\/sup>) = 770<\/p>\n\n\n\n

441 \u2013 r2<\/sup> = 770 \u00d7 (7\/22) = 35 \u00d7 7 = 245<\/p>\n\n\n\n

r2<\/sup> = 441 \u2013 245<\/p>\n\n\n\n

r2<\/sup> = 196<\/p>\n\n\n\n

r = \u221a196<\/p>\n\n\n\n

\u2234 r = 14 cm<\/p>\n\n\n\n


8. A copper wire when bent in the form of a square encloses an area of 121 cm2<\/sup>. If the same wire is bent into the form of a circle, find the area of the circle.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Area of the square = 121 cm2<\/sup><\/p>\n\n\n\n

So, side = \u221a121 = 11 cm<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Perimeter = 4 a = 4 \u00d7 11= 44 cm<\/p>\n\n\n\n

And, circumference of the circle = 44 cm<\/p>\n\n\n\n

\u2234 Radius = (44 x 7)\/ (2 x 22) = 7cm<\/p>\n\n\n\n

Therefore, area of the circle = \u03c0r2<\/sup> = (7)2<\/sup><\/p>\n\n\n\n

= 22\/7 \u00d7 7 \u00d7 7<\/p>\n\n\n\n

<\/strong>
= 154 cm2<\/sup><\/p>\n\n\n\n


9. From the given figure, find
(i) the area of \u2206 ABC
(ii) length of BC
(iii) the length of altitude from A to BC<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We have,<\/p>\n\n\n\n

Base = 3 cm and height = 4 cm.<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Area = \u00bd \u00d7 base \u00d7 height<\/p>\n\n\n\n

= \u00bd \u00d7 3 \u00d7 4<\/p>\n\n\n\n

= 6 cm2<\/sup><\/p>\n\n\n\n

(ii) By Pythagoras theorem, we have<\/p>\n\n\n\n

BC2<\/sup> = AB2<\/sup> + AC2<\/sup><\/p>\n\n\n\n

BC2<\/sup> = (3)2<\/sup> + (4)2<\/sup><\/p>\n\n\n\n

= 9 + 16 = 25<\/p>\n\n\n\n

\u21d2 BC = \u221a25 cm = 5 cm<\/p>\n\n\n\n

(iii) Now,<\/p>\n\n\n\n

Base = BC = 5 cm, h = AD =?<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Area = \u00bd \u00d7 b \u00d7 h<\/p>\n\n\n\n

6 = \u00bd \u00d7 5 \u00d7 h [\u2235 Area = 6 cm2<\/sup> as in part (i)]<\/p>\n\n\n\n

\u21d2 h = 12\/6 = 2.4 cm.<\/p>\n\n\n\n


10. A rectangular garden 80 m by 40 m is divided into four equal parts by two cross-paths 2.5 m wide. Find
(i) the area of the cross-paths.
(ii) the area of the unshaded portion.

Solution:<\/strong><\/p>\n\n\n\n

\"ML
<\/strong>
Given,<\/p>\n\n\n\n

Length of rectangular garden = 80 m<\/p>\n\n\n\n

and breadth = 40 m<\/p>\n\n\n\n

Width of crossing path 2.5 m
<\/strong>
So,
<\/strong>
Area of length wise path = 80 \u00d7 2.5 = 200 m2<\/sup><\/p>\n\n\n\n

and<\/p>\n\n\n\n

Area of breadth wise path = 40 \u00d7 2.5 = 100 m2<\/sup><\/p>\n\n\n\n

(i) Total area of both paths<\/p>\n\n\n\n

= 200 + 100 \u2013 2.5 \u00d7 2.5 m2<\/sup><\/p>\n\n\n\n

= 300 \u2013 6.25 = 293.75 m2<\/sup><\/p>\n\n\n\n

(ii) Area of unshaded portion<\/p>\n\n\n\n

= Area of garden \u2013 Area of paths<\/p>\n\n\n\n

= 80 \u00d7 40 \u2013 293.75 m2<\/sup><\/p>\n\n\n\n

= 3200 \u2013 293.75 m2<\/sup><\/p>\n\n\n\n

= 2906.25 m2<\/sup><\/p>\n\n\n\n


11. In the given figure, ABCD is a rectangle. Find the area of the shaded region.

Solution:<\/strong><\/p>\n\n\n\n

In the given figure, we have<\/p>\n\n\n\n

Length of rectangle = 18 cm and breadth = 12 cm<\/p>\n\n\n\n

\u2234 Area = l \u00d7 b = 18 \u00d7 12 cm2<\/sup> = 216 cm2<\/sup><\/p>\n\n\n\n

Area of triangle I = \u00bd \u00d7 12 \u00d7 10 = 60 cm2<\/sup><\/p>\n\n\n\n

Area of triangle III = \u00bd \u00d7 18 \u00d7 7 = 63 cm2<\/sup><\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

Area of shaded portion<\/p>\n\n\n\n

= Area of rectangle \u2013 Area of 3 triangles<\/p>\n\n\n\n

= 216 \u2013 (60 + 63 + 20)<\/p>\n\n\n\n

= 216 \u2013 143 cm2<\/p>\n\n\n\n

= 73 cm2<\/sup><\/p>\n\n\n\n


12. In the adjoining figure, ABCD is a square grassy lawn of area 729 m2<\/sup>. A path of uniform width runs all around it. If the area of the path is 295 m2<\/sup>, find
(i) the length of the boundary of the square field enclosing the lawn and the path.
(ii) the width of the path.

Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
\"MLGiven,<\/p>\n\n\n\n

Area of square ABCD = 729 m2<\/sup><\/p>\n\n\n\n

So, its side = \u221a729 = 27 m<\/p>\n\n\n\n

Let\u2019s take the width of path = x m<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Side of outer field = 27 + x + x = (27 + 2x) m<\/p>\n\n\n\n

And, area of square PQRS = (27 + 2x)2<\/sup> m2<\/sup><\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of PQRS \u2013 Area of ABCD = Area of path<\/p>\n\n\n\n

\u21d2 (27 + 2x)2<\/sup> m2<\/sup> \u2013 729 m2<\/sup> = 295 m2<\/sup><\/p>\n\n\n\n

729 + 4x2<\/sup> + 108x \u2013 729 = 295<\/p>\n\n\n\n

4x2<\/sup> + 108x \u2013 295 = 0<\/p>\n\n\n\n

By using the quadratic formula, we have<\/p>\n\n\n\n

a = 4, b = 108 and c = -295<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Hence,<\/p>\n\n\n\n

Width of the path is 2.5 m<\/p>\n\n\n\n

Now, side of square field PQRS = 27 + 2x<\/p>\n\n\n\n

= (27 + 2 \u00d7 2.5) m<\/p>\n\n\n\n

= 32 m<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Length of boundary = 4 \u00d7 side = 32 \u00d7 4 = 128 m<\/p>\n\n\n\n

Exercise 18.2<\/h4>\n\n\n\n

1. Each sides of a rhombus is 13 cm and one diagonal is 10 cm. Find
(i) the length of its other diagonal
(ii) the area of the rhombus
Solution:<\/strong><\/p>\n\n\n\n

\"ML
<\/strong>
(i) Given,<\/p>\n\n\n\n

Side of rhombus = 13 cm.<\/p>\n\n\n\n

Length of diagonal AC = 10 cm.<\/p>\n\n\n\n

\u2234 OC = 5 cm.<\/p>\n\n\n\n

Since, the diagonals of rhombus bisect each other at right angles<\/p>\n\n\n\n

So, \u2206BOC is rt. angled.<\/p>\n\n\n\n

Then, by Pythagoras Theorem we have<\/p>\n\n\n\n

BC2<\/sup> = OC2<\/sup> + OB2<\/sup><\/p>\n\n\n\n

132<\/sup> = 52<\/sup> + OB2<\/sup><\/p>\n\n\n\n

OB2<\/sup> = 169 \u2013 25 = 144<\/p>\n\n\n\n

\u21d2 OB = \u221a144 = 12 cm<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Diagonal BD = 2 \u00d7 OB = 2 \u00d7 12 = 24 cm<\/p>\n\n\n\n

(ii) Area of rhombus = \u00bd \u00d7 d1<\/sub> \u00d7 d2<\/sub><\/p>\n\n\n\n

= \u00bd \u00d7 10 \u00d7 24 = 120cm2<\/sup><\/p>\n\n\n\n


2. The cross-section ABCD of a swimming pool is a trapezium. Its width AB = 14 m, depth at the shallow end is 1-5 m and at the deep end is 8 m. Find the area of the cross-section.

Solution:<\/strong><\/p>\n\n\n\n

Here, AD and BC are the two parallel sides of trapezium<\/p>\n\n\n\n

And, distance between them is 14 m.<\/p>\n\n\n\n

\u2234 Area of trapezium = \u00bd (1\u00b75 + 8) \u00d7 14<\/p>\n\n\n\n

= \u00bd \u00d7 9\u00b75 \u00d7 14<\/p>\n\n\n\n

= 66 \u00d7 5 m2<\/sup><\/p>\n\n\n\n


3. The area of a trapezium is 360 m2<\/sup>, the distance between two parallel sides is 20 m and one of the parallel side is 25 m. Find the other parallel side.
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
\"MLGiven,<\/p>\n\n\n\n

Area of a trapezium = 360 m2<\/sup><\/p>\n\n\n\n

Distance between two parallel lines = 20 m<\/p>\n\n\n\n

One parallel side = 25 m<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Let\u2019s assume the second parallel side to be x m<\/p>\n\n\n\n

So, Area = (25 + x) \u00d7 20<\/p>\n\n\n\n

\u21d2 360 = (25 + x) \u00d7 20<\/p>\n\n\n\n

\u2234 x = 36 \u2013 25 = 11 m<\/p>\n\n\n\n

Therefore, the second parallel side is 11 m.<\/p>\n\n\n\n


4. Find the area of a rhombus whose side is 6.5 cm and altitude is 5 cm. If one of its diagonal is 13 cm long, find the length of other diagonal.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Side of rhombus = 6.5 cm<\/p>\n\n\n\n

And altitude = 5 cm<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Area of a rhombus = Side \u00d7 Altitude = 6.5 \u00d7 5 = 32.5 cm2<\/sup><\/p>\n\n\n\n

We have, one diagonal = 13 cm<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Length of other diagonal = (2 x Area)\/ One diagonal<\/p>\n\n\n\n

= (32.5 x 2)\/ 13<\/p>\n\n\n\n

= 5 cm<\/p>\n\n\n\n


5. From the given diagram, calculate
(i) the area of trapezium ACDE
(ii) the area of parallelogram ABDE
(iii) the area of triangle BCD.

Solution:<\/strong><\/p>\n\n\n\n

(i) Area of trapezium ACDE = \u00bd \u00d7 (AC + DE) \u00d7 h<\/p>\n\n\n\n

= \u00bd \u00d7 (13 + 7) \u00d7 6.5<\/p>\n\n\n\n

= \u00bd \u00d7 20 \u00d7 6.5<\/p>\n\n\n\n

= 65 m2<\/sup><\/p>\n\n\n\n

(ii) Area of parallelogram ABDE = \u00bd \u00d7 b \u00d7 h<\/p>\n\n\n\n

= \u00bd \u00d7 6 \u00d7 6.5<\/p>\n\n\n\n

= 15.5 m2<\/sup><\/p>\n\n\n\n

(iii) Area of \u2206BCD = \u00bd \u00d7 base \u00d7 height<\/p>\n\n\n\n

= \u00bd \u00d7 BC \u00d7 height<\/p>\n\n\n\n

= \u00bd \u00d7 6 \u00d7 6.5 [\u2235 BC = AC \u2013 AB = 13 \u2013 7 = 6 m]<\/p>\n\n\n\n

= 19.5 m2<\/sup><\/p>\n\n\n\n


6. The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Base of triangle = 24.8 cm and altitude = 16.5 cm<\/p>\n\n\n\n

Area = \u00bd \u00d7 base \u00d7 altitude<\/p>\n\n\n\n

= \u00bd \u00d7 24.8 \u00d7 16.5 cm2<\/sup><\/p>\n\n\n\n

= 204.6 cm2<\/sup><\/p>\n\n\n\n

Now, Area of \u2206 = Area of rhombus<\/p>\n\n\n\n

But, area of rhombus = 204.6 cm2<\/sup><\/p>\n\n\n\n

Length of one diagonal = 22 cm<\/p>\n\n\n\n

Area of rhombus = (First diagonal \u00d7 Second diagonal)<\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

Second diagonal = (2 \u00d7 Area)\/ First diagonal<\/p>\n\n\n\n

= (204.6 \u00d7 2)\/ 22<\/p>\n\n\n\n

= 18.6 cm<\/p>\n\n\n\n


7. The perimeter of a trapezium is 52 cm. If its non-parallel sides are 10 cm each and its altitude is 8 cm, find the area of the trapezium.
Solution:<\/strong><\/p>\n\n\n\n

\"ML
<\/strong>
Given,<\/p>\n\n\n\n

Perimeter of a trapezium = 52 cm<\/p>\n\n\n\n

Length of each non-parallel side = 10 cm<\/p>\n\n\n\n

Altitude DL = 8 cm
<\/strong>
Now,
<\/strong>
In right \u2206DAL, by Pythagoras Theorem we have<\/p>\n\n\n\n

DA2<\/sup> = DL2<\/sup> + AL2<\/sup><\/p>\n\n\n\n

(10)2<\/sup> = (8)2<\/sup> + AL2<\/sup><\/p>\n\n\n\n

100 = 64 + AL2<\/sup><\/p>\n\n\n\n

AL2<\/sup> = 100 \u2013 64 = 36 = (6)2<\/sup><\/p>\n\n\n\n

\u2234 AL = 6 cm<\/p>\n\n\n\n

Similarly,<\/p>\n\n\n\n

BM = 6 cm and DC = LM<\/p>\n\n\n\n

Also, we have<\/p>\n\n\n\n

Perimeter = AB + BC + CD + DA<\/p>\n\n\n\n

and CD = DA<\/p>\n\n\n\n

So, CD + DA = 2DA<\/p>\n\n\n\n

But,<\/p>\n\n\n\n

AB + CD = Perimeter \u2013 2 AD<\/p>\n\n\n\n

= 52 \u2013 2 \u00d7 10<\/p>\n\n\n\n

= 52 \u2013 20<\/p>\n\n\n\n

= 32 cm<\/p>\n\n\n\n

Thus, area of trapezium = \u00bd \u00d7 (sum of parallel sides) \u00d7 altitude<\/p>\n\n\n\n

= \u00bd \u00d7 32 \u00d7 8<\/p>\n\n\n\n

= 128 cm2<\/sup><\/p>\n\n\n\n


8. The area of a trapezium is 540 cm2<\/sup>. If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the lengths of parallel sides.
Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s assume the two parallel sides of trapezium to be 7x and 5x.<\/p>\n\n\n\n

Height = 18 cm<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of trapezium = \u00bd \u00d7 [Sum of || gm sides \u00d7 height]<\/p>\n\n\n\n

\u21d2 540 = \u00bd \u00d7 (7x + 5x) \u00d7 18<\/p>\n\n\n\n

540 = \u00bd \u00d7 12x \u00d7 18<\/p>\n\n\n\n

540 = 108x<\/p>\n\n\n\n

x = 540\/108<\/p>\n\n\n\n

x = 5 cm<\/p>\n\n\n\n

Hence, the two parallel sides are:<\/p>\n\n\n\n

7x = 7 \u00d7 5 = 35 cm and 5x = 5 \u00d7 5 = 25 cm<\/p>\n\n\n\n


9. Calculate the area enclosed by the given shapes. All measurements are in cm.

Solution:<\/strong><\/p>\n\n\n\n

(i) Firstly,<\/p>\n\n\n\n

Area of trapezium ABCD<\/p>\n\n\n\n

= (Sum of opposite ||gm sides) \u00d7 height<\/p>\n\n\n\n

\"ML= [(AB + CD) \u00d7 (AF + FD)]<\/p>\n\n\n\n

= [(AB + CD) \u00d7 (AF + FD)<\/p>\n\n\n\n

= [(5 + 3) \u00d7 (5 + 4)]<\/p>\n\n\n\n

= (5 + 3) \u00d7 9<\/p>\n\n\n\n

= 36 cm2<\/sup><\/p>\n\n\n\n

Secondly,<\/p>\n\n\n\n

Area of rectangle GAFE = Length \u00d7 Breadth<\/p>\n\n\n\n

= 2 \u00d7 5 = 10 cm2<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Total area of the figure = Area of trapezium ABCD + Area of rectangle GAFE<\/p>\n\n\n\n

= (36 + 10) cm2<\/sup><\/p>\n\n\n\n

= 46 cm2+<\/sup><\/p>\n\n\n\n

(ii) It\u2019s seen that,<\/p>\n\n\n\n

\"MLArea of given figure = Area of rect. ABCD + Area of || gm BIHJ + Area of rectangle EFGH<\/p>\n\n\n\n

Area of rectangle ABCD = Length \u00d7 Breadth<\/p>\n\n\n\n

= AD \u00d7 DC<\/p>\n\n\n\n

= 9 \u00d7 2 = 18 cm2<\/sup><\/p>\n\n\n\n

And,<\/p>\n\n\n\n

Area of rectangle EFGH = Length \u00d7 Breadth<\/p>\n\n\n\n

= (EJ + JH) \u00d7 EF<\/p>\n\n\n\n

= (7 + 2) \u00d7 2<\/p>\n\n\n\n

= 9 \u00d7 2 = 18 cm2<\/sup><\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of parallelogram BIHJ = 2 \u00d7 5 = 10 cm2<\/sup>[Since, distance between BI and HJ = 9 \u2013 2 \u2013 2 = 5 cm]<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Total area of the figure = (18 + 18 + 10) cm2<\/sup> = 46 cm2<\/sup><\/p>\n\n\n\n

10. From the adjoining sketch, calculate
(i) the length AD
(ii) the area of trapezium ABCD
(iii) the area of triangle BCD

Solution:<\/strong><\/p>\n\n\n\n

(i) In right angled \u2206 ABD, by Pythagoras Theorem we have<\/p>\n\n\n\n

BD2<\/sup> = AD2<\/sup> + AB2<\/sup><\/p>\n\n\n\n

\u21d2 AD2<\/sup> = BD2<\/sup> \u2013 AB2<\/sup><\/p>\n\n\n\n

= (41)2<\/sup> \u2013 (40)2 <\/sup><\/p>\n\n\n\n

= 1681 \u2013 1600<\/p>\n\n\n\n

= 81<\/p>\n\n\n\n

\u2234 AD = \u221a81 = 9 cm<\/p>\n\n\n\n

(ii) Area of trapezium ABCD<\/p>\n\n\n\n

= (Sum of opposite || gm lines) \u00d7 height<\/p>\n\n\n\n

= (AB + CD) \u00d7 AD<\/p>\n\n\n\n

= (40+ 15) \u00d7 9<\/p>\n\n\n\n

= 247.5 cm2<\/sup><\/p>\n\n\n\n

(iii) Area of triangle BCD = Area of trapezium ABCD \u2013 Area of \u2206 ABD<\/p>\n\n\n\n

= (247.5 \u2013 \u00d7 40 \u00d7 9) cm2<\/sup><\/p>\n\n\n\n

= (247.5 \u2013 180) cm2<\/sup><\/p>\n\n\n\n

= 67\u00b75 cm2<\/sup><\/p>\n\n\n\n


11. Diagram of the adjacent picture frame has outer dimensions = 28 cm \u00d7 32 cm and inner dimensions 20 cm \u00d7 24 cm. Find the area of each section of the frame, if the width of each section is same.

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Outer length of the frame = 32 cm and outer breadth = 28 cm<\/p>\n\n\n\n

Inner length = 24 cm and outer breadth = 20 cm<\/p>\n\n\n\n

So, width of the frame = (32 \u2013 24)\/ 2 = 4 cm<\/p>\n\n\n\n

\u21d2 Height = 4 cm<\/p>\n\n\n\n

Now, area of each portion of length side<\/p>\n\n\n\n

= \u00bd \u00d7 (24 + 32) \u00d7 4<\/p>\n\n\n\n

= \u00bd \u00d7 56 \u00d7 4<\/p>\n\n\n\n

= 112 cm2<\/sup><\/p>\n\n\n\n

And,<\/p>\n\n\n\n

Area of each portion of breadth side<\/p>\n\n\n\n

= \u00bd \u00d7 (20 + 28) \u00d7 4<\/p>\n\n\n\n

= \u00bd \u00d7 48 \u00d7 4<\/p>\n\n\n\n

= 96 cm2<\/sup><\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Area each section are 112 cm2<\/sup>, 96 cm2<\/sup>, 112 cm2<\/sup>, 96 cm2<\/sup>.<\/p>\n\n\n\n


12. In the given quadrilateral ABCD, \u2220BAD = 90\u00b0 and \u2220BDC = 90\u00b0. All measurements are in centimetres. Find the area of the quadrilateral ABCD.

Solution:<\/strong><\/p>\n\n\n\n

In right angled triangle ABD, by Pythagoras Theorem we have<\/p>\n\n\n\n

BD2<\/sup> = AB2<\/sup> + AD2 <\/sup>= (6)2<\/sup> + (8)2<\/sup><\/p>\n\n\n\n

= 36 + 64<\/p>\n\n\n\n

= 100 cm2<\/sup><\/p>\n\n\n\n

\u2234 BD = \u221a100 = 10 cm<\/p>\n\n\n\n

Now, Area of \u2206ABD = \u00bd \u00d7 b \u00d7 h<\/p>\n\n\n\n

= \u00bd \u00d7 6 \u00d7 8<\/p>\n\n\n\n

= 24cm2<\/sup> \u2026(i)<\/p>\n\n\n\n

In \u2206 BDC, we have<\/p>\n\n\n\n

BD = 10 cm, BC = 26 cm<\/p>\n\n\n\n

DC = ?<\/p>\n\n\n\n

By Pythagoras theorem,<\/p>\n\n\n\n

BC2<\/sup> = BD2<\/sup> + DC2<\/sup><\/p>\n\n\n\n

(26)2<\/sup> = (10)2<\/sup> + DC2<\/sup><\/p>\n\n\n\n

676 \u2013 100 = DC2<\/sup><\/p>\n\n\n\n

\u21d2 DC = \u221a576 = 24 cm.<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of \u2206 BDC = \u00bd \u00d7 b \u00d7 h<\/p>\n\n\n\n

= \u00bd \u00d7 24 \u00d7 10<\/p>\n\n\n\n

= 12 cm2<\/sup> \u2026(ii)<\/p>\n\n\n\n

Adding (i) and (ii), we get<\/p>\n\n\n\n

Area of \u2206ABD + Area of \u2206BDC = (24 + 120) cm2<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Area of quadrilateral ABCD = 144 cm2<\/sup><\/p>\n\n\n\n


13. Top surface of a raised platform is in the shape of a regular octagon as shown in the given figure. Find the area of the octagonal surface.

Solution:<\/strong><\/p>\n\n\n\n

The raised surface of platform is in the shape of regular octagon ABCDEFGH of each side = 8 cm.<\/p>\n\n\n\n

Join HC.<\/p>\n\n\n\n

GD = HC = 15 cm, FL = AM = 6 cm<\/p>\n\n\n\n

Now, in each trapezium parallel sides are 15 cm and 6 cm and height = 6 cm<\/p>\n\n\n\n

So, Area of each trapezium FEDG = \u00bd (GD + FE) \u00d7 FL<\/p>\n\n\n\n

= \u00bd (15 + 8) \u00d7 6<\/p>\n\n\n\n

= 23 \u00d7 3 cm2<\/sup><\/p>\n\n\n\n

= 69 cm2<\/sup><\/p>\n\n\n\n

Also,<\/p>\n\n\n\n

Area of trapezium FEDG = Area of trapezium ABCH = 69 cm2<\/sup><\/p>\n\n\n\n

And area of rectangle HCDG = HC \u00d7 CD<\/p>\n\n\n\n

= 15 \u00d7 8<\/p>\n\n\n\n

= 120 cm2<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Total area = Area of trapezium FEDG + Area of trapezium ABCH + Area of rectangle HCDG.<\/p>\n\n\n\n

= 69 + 69 + 120<\/p>\n\n\n\n

= 258 cm2<\/sup><\/p>\n\n\n\n


14. There is a pentagonal shaped park as shown in the following figure:
For finding its area Jaspreet and Rahul divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?
Solution:<\/strong><\/p>\n\n\n\n

The pentagonal shaped park is shown in the given figure.<\/p>\n\n\n\n

In which DL \u22a5 CE and is produced to M.<\/p>\n\n\n\n

So, DM = 32 m<\/p>\n\n\n\n

LM = CB = 18 m<\/p>\n\n\n\n

\u2234 DL = 32 \u2013 18 = 14 m<\/p>\n\n\n\n

(i) According to Jaspreet\u2019s the figure is divided into two equal trapezium in area: DEAM and DCBM<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of trapezium DEAM = \u00bd (AE + DM) \u00d7 AM<\/p>\n\n\n\n

= \u00bd (32 + 18) \u00d7 9<\/p>\n\n\n\n

= (50 x 9)\/ 2<\/p>\n\n\n\n

= 225m2<\/sup><\/p>\n\n\n\n

(ii) According to Rahul\u2019s the figure is divided into shapes: one square and on isosceles triangle.<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Area of square ABCE = (Side)2 <\/sup><\/p>\n\n\n\n

= (18)2<\/sup><\/p>\n\n\n\n

= 324 m2<\/sup><\/p>\n\n\n\n

And, area of isosceles \u2206EDC = \u00bd \u00d7 EC \u00d7 DC<\/p>\n\n\n\n

= \u00bd \u00d7 18 \u00d7 14<\/p>\n\n\n\n

= 126 m2<\/sup><\/p>\n\n\n\n

\u2234 Total area = 225 \u00d7 2 = 450 m2<\/sup><\/p>\n\n\n\n

The third way to find out the area of given figure is as follow:<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Here, DL \u22a5 ED and DL = 14 m<\/p>\n\n\n\n

Area of \u2206DEC = \u00bd \u00d7 EC \u00d7 LD<\/p>\n\n\n\n

= \u00bd \u00d7 18 \u00d7 14<\/p>\n\n\n\n

= 126 m2<\/sup><\/p>\n\n\n\n

Area of \u2206AEB = \u00bd \u00d7 AB \u00d7 AE<\/p>\n\n\n\n

= \u00bd \u00d7 18 \u00d7 18<\/p>\n\n\n\n

= 162 m2<\/sup><\/p>\n\n\n\n

Area of \u2206BEC = \u00bd \u00d7 BC \u00d7 EC<\/p>\n\n\n\n

= \u00bd \u00d7 18 \u00d7 18 = 162 m2<\/sup><\/p>\n\n\n\n

Hence, area of pentagon ABCDE = Area \u2206DEC + Area of \u2206AEB + Area of \u2206BEC<\/p>\n\n\n\n

= (126 + 162 + 162) m2<\/sup><\/p>\n\n\n\n

= 450 m2<\/sup><\/p>\n\n\n\n


15. In the diagram, ABCD is a rectangle of size 18 cm by 10 cm. In \u2206 BEC, \u2220E = 90\u00b0 and EC = 8 cm. Find the area enclosed by the pentagon ABECD.

Solution:<\/strong><\/p>\n\n\n\n

Area of rectangle ABCD = Length \u00d7 Breadth<\/p>\n\n\n\n

= 18 \u00d7 10<\/p>\n\n\n\n

= 180 cm2<\/sup><\/p>\n\n\n\n

In right angled \u2206 BEC,<\/p>\n\n\n\n

By Pythagoras theorem, we have<\/p>\n\n\n\n

BC2<\/sup> = CE2<\/sup> + BE2<\/sup><\/p>\n\n\n\n

(10)2<\/sup> = 82<\/sup> + BE2<\/sup><\/p>\n\n\n\n

BE2<\/sup> = 100 \u2013 64 = 36<\/p>\n\n\n\n

\u21d2 BE = \u221a36 = 6 cm.<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Area of rt. \u2206 BEC = \u00bd \u00d7 6 \u00d7 8<\/p>\n\n\n\n

= 24cm2<\/sup><\/p>\n\n\n\n

Area of pentagon ABECD = Area of rectangle \u2013 area of \u2206<\/p>\n\n\n\n

= (180 \u2013 24) cm2<\/sup><\/p>\n\n\n\n

= 156 cm2<\/sup><\/p>\n\n\n\n

16. Polygon ABCDE is divided into parts as shown in the given figure. Find its area if AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.

Solution:<\/strong><\/p>\n\n\n\n

In the given figure, ABCDE, AD = 8 cm, AH = 6 cm, AG = 4 cm,<\/p>\n\n\n\n

AF = 3cm \u22a5 BF = 2 cm CH = 3 cm and \u22a5 EG = 2.5 cm<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

The given figure, consists of 3 triangles and one trapezium.<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of \u2206AED = \u00bd \u00d7 AD \u00d7 GE<\/p>\n\n\n\n

= \u00bd \u00d7 8 \u00d7 2.5<\/p>\n\n\n\n

= 10 cm2<\/sup><\/p>\n\n\n\n

Area of \u2206ABF = \u00bd \u00d7 AF \u00d7 BF<\/p>\n\n\n\n

= \u00bd \u00d7 3 \u00d7 2<\/p>\n\n\n\n

= 3 cm2<\/sup><\/p>\n\n\n\n

Area of \u2206CDH = \u00bd \u00d7 HD \u00d7 CH<\/p>\n\n\n\n

= \u00bd \u00d7 (AD \u2013 AH) \u00d7 3<\/p>\n\n\n\n

= \u00bd \u00d7 (8 \u2013 6) \u00d7 3<\/p>\n\n\n\n

= \u00bd \u00d7 2 \u00d7 3<\/p>\n\n\n\n

= 3 cm2<\/sup><\/p>\n\n\n\n

Area of trapezium BFHC = \u00bd \u00d7 (BF + CH) \u00d7 FH<\/p>\n\n\n\n

= \u00bd \u00d7 (2 + 3) \u00d7 (AH \u2013 AF)<\/p>\n\n\n\n

= \u00bd \u00d7 5 \u00d7 (6 \u2013 3)<\/p>\n\n\n\n

= \u00bd \u00d7 5 \u00d7 3<\/p>\n\n\n\n

= 7.5 cm2<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Total area of the figure = Area of \u2206AED + Area of \u2206ABF + Area of \u2206CDH + Area of trapezium BFHC<\/p>\n\n\n\n

= 10 + 3 + 3 + 7.5<\/p>\n\n\n\n

= 23.5 cm2<\/sup><\/p>\n\n\n\n


17. Find the area of polygon PQRSTU shown in 1 the given figure, if PS = 11 cm, PY = 9 cm, PX = 8 cm, PW = 5 cm, PV = 3 cm, QV = 5 cm, UW = 4 cm, RX = 6 cm, TY = 2 cm.

Solution:<\/strong><\/p>\n\n\n\n

In the figure PQRSTU, we have<\/p>\n\n\n\n

PS = 11 cm, PY = 9 cm, PX = 8 cm, PW = 5 cm, PV = 3 cm, QV = 5 cm, UW = 4 cm, RX = 6 cm and TY = 2 cm<\/p>\n\n\n\n

And,<\/p>\n\n\n\n

The figure consists of 4 triangle and 2 trapeziums<\/p>\n\n\n\n

From the figure its seen that,<\/p>\n\n\n\n

VX = PX \u2013 PV<\/p>\n\n\n\n

= 8 \u2013 3<\/p>\n\n\n\n

= 5 cm<\/p>\n\n\n\n

XS = PS \u2013 PX<\/p>\n\n\n\n

= 11 \u2013 8<\/p>\n\n\n\n

= 3 cm<\/p>\n\n\n\n

YS = PS \u2013 PY<\/p>\n\n\n\n

= 11 \u2013 9<\/p>\n\n\n\n

= 2 cm<\/p>\n\n\n\n

WY = PY \u2013 PW<\/p>\n\n\n\n

= 9 \u2013 5<\/p>\n\n\n\n

= 4 cm<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area \u2206PQV = \u00bd \u00d7 PV \u00d7 QV<\/p>\n\n\n\n

= \u00bd \u00d7 3 \u00d7 5 = 15\/2<\/p>\n\n\n\n

= 7.5 cm2<\/sup><\/p>\n\n\n\n

Area of \u2206RXS = \u00bd \u00d7 XS \u00d7 RX<\/p>\n\n\n\n

= \u00bd \u00d7 3 \u00d7 6<\/p>\n\n\n\n

= 9 cm2<\/sup><\/p>\n\n\n\n

Area of \u2206PUW = \u00bd \u00d7 PW \u00d7 UW<\/p>\n\n\n\n

= \u00bd \u00d7 5 \u00d7 4<\/p>\n\n\n\n

= 10 cm2<\/sup><\/p>\n\n\n\n

Area \u2206YTS = \u00bd \u00d7 YS \u00d7 TY<\/p>\n\n\n\n

= \u00bd \u00d7 2 \u00d7 2<\/p>\n\n\n\n

= 2 cm2<\/sup><\/p>\n\n\n\n

Area of trapezium \u2206VX R = \u00bd \u00d7 (QV + RX) \u00d7 VX<\/p>\n\n\n\n

= \u00bd \u00d7 (5 + 6) \u00d7 5<\/p>\n\n\n\n

= \u00bd \u00d7 11 \u00d7 5 cm2<\/sup><\/p>\n\n\n\n

= 55\/7<\/p>\n\n\n\n

= 27.5 cm2<\/sup><\/p>\n\n\n\n

Area of trapezium WUTY = \u00bd \u00d7 (UW + TY) \u00d7 WY<\/p>\n\n\n\n

= \u00bd \u00d7 (4 + 2) \u00d7 4<\/p>\n\n\n\n

= \u00bd \u00d7 6 \u00d7 4<\/p>\n\n\n\n

= 12 cm2<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Area of the figure = (7.5 + 9 + 10 + 2 + 27.5 + 12) cm2<\/sup> = 68 cm2<\/sup>.<\/p>\n\n\n\n

Exercise 18.3<\/h4>\n\n\n\n

1. The volume of a cube is 343 cm3<\/sup>, find the length of an edge of cube.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Volume of a cube = 343 cm3<\/sup><\/p>\n\n\n\n

Let\u2019s consider \u2018a\u2019 to be the edge of cube, then<\/p>\n\n\n\n

V = a3<\/sup> = 343 = (7)3<\/sup><\/p>\n\n\n\n

\u2234 a = 7 cm<\/p>\n\n\n\n


2. Fill in the following blanks:<\/strong><\/p>\n\n\n\n

<\/td>Volume of cuboid<\/td>Length<\/td>Breadth<\/td>Height<\/td><\/tr>
(i)<\/td>90 cm3<\/sup><\/td>\u2013<\/td>5 cm<\/td>3 cm<\/td><\/tr>
(ii)<\/td>\u2013<\/td>15 cm<\/td>8 cm<\/td>7 cm<\/td><\/tr>
(iii)<\/td>62.5 m3<\/sup><\/td>10 cm<\/td>5 cm<\/td>\u2013<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Volume of cuboid = length x Breadth x Height<\/p>\n\n\n\n

(i) 90 cm3<\/sup> = length x 5 cm x 3 cm<\/p>\n\n\n\n

Length = 90\/(5 x 3) = 90\/15 = 6 cm<\/p>\n\n\n\n

(ii) Volume = 15 cm x 8 cm x 7 cm<\/p>\n\n\n\n

= 840 cm3<\/sup><\/p>\n\n\n\n

(iii) 62.5 m3<\/sup> = 10 m x 5 m x height<\/p>\n\n\n\n

Height = 62.5\/(10 x 5) = 1.25 m<\/p>\n\n\n\n

<\/td>Volume of cuboid<\/td>Length<\/td>Breadth<\/td>Height<\/td><\/tr>
(i)<\/td>90 cm3<\/sup><\/td>6 cm<\/td>5 cm<\/td>3 cm<\/td><\/tr>
(ii)<\/td>840 cm3<\/sup><\/td>15 cm<\/td>8 cm<\/td>7 cm<\/td><\/tr>
(iii)<\/td>62.5 m3<\/sup><\/td>10 cm<\/td>5 cm<\/td>1.25 m<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n


3. Find the height of a cuboid whose volume is 312 cm3<\/sup> and base area is 26 cm2<\/sup>.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Volume of a cuboid = 312 cm3<\/sup><\/p>\n\n\n\n

Base area = l \u00d7 b = 26 cm2<\/sup><\/p>\n\n\n\n

\u2234 Height= Volume\/Base area = 312\/26 = 12cm<\/p>\n\n\n\n


4. A godown is in the form of a cuboid of measures 55 m \u00d7 45 m \u00d7 30 m. How many cuboidal boxes can be stored in it if the volume of one box is 1.25 m3<\/sup>?
Solution:<\/strong><\/p>\n\n\n\n

Given,
<\/strong>
Length of a godown (l) = 55 m<\/p>\n\n\n\n

Breadth (b) = 45 m<\/p>\n\n\n\n

Height (h) = 30 m<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Volume = l \u00d7 b \u00d7 h<\/p>\n\n\n\n

= (55 \u00d7 45 \u00d7 30) m3<\/sup><\/p>\n\n\n\n

= 74250 m3<\/sup><\/p>\n\n\n\n

Also given, volume of one box = 1.25 m3<\/sup><\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

Number of boxes = 74250\/1.25 = 59400 boxes<\/p>\n\n\n\n


5. A rectangular pit 1.4 m long, 90 cm broad and 70 cm deep was dug and 1000 bricks of base 21 cm by 10.5 cm were made from the earth dug out. Find the height of each brick.
Solution:<\/strong><\/p>\n\n\n\n

Here l = 1.4 m = 140 cm, b = 90 cm and h = 70 cm<\/p>\n\n\n\n

Volume of rectangular pit = l \u00d7 b \u00d7 h<\/p>\n\n\n\n

= (140 \u00d7 90 \u00d7 70) cm3<\/sup><\/p>\n\n\n\n

= 882000 cm3<\/sup><\/p>\n\n\n\n

Volume of brick = 21 \u00d7 10.5 \u00d7 h<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Number of bricks = Volume of pit\/ Volume of brick<\/p>\n\n\n\n

1000 = 882000\/ (21 \u00d7 10.5 \u00d7 h)<\/p>\n\n\n\n

h = 882000\/ (21 \u00d7 10.5 \u00d7 1000)<\/p>\n\n\n\n

= 4 cm<\/p>\n\n\n\n

Thus, the height of each brick is 4 cm.<\/p>\n\n\n\n


6. If each edge of a cube is tripled, then find how many times will its volume become?
Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s consider the edge of a cube to be x<\/p>\n\n\n\n

Then, it\u2019s volume = x3<\/sup><\/p>\n\n\n\n

Now, if the edge is tripled<\/p>\n\n\n\n

Edge = 3x<\/p>\n\n\n\n

So, volume = (3x)3<\/sup> = 27x3<\/sup><\/p>\n\n\n\n

\u2234 Its volume is 27 times the volume of the given cube.<\/p>\n\n\n\n


7. A milk tank is in the form of cylinder whose radius is 1.4 m and height is 8 m. Find the quantity of milk in litres that can be stored in the tank.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Radius of the milk cylindrical tank = 1.4 m and height (h) = 8 m<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Volume of milk in the tank = \u03c0r2<\/sup>h<\/p>\n\n\n\n

= (22\/7) \u00d7 1.4 \u00d7 1.4 \u00d7 8 m3<\/sup><\/p>\n\n\n\n

= 49.28 m3<\/sup><\/p>\n\n\n\n

= 49.28 \u00d7 1000 litres<\/p>\n\n\n\n

= 49280 litres<\/p>\n\n\n\n

Therefore, the quantity of the tank is 49280 litres.<\/p>\n\n\n\n


8. A closed box is made of 2 cm thick wood with external dimension 84 cm \u00d7 75 cm \u00d7 64 cm. Find the volume of the wood required to make the box.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Thickness of the wood used in a closed box = 2 cm<\/p>\n\n\n\n

External length of box (L) = 84 cm<\/p>\n\n\n\n

Breadth (b) = 75 cm and height (h) = 64 cm<\/p>\n\n\n\n

So, internal length (l) = 84 \u2013 (2 \u00d7 2)<\/p>\n\n\n\n

= 84 \u2013 4<\/p>\n\n\n\n

= 80 cm<\/p>\n\n\n\n

Breadth (b) = 75 \u2013 (2 \u00d7 2)<\/p>\n\n\n\n

= 75 \u2013 4<\/p>\n\n\n\n

= 71 cm<\/p>\n\n\n\n

and height (h) = 64 \u2013 (2 \u00d7 2)<\/p>\n\n\n\n

= 64 \u2013 4<\/p>\n\n\n\n

= 60 cm<\/p>\n\n\n\n

Hence, Volume of wood used = 84 \u00d7 75 \u00d7 64 \u2013 80 \u00d7 71 \u00d7 60 cm3<\/sup><\/p>\n\n\n\n

= 403200 \u2013 340800 cm3<\/sup><\/p>\n\n\n\n

= 62400 cm3<\/sup><\/p>\n\n\n\n


9. Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Ratio in diameters of two cylindrical jars = 3 : 4<\/p>\n\n\n\n

But their volumes are same.<\/p>\n\n\n\n

Let\u2019s assume h1<\/sub> and h2<\/sub> to be the heights of the two jars respectively.<\/p>\n\n\n\n

Let radius of the first jar (r1<\/sub>) = 3x\/2<\/p>\n\n\n\n

and radius of the second jar (r2<\/sub>) = 4x\/2<\/p>\n\n\n\n

Then according to the condition in the problem, we have<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n
\"ML<\/figure>\n\n\n\n

Therefore, the ratio in their heights =16 : 9<\/p>\n\n\n\n


10. The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder?
Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s consider the radius of a cylinder to be r<\/p>\n\n\n\n

And height = h<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Volume = \u03c0r2<\/sup>h<\/p>\n\n\n\n

Now, its radius is halved and height is doubled, then<\/p>\n\n\n\n

Volume = \u03c0(r\/2)2 <\/sup>\u00d7 (2h)<\/p>\n\n\n\n

= \u03c0r2<\/sup>h\/ 2<\/p>\n\n\n\n

Thus, the ratio in the volumes of the new cylinder to old one is<\/p>\n\n\n\n

= \u03c0r2<\/sup>h\/ 2 : \u03c0r2<\/sup>h<\/p>\n\n\n\n

= 1 : 2<\/p>\n\n\n\n


11. A rectangular piece of tin of size 30 cm \u00d7 18 cm is rolled in two ways, once along its length (30 cm) and once along its breadth. Find the ratio of volumes of two cylinders so formed.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Size of rectangular tin plate = 30 cm \u00d7 18 cm<\/p>\n\n\n\n

(i) When rolled along its length (30 cm),<\/p>\n\n\n\n

Then, the circumference of the circle so formed = 30 cm<\/p>\n\n\n\n

Radius(r1<\/sub>) = C\/2\u03c0 = (30 x 7)\/ (2 x 22) = 105\/22 cm<\/p>\n\n\n\n

And height (h1<\/sub>) = 18 cm<\/p>\n\n\n\n

Then, volume = \u03c0r1<\/sub>2<\/sup>h1<\/sub> = \u03c0 x (105\/22)2<\/sup> x (18) cm3<\/sup><\/p>\n\n\n\n

If it is rolled along its breadth (18 cm) then,<\/p>\n\n\n\n

Circumference = 18 cm<\/p>\n\n\n\n

So, radius (r2<\/sub>) = C\/2\u03c0 = (18 x 7)\/ (2 x 22) = 63\/22 cm<\/p>\n\n\n\n

And height (h2<\/sub>) = 30 cm<\/p>\n\n\n\n

Then, volume = \u03c0r2<\/sub>2<\/sup>h2<\/sub> = \u03c0 x (63\/22)2<\/sup> x (30) cm3<\/sup><\/p>\n\n\n\n

Now, ratio between the two volumes<\/p>\n\n\n\n

= \u03c0 x (105\/22)2<\/sup> x (18) : \u03c0 x (63\/22)2<\/sup> x (30)<\/p>\n\n\n\n

= (105\/22)2<\/sup> x (18) : (63\/22)2<\/sup> x (30)<\/p>\n\n\n\n

= 5 : 3<\/p>\n\n\n\n


12. Water flows through a cylindrical pipe of internal diameter 7 cm at 5 m per sec. Calculate
(i) the volume in litres of water discharged by the pipe in one minute.
(ii) the time in minutes, the pipe would take to fill an empty rectangular tank of size 4 m \u00d7 3 m \u00d7 2.31 m.
Solution:<\/strong><\/p>\n\n\n\n

Given,
<\/strong>
Speed of water flow through cylindrical pipe = 5 m\/sec.<\/p>\n\n\n\n

Internal diameter of the pipe = 7 cm<\/p>\n\n\n\n

So, radius (r) = 7\/2 cm<\/p>\n\n\n\n

Now, length of water flow in 1 minutes (h) = 5 \u00d7 60 = 300 m<\/p>\n\n\n\n

\u2234 Volume of water = \u03c0r2<\/sup>h<\/p>\n\n\n\n

= 22\/7 x 7\/2 x 7\/2 x 300 x 100 cm3<\/sup><\/p>\n\n\n\n

= 1155000 cm3<\/sup> = 1155 litres<\/p>\n\n\n\n

(i) the volume of water = 1155000 cm3<\/sup><\/p>\n\n\n\n

Volume of rectangular tank of size = 4m \u00d7 3m \u00d7 2.31m<\/p>\n\n\n\n

= 27.72 m3<\/sup><\/p>\n\n\n\n

Also given, speed of water = 4 m\/sec.<\/p>\n\n\n\n

Radius of pipe = 7\/2 cm<\/p>\n\n\n\n

Volume of water in 1 sec = 22\/7 x 7\/2 x 7\/2 x 5 x 100 cm3<\/sup><\/p>\n\n\n\n

= 19250 cm3<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

(ii) Time taken to empty the tank = 27.72 m3<\/sup>\/ 19250<\/p>\n\n\n\n

= (2772 x 100 x 100 x 100)\/(100 x 19250) sec<\/p>\n\n\n\n

= 1440 sec<\/p>\n\n\n\n

= 1440\/60 = 24 minutes<\/p>\n\n\n\n


13. Two cylindrical vessels are filled with milk. The radius of one vessel is 15 cm and height is 40 cm, and the radius of other vessel is 20 cm and height is 45 cm. Find the radius of another cylindrical vessel of height 30 cm which may just contain the milk which is in the two given vessels.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Radius of one cylinder (r1<\/sub>) = 15 cm<\/p>\n\n\n\n

And height (h1<\/sub>) = 40 cm<\/p>\n\n\n\n

Radius of second cylinder (r2<\/sub>) = 20 cm<\/p>\n\n\n\n

And height (h2<\/sub>) = 45 cm<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Volume of first cylinder = \u03c0r1<\/sub>2<\/sup>h1<\/sub><\/p>\n\n\n\n

= 22\/7 x 15 x 15 x 40 cm3<\/sup><\/p>\n\n\n\n

= 198000\/7 cm3<\/sup><\/p>\n\n\n\n

And,<\/p>\n\n\n\n

Volume of second cylinder = \u03c0r2<\/sub>2<\/sup>h2<\/sub><\/p>\n\n\n\n

= 22\/7 x 20 x 20 x 45<\/p>\n\n\n\n

= 396000\/7 cm3<\/sup><\/p>\n\n\n\n

So, total volume = (198000\/7 + 396000\/7) cm3<\/sup><\/p>\n\n\n\n

= 594000\/7 cm3<\/sup><\/p>\n\n\n\n

Now, volume of third cylinder = 594000\/7 cm3<\/sup><\/p>\n\n\n\n

And height = 30 cm<\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

= \u221a900 = 30 cm<\/p>\n\n\n\n

\u2234 Radius of the third cylinder = 30 cm<\/p>\n\n\n\n


14. A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m3<\/sup>.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Height of pole (h) = 7 m<\/p>\n\n\n\n

Diameter = 20 cm<\/p>\n\n\n\n

So, radius (r) = 20\/2 = 10 cm = 10\/100 = 1\/10 m<\/p>\n\n\n\n

And,<\/p>\n\n\n\n

Volume = \u03c0r2<\/sup>h<\/p>\n\n\n\n

= 22\/7 x 1\/10 x 1\/10 x 7 m3<\/sup><\/p>\n\n\n\n

= 22\/100 m3<\/sup><\/p>\n\n\n\n

Weight of wood = 225 kg per m3<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Total weight = 225 \u00d7 (22\/100) = 99\/2 = 49.5 kg<\/p>\n\n\n\n


15. A cylinder of maximum volume is cut from a wooden cuboid of length 30 cm and cross-section a square of side 14 cm. Find the volume of the cylinder and the volume of the wood wasted.
Solution:<\/strong><\/p>\n\n\n\n

A cylinder of the maximum volume is cut from a wooden cuboid of length 30 cm and cross-section a square side 14 cm.<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So,<\/p>\n\n\n\n

Diameter of the cylinder = 14 cm<\/p>\n\n\n\n

\u21d2 Radius (r) = 14\/2 = 7 cm<\/p>\n\n\n\n

and height (h) = 30 cm<\/p>\n\n\n\n

Volume of cuboid = 30 \u00d7 14 \u00d7 14 = 5880 cm3<\/sup><\/p>\n\n\n\n

Volume of cylinder = \u03c0r2<\/sup>h<\/p>\n\n\n\n

= 22\/7 \u00d7 7 \u00d7 7 \u00d7 30<\/p>\n\n\n\n

= 4620 cm3<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

The wastage of wood = 5880 \u2013 4620 = 1260 cm3<\/sup><\/p>\n\n\n\n

Exercise 18.4<\/h4>\n\n\n\n

1. The surface area of a cube is 384 cm2<\/sup>. Find
(i) the length of an edge
(ii) volume of the cube.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Surface area of a cube = 384 cm2<\/sup><\/p>\n\n\n\n

(i) Surface area of cube = 6(side)2<\/sup><\/p>\n\n\n\n

Hence, edge (side) = \u221a(surface area\/6)<\/p>\n\n\n\n

= \u221a(384\/6)<\/p>\n\n\n\n

= \u221a64<\/p>\n\n\n\n

= 8 cm<\/p>\n\n\n\n

(ii) Volume = (Edge)3<\/sup> = (8)3 <\/sup>= 8 \u00d7 8 \u00d7 8 cm3<\/sup> = 512 cm3<\/sup><\/p>\n\n\n\n


2. Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of n.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Radius of a solid cylinder (r) = 5 cm<\/p>\n\n\n\n

Height (h) = 10 cm<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Total surface area = 2\u03c0rh + 2\u03c0r2<\/sup><\/p>\n\n\n\n

= 2r\u03c0(h + r)<\/p>\n\n\n\n

= 2\u03c0 \u00d7 5(10 + 5)<\/p>\n\n\n\n

= \u03c0 \u00d7 10 \u00d7 15<\/p>\n\n\n\n

= 150\u03c0 cm2<\/sup><\/p>\n\n\n\n

\"ML
3. An aquarium is in the form of a cuboid whose external measures are 70 cm \u00d7 28 cm \u00d7 35 cm. The base, side faces and back face are to be covered with coloured paper. Find the area of the paper needed.
Solution:<\/strong><\/p>\n\n\n\n

Given, a cuboid shaped aquarium<\/p>\n\n\n\n

Length (l) = 70 cm<\/p>\n\n\n\n

Breadth (b) = 28 cm<\/p>\n\n\n\n

and height (h) = 35 cm<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of base = 70 \u00d7 28 cm3<\/sup><\/p>\n\n\n\n

= 1960 cm3<\/sup><\/p>\n\n\n\n

Area of side face = (28 \u00d7 35) \u00d7 2 cm2 <\/sup><\/p>\n\n\n\n

= 1960 cm2<\/sup><\/p>\n\n\n\n

Area of back face = 70 \u00d7 35 cm2<\/sup><\/p>\n\n\n\n

= 2450 cm<\/p>\n\n\n\n

Thus, the total area = 1960 + 1960 + 2450 = 6370 cm2<\/sup><\/p>\n\n\n\n

Hence, area of paper required is 6370 cm2<\/sup>.<\/p>\n\n\n\n


4. The internal dimensions of rectangular hall are 15 m \u00d7 12 m \u00d7 4 m. There are 4 windows each of dimension 2 m \u00d7 1.5 m and 2 doors each of dimension 1.5 m \u00d7 2.5 m. Find the cost of white washing all four walls of the hall, if the cost of white washing is \u20b95 per m2<\/sup>. What will be the cost of white washing if the ceiling of the hall is also white washed?
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Internal dimension of rectangular hall = 15m \u00d7 12 m \u00d7 4 m<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of 4-walls = 2(l + b) \u00d7 h<\/p>\n\n\n\n

= 2(15 + 12) \u00d7 4<\/p>\n\n\n\n

= 2 \u00d7 27 \u00d7 4 m2<\/sup><\/p>\n\n\n\n

= 216 m2<\/sup><\/p>\n\n\n\n

Area of 4 windows of size = (2 \u00d7 1.5) \u00d7 4 = 12 m2<\/sup><\/p>\n\n\n\n

Area of 2 door of size = 2 \u00d7 (1.5 \u00d7 2.5) = 7.5 m2<\/sup><\/p>\n\n\n\n

So, area of remaining hall = 216 \u2013 (12 + 7.5) = 216 \u2013 19.5 m2<\/sup> = 196.5 m2<\/sup><\/p>\n\n\n\n

And,<\/p>\n\n\n\n

Cost of white washing the walls all four halls of the house is at the rate of \u20b95 per m2
<\/sup>
= 196.5 \u00d7 5 = \u20b9982.50<\/p>\n\n\n\n

Area of ceiling = l \u00d7 b = 15 \u00d7 12 = 180 m2<\/sup><\/p>\n\n\n\n

Cost of white washing = 180 \u00d7 5 = \u20b9900<\/p>\n\n\n\n

Therefore, the total cost for white washing = \u20b9982.50 + 900.00<\/p>\n\n\n\n

= \u20b91882.50<\/p>\n\n\n\n


5. A swimming pool is 50 m in length, 30 m in breadth and 2.5 m in depth. Find the cost of cementing its floor and walls at the rate of \u20b927 per square metre.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Length of swimming pool = 50 m<\/p>\n\n\n\n

Breadth of swimming pool = 30 m<\/p>\n\n\n\n

Depth (Height) of swimming pool = 2\u00b75 m<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of floor = 50 \u00d7 30 = 1500 m2<\/sup><\/p>\n\n\n\n

Area of four walls = 2 (50 + 30) \u00d7 2.5 = 160 \u00d7 2.5 = 400 m2<\/sup><\/p>\n\n\n\n

So, the area to be cemented = 1500 m2<\/sup> + 400 m2<\/sup> = 1900 m2<\/sup><\/p>\n\n\n\n

Cost of cementing 1m2<\/sup> = \u20b927<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Cost of cementing 1900m2 <\/sup>= \u20b927 \u00d7 1900 = \u20b951300<\/p>\n\n\n\n


6. The floor of a rectangular hall has a perimeter 236 m. Its height is 4\u00b75 m. Find the cost of painting its four walls (doors and windows be ignored) at the rate of Rs. 8.40 per square metre.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Perimeter of Hall = 236 m.<\/p>\n\n\n\n

Height = 4.5 m<\/p>\n\n\n\n

Perimeter = 2 (l + b) = 236 m<\/p>\n\n\n\n

Area of four walls = 2 (l + b) \u00d7 h<\/p>\n\n\n\n

= 236 \u00d7 4.5<\/p>\n\n\n\n

= 1062 m2<\/sup><\/p>\n\n\n\n

We have, cost of painting 1 m2<\/sup> = \u20b98.40<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Cost of painting 1062 m2<\/sup> = \u20b98.40 \u00d7 1062 = \u20b98920.80<\/p>\n\n\n\n


7. A cuboidal fish tank has a length of 30 cm, a breadth of 20 cm and a height of 20 cm. The tank is placed on a horizontal table and it is three-quarters full of water. Find the area of the tank which is in contact with water.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Length of tank = 30 cm<\/p>\n\n\n\n

Breadth of tank = 20 cm<\/p>\n\n\n\n

Height of tank = 20 cm<\/p>\n\n\n\n

As the tank is three-quarters full of water<\/p>\n\n\n\n

So, the height of water in the tank = (20 x 3)\/4 = 15 cm<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Area of the tank in contact with the water = Area of floor of Tank + Area of 4 walls upto 15 cm<\/p>\n\n\n\n

= 30 \u00d7 20 + 2 (30 + 20) \u00d7 15<\/p>\n\n\n\n

= 600 + 2 \u00d7 50 \u00d7 15<\/p>\n\n\n\n

= 600 + 1500 = 2100 cm2<\/sup><\/p>\n\n\n\n


8. The volume of a cuboid is 448 cm3<\/sup>. Its height is 7 cm and the base is a square. Find
(i) a side of the square base
(ii) surface area of the cuboid.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Volume of a cuboid = 448 cm3<\/sup><\/p>\n\n\n\n

Height = 7 cm<\/p>\n\n\n\n

So, area of base = 448\/7 = 64 cm2<\/sup><\/p>\n\n\n\n

Thus, the base is a square.<\/p>\n\n\n\n

(i) Side of square base = \u221a64 = 8 cm<\/p>\n\n\n\n

(ii) Surface area of the cuboid = 2 [lb + bh + hl]<\/p>\n\n\n\n

= 2[8 \u00d7 8 + 8 \u00d7 7 + 7 \u00d7 8] cm2<\/sup><\/p>\n\n\n\n

= 2[64 + 56 + 56]<\/p>\n\n\n\n

= 2 \u00d7 176 = 352 cm2<\/sup><\/p>\n\n\n\n


9. The length, breadth and height of a rectangular solid are in the ratio 5 : 4 : 2. If its total surface area is 1216 cm2<\/sup>, find the volume of the solid.
Solution:<\/strong><\/p>\n\n\n\n

Given that the ratio in length, breadth and height of a rectangular solid = 5 : 4 : 2<\/p>\n\n\n\n

Total surface area =1216 cm2<\/sup><\/p>\n\n\n\n

Let\u2019s assume the length = 5x, breadth = 4x and height = 2x<\/p>\n\n\n\n

Total surface area = 2[5x \u00d7 4x + 4x \u00d7 2x + 2x \u00d7 5x]<\/p>\n\n\n\n

= 2[20x2<\/sup> + 8x2<\/sup> + 10x2<\/sup> ]<\/p>\n\n\n\n

= 2 \u00d7 38x2<\/sup><\/p>\n\n\n\n

= 76x2<\/sup><\/p>\n\n\n\n

So, 76x2<\/sup> = 1216<\/p>\n\n\n\n

\u21d2 x2<\/sup> = 1216\/76 = 16 = (4)2<\/sup><\/p>\n\n\n\n

\u2234 x = 4<\/p>\n\n\n\n

Hence, the dimensions of the rectangular solid are<\/p>\n\n\n\n

Length = 5 \u00d7 4 = 20 cm<\/p>\n\n\n\n

Breadth = 4 \u00d7 4 = 16 cm<\/p>\n\n\n\n

Height = 2 \u00d7 4 = 8 cm<\/p>\n\n\n\n

and volume = lbh = 20 \u00d7 16 \u00d7 8 = 2560 cm3<\/sup><\/p>\n\n\n\n


10. A rectangular room is 6 m long, 5 m wide and 3.5 m high. It has 2 doors of size 1\u00b71 m by 2 m and 3 windows of size 1.5 m by 1.4 m. Find the cost of whitewashing the walls and the ceiling of the room at the rate of \u20b95.30 per square metre.
Solution:<\/strong><\/p>\n\n\n\n

Given, Length of room = 6 m<\/p>\n\n\n\n

Breadth of room = 5 m<\/p>\n\n\n\n

Height of room = 3.5 m<\/p>\n\n\n\n

So, Area of four walls = 2 (l + b) \u00d7 h<\/p>\n\n\n\n

= 2 (6 + 5) \u00d7 3.5<\/p>\n\n\n\n

= 77 m2<\/sup><\/p>\n\n\n\n

Area of 2 doors and 3 windows = (2 \u00d7 1.1 \u00d7 2 + 3 \u00d7 1.5 \u00d7 1.4)<\/p>\n\n\n\n

= (44 + 6.3) m2<\/sup><\/p>\n\n\n\n

= 10.7 m2<\/sup><\/p>\n\n\n\n

Area of ceiling = l \u00d7 b = 6 \u00d7 5 = 30 m2<\/sup><\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

Total area for white washing = (77 \u2013 10.7 + 30) m2<\/sup> = 96.3 m2<\/sup><\/p>\n\n\n\n

Hence, the cost of white washing = \u20b9(96.3 \u00d7 5.30) = \u20b9510.39<\/p>\n\n\n\n


11. A cuboidal block of metal has dimensions 36 cm by 32 cm by 0\u00b725 m. It is melted and recast into cubes with an edge of 4 cm.
(i) How many such cubes can be made?
(ii) What is the cost of silver coating the surfaces of the cubes at the rate of \u20b90\u00b775 per square centimetre?
Solution:<\/strong><\/p>\n\n\n\n

(i) Given, Length of cuboid = 36 cm<\/p>\n\n\n\n

Breadth of cuboid = 32 cm<\/p>\n\n\n\n

Height of cuboid = 0\u00b725 \u00d7 100 = 25 cm<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Volume of cuboid = lbh<\/p>\n\n\n\n

= (36 \u00d7 32 \u00d7 25) cm3<\/sup><\/p>\n\n\n\n

= 28800 cm3<\/sup><\/p>\n\n\n\n

Volume of cube = (side)2 <\/sup><\/p>\n\n\n\n

= (4)2<\/sup><\/p>\n\n\n\n

= 64 cm2<\/sup><\/p>\n\n\n\n

Hence, the number of cubes recasting from cuboid = 28800\/64 = 450<\/p>\n\n\n\n

(ii) Surface area of 1 cube = 6 \u00d7 a2 <\/sup><\/p>\n\n\n\n

= 6 \u00d7 16<\/p>\n\n\n\n

= 96 cm2<\/sup><\/p>\n\n\n\n

So, the surface area of 450 cubes = 96 \u00d7 450<\/p>\n\n\n\n

= 43200 cm2<\/sup><\/p>\n\n\n\n

Hence, the cost of silver coating on cubes = \u20b90.75 \u00d7 43200<\/p>\n\n\n\n

= \u20b932400<\/p>\n\n\n\n


12. Three cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and recast into a single cube, find the cost of coating the surface of the new cube with gold at the rate of \u20b93.50 per square centimetre?
Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s consider a cm to be the edge of new cube<\/p>\n\n\n\n

Then, according to given conditions in the question<\/p>\n\n\n\n

a3<\/sup> = 33<\/sup> + 43<\/sup> + 53<\/sup><\/p>\n\n\n\n

= 27 + 64 + 125<\/p>\n\n\n\n

= 216 cm3<\/sup><\/p>\n\n\n\n

a =
\"ML<\/p>\n\n\n\n

\u2234 a = 6 cm<\/p>\n\n\n\n

So, the surface area of new cube = 6 \u00d7 (side)2 <\/sup><\/p>\n\n\n\n

= 6 \u00d7 (6)2<\/sup><\/p>\n\n\n\n

= 216 cm2<\/sup><\/p>\n\n\n\n

Hence, the cost of coating the surface of new cube = \u20b93.50 \u00d7 216 = \u20b9156<\/p>\n\n\n\n


13. The curved surface area of a hollow cylinder is 4375 cm2<\/sup>, it is cut along its height and formed a rectangular sheet of width 35 cm. Find the perimeter of the rectangular sheet.
Solution:<\/strong><\/p>\n\n\n\n

Given, curved surface area of a hollow cylinder = 4375 cm2<\/sup><\/p>\n\n\n\n

By cutting it from the height,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It becomes a rectangular sheet whose width = 35 cm<\/p>\n\n\n\n

So, the height of cylinder = 35 cm<\/p>\n\n\n\n

And, length of sheet = Area\/Height<\/p>\n\n\n\n

= 4375\/35<\/p>\n\n\n\n

= 125 cm<\/p>\n\n\n\n

Hence, Perimeter of the sheet = 2(l + b)<\/p>\n\n\n\n

= 2 \u00d7 (125 + 35)<\/p>\n\n\n\n

= 2 \u00d7 160<\/p>\n\n\n\n

= 320 cm<\/p>\n\n\n\n


14. A road roller has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must take in order to level a playground of size 120 m \u00d7 44 m.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Diameter of a road roller = 0.7 m = 70 cm<\/p>\n\n\n\n

So, radius (r) = 70\/2 cm = 35 cm = 35\/100 m<\/p>\n\n\n\n

and width (h) = 1.2 m<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Curved surface area = 2\u03c0rh<\/p>\n\n\n\n

= (2 \u00d7 22\/7 \u00d7 35\/100 \u00d7 1.2) m2<\/sup><\/p>\n\n\n\n

= 264\/100 m2<\/sup><\/p>\n\n\n\n

Area of playground = 120 m \u00d7 44 m<\/p>\n\n\n\n

= 120 \u00d7 44 m2<\/sup><\/p>\n\n\n\n

= 5280 m2<\/sup><\/p>\n\n\n\n

Hence, the number of revolution made by the road roller = (5280\/264) \u00d7 100<\/p>\n\n\n\n

= 2000 revolutions<\/p>\n\n\n\n


15. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Diameter of cylindrical container = 14 cm<\/p>\n\n\n\n

So, radius (r) = 14\/2 = 7 cm<\/p>\n\n\n\n

And, height (h) = 20 cm<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Width of label = 2 0 \u2013 (2 + 2) cm = 20 \u2013 4 = 16 cm<\/p>\n\n\n\n

Hence, area of label = 2\u03c0rh<\/p>\n\n\n\n

= 2 \u00d7 (22\/7) \u00d7 7 \u00d7 16<\/p>\n\n\n\n

= 704 cm2<\/sup><\/p>\n\n\n\n


16. The sum of the radius and height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm2<\/sup>. Find the height and the volume of the cylinder.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Sum of height and radius of a cylinder = 37 cm<\/p>\n\n\n\n

Total surface area = 1628 cm2<\/sup><\/p>\n\n\n\n

Let\u2019s consider the radius to be r<\/p>\n\n\n\n

Then, height = (37 \u2013 r) cm<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Total surface area = 2\u03c0(h + r)<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

\u21d2 r = 7 cm<\/p>\n\n\n\n

Height = 37 \u2013 7 = 30 cm<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Volume = \u03c0r2<\/sup>h<\/p>\n\n\n\n

= (22\/7) \u00d7 7 \u00d7 7 \u00d7 30 cm3 <\/sup><\/p>\n\n\n\n

= 4620 cm3<\/sup><\/p>\n\n\n\n


17. The ratio between the curved surface and total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm3<\/sup>.
Solution:<\/strong><\/p>\n\n\n\n

Given that ratio between curved surface and total surface area of a cylinder = 1 : 2<\/p>\n\n\n\n

Total surface area = 616 cm2<\/sup><\/p>\n\n\n\n

So, curved surface area = 616\/2 = 308 cm2<\/sup><\/p>\n\n\n\n

And,<\/p>\n\n\n\n

Area of two circular faces = 616 \u2013 308 = 308 cm2<\/sup><\/p>\n\n\n\n

Area of one circular face = 308\/2 = 154 cm2<\/sup><\/p>\n\n\n\n

Now, let\u2019s consider the radius to be r<\/p>\n\n\n\n

\u03c0r2<\/sup> = 154<\/p>\n\n\n\n

(22\/7) \u00d7 r2<\/sup> = 154<\/p>\n\n\n\n

r2<\/sup> = (154 \u00d7 7)\/ 22 = 49<\/p>\n\n\n\n

\u21d2 r = 49 = \u221a7 cm<\/p>\n\n\n\n

Hence, the volume = \u03c0r2<\/sup>h = (22\/7) \u00d7 7 \u00d7 7 \u00d7 7 = 1078 cm2<\/sup><\/p>\n\n\n\n


18. The given figure shown a metal pipe 77 cm long. The inner diameter of cross section is 4 cm and the outer one is 4.4 cm.
Find its
(i) inner curved surface area
(ii) outer curved surface area
(iii) total surface area.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Length of metal pipe (h) = 77 cm<\/p>\n\n\n\n

Inner diameter = 4 cm<\/p>\n\n\n\n

and outer diameter = 4.4 cm<\/p>\n\n\n\n

So, inner radius (r) = 4\/2 = 2 cm<\/p>\n\n\n\n

And outer radius (R) = 4.4\/2 = 2.2 cm<\/p>\n\n\n\n

(i) Inner curved surface area = 2\u03c0rh<\/p>\n\n\n\n

= 2 \u00d7 22\/7 \u00d7 2 \u00d7 77 cm2<\/sup><\/p>\n\n\n\n

= 968 cm2<\/sup><\/p>\n\n\n\n

(ii) Outer surface area = 2\u03c0RH<\/p>\n\n\n\n

= 2 \u00d7 22\/7 \u00d7 2.2 \u00d7 77 cm2<\/sup><\/p>\n\n\n\n

= 1064.8 cm2<\/sup><\/p>\n\n\n\n

(iii) Surface area of upper and lower rings = 2[\u03c0R2<\/sup> \u2013 \u03c0r2<\/sup>]<\/p>\n\n\n\n

= 2 \u00d7 22\/7 (2.22<\/sup> \u2013 22<\/sup>) cm2<\/sup><\/p>\n\n\n\n

= 44\/7 \u00d7 4.2 \u00d7 0.2<\/p>\n\n\n\n

= 5.28 cm2<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Total surface area = (968 + 1064.8 + 5.28) cm2<\/sup> = 2038.08 cm2<\/sup><\/p>\n\n\n\n

Check Your Progress<\/p>\n\n\n\n

1. A square field of side 65 m and rectangular field of length 75 m have the same perimeter. Which field has a larger area and by how much?
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Side of a square field = 65 m<\/p>\n\n\n\n

So, perimeter = 4 \u00d7 Side<\/p>\n\n\n\n

= 4 \u00d7 65<\/p>\n\n\n\n

= 260 m<\/p>\n\n\n\n

Now, perimeter of a rectangular field = 260 m<\/p>\n\n\n\n

And given, length (l) = 75 m<\/p>\n\n\n\n

Perimeter of rectangle = 2(l + b)<\/p>\n\n\n\n

\u21d2 260 = 2(75 + b)<\/p>\n\n\n\n

260 = 150 + 2b<\/p>\n\n\n\n

2b = 260 \u2013 150<\/p>\n\n\n\n

b = 110\/2 = 55 m<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Area of square field = (side)2<\/sup><\/p>\n\n\n\n

= (65)2<\/sup> m2 <\/sup><\/p>\n\n\n\n

= 4225 m2<\/sup><\/p>\n\n\n\n

And, area of rectangular field = l \u00d7 b<\/p>\n\n\n\n

= 75 \u00d7 55 m<\/p>\n\n\n\n

= 4125 m2<\/sup><\/p>\n\n\n\n

Hence, it is clear that area of square field is greater.<\/p>\n\n\n\n

Difference = 4225 \u2013 4125<\/p>\n\n\n\n

= 100 m2<\/sup><\/p>\n\n\n\n


2. The shape of a top surface of table is a trapezium. Find the area if its parallel sides are 1.5 m and 2.5 m and perpendicular distance between them is 0.8 m.
Solution:<\/strong><\/p>\n\n\n\n

\"ML
<\/strong>
Shape of the top of a table is trapezium and parallel sides are 1.5 m and 2.5 m and the perpendicular distance between them = 0.8m<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Area = \u00bd \u00d7 (Sum of parallel sides) \u00d7 height<\/p>\n\n\n\n

= \u00bd \u00d7 (1.5 + 2.5) \u00d7 0.8<\/p>\n\n\n\n

= \u00bd \u00d7 4 \u00d7 0.8 m2<\/sup><\/p>\n\n\n\n

= 1.6 m2<\/sup><\/p>\n\n\n\n


3. The length and breadth of a hall of a school are 26 m and 22 m respectively. If one student requires 1.1 sq. m area, then find the maximum number of students to be seated in this hall.
Solution:<\/strong><\/p>\n\n\n\n

Given, length of a school hall (l) = 26 m and breadth (b) = 22 m<\/p>\n\n\n\n

So, area = l \u00d7 b<\/p>\n\n\n\n

= 26 \u00d7 22 m2<\/sup><\/p>\n\n\n\n

= 572 m2<\/sup><\/p>\n\n\n\n

One student requires 1.1 sq. m area<\/p>\n\n\n\n

Hence, number of students = 572\/1.1<\/p>\n\n\n\n

= (572 \u00d7 10)\/11<\/p>\n\n\n\n

= 520 students<\/p>\n\n\n\n


4. It costs \u20b9936 to fence a square field at \u20b97.80 per metre. Find the cost of levelling the field at \u20b92.50 per square metre.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Cost of fencing the square field at \u20b97.80 per metre = \u20b9936.<\/p>\n\n\n\n

So, total fence required will be = 936\/7.80 = 120<\/p>\n\n\n\n

Thus, the perimeter of the field = 120 m<\/p>\n\n\n\n

\u21d2 4 \u00d7 Side = 120 m [Since, it\u2019s a square field]<\/p>\n\n\n\n

\u21d2 Side = 120\/4<\/p>\n\n\n\n

\u2234 Side = 30 m<\/p>\n\n\n\n

Hence, Area of square field = (30)2<\/sup> = 900 m2<\/sup><\/p>\n\n\n\n

= 900 \u00d7 2.50 = \u20b92250<\/p>\n\n\n\n


5. Find the area of the shaded portion in the following figures all measurements are given in cm.

Solution:<\/strong><\/p>\n\n\n\n

(i) Outer length = 30 cm<\/p>\n\n\n\n

Breadth = 10 cm<\/p>\n\n\n\n

Side of each rectangle of the corner (l) = (30 \u2013 18)\/ 2 = 6 cm<\/p>\n\n\n\n

and b = (10 \u2013 6)\/2 = 4\/2 = 2 cm<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Area of 4 comer = (6 \u00d7 2) \u00d7 4<\/p>\n\n\n\n

= 48 cm2<\/sup><\/p>\n\n\n\n

And area of inner rectangle = 18 \u00d7 6<\/p>\n\n\n\n

= 108 cm2<\/sup><\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Area of shaded portion = 108 + 48 = 156cm2<\/sup><\/p>\n\n\n\n

(ii) Area of rectangle I = 4 \u00d7 2 = 8 cm2<\/sup><\/p>\n\n\n\n

Area of rectangle II = 4 \u00d7 1 = 4 cm2<\/sup><\/p>\n\n\n\n

Area of rectangle III = 6 \u00d7 l = 6 cm2<\/sup><\/p>\n\n\n\n

and area of square IV = 1 \u00d7 1 = 1 Cm2<\/sup><\/p>\n\n\n\n

\u2234 Total area of shaded portion = 8 + 4 + 6 + 1 = 19 cm2<\/sup><\/p>\n\n\n\n


6. Area of a trapezium is 160 sq. cm. Lengths of parallel sides are in the ratio 1:3. If smaller of the parallel sides is 10 cm in length, then find the perpendicular distance between them.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Area of trapezium = 160 cm2<\/sup><\/p>\n\n\n\n

Ratio of the length of its parallel sides = 1 : 3<\/p>\n\n\n\n

Smaller parallel side = 10 cm<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Length of greater side = 10 \u00d7 3 = 30 cm<\/p>\n\n\n\n

Now, distance between them = h<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n


7. The area of a trapezium is 729 cm2<\/sup> and the distance between two parallel sides is 18 cm. If one of its parallel sides is 3 cm shorter than the other parallel side, find the lengths of its parallel sides.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Area of a trapezium = 729 cm2<\/sup><\/p>\n\n\n\n

Distance between two parallel sides (Altitude) = 18 cm<\/p>\n\n\n\n

So, the sum of parallel sides = (Area \u00d7 2)\/ Altitude<\/p>\n\n\n\n

= (729 \u00d7 2)\/ 18<\/p>\n\n\n\n

= 81 cm<\/p>\n\n\n\n

One parallel side is shorter than the second by 3 cm<\/p>\n\n\n\n

Let the longer side be taken as x<\/p>\n\n\n\n

Then, shorter side = x \u2013 3<\/p>\n\n\n\n

According to the question, we have<\/p>\n\n\n\n

\u21d2 x + x \u2013 3 = 81<\/p>\n\n\n\n

2x = 81 + 3 = 84<\/p>\n\n\n\n

x = 84\/2 = 42<\/p>\n\n\n\n

Hence, the longer side = 42 cm and shorter side = 42 \u2013 3 = 39 cm<\/p>\n\n\n\n


8. Find the area of the polygon given in the figure:

Solution:<\/strong><\/p>\n\n\n\n

In the given figure,<\/p>\n\n\n\n

AC = 60 m, AH = 46 m, AF = 16 m, EF = 24 m, DH = 14 m, BG = 16 m<\/p>\n\n\n\n

\u2234 FH = AH \u2013 AF = 46 \u2013 16 = 30<\/p>\n\n\n\n

And, HC = AC \u2013 AH = 60 \u2013 46 = 14<\/p>\n\n\n\n

In the figure, there are 3 triangles and one trapezium.<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Area of \u2206ABC = \u00bd AC \u00d7 BG<\/p>\n\n\n\n

= \u00bd \u00d7 60 \u00d7 16<\/p>\n\n\n\n

= 480 m2<\/sup><\/p>\n\n\n\n

Area of \u2206AEF = \u00bd AF \u00d7 EF<\/p>\n\n\n\n

= \u00bd \u00d7 16 \u00d7 24<\/p>\n\n\n\n

= 192 m2<\/sup><\/p>\n\n\n\n

Area of \u2206DHC = \u00bd HC \u00d7 DH<\/p>\n\n\n\n

= \u00bd \u00d7 14 \u00d7 14<\/p>\n\n\n\n

= 98 m2<\/sup><\/p>\n\n\n\n

Area of trapezium EFHD = \u00bd (EF + DH) \u00d7 FH<\/p>\n\n\n\n

= \u00bd (24 + 14) \u00d7 30<\/p>\n\n\n\n

= \u00bd \u00d7 38 \u00d7 30<\/p>\n\n\n\n

= 570 m2<\/sup><\/p>\n\n\n\n

Therefore, total area of the figure = Area of \u2206ABC + area \u2206AEF + area \u2206DHC + area trapezium EFHD<\/p>\n\n\n\n

= 480 + 192 + 98 + 570<\/p>\n\n\n\n

= 1340 m2<\/sup><\/p>\n\n\n\n


9. The diagonals of a rhombus are 16 m and 12 m, find:
(i) its area
(ii) length of a side
(iii) perimeter.
Solution:<\/strong><\/p>\n\n\n\n

Diagonals of a rhombus are d1<\/sub> = 16 cm and d2<\/sub> = 12 cm

<\/strong>
(i) Area = (d1<\/sub> \u00d7 d2<\/sub>) \/ 2<\/p>\n\n\n\n

= (16 \u00d7 12)\/ 2<\/p>\n\n\n\n

= 96 cm2<\/sup><\/p>\n\n\n\n

(ii) As the diagonals of rhombus bisect each other at right angles, we have<\/p>\n\n\n\n

AO = OC and BO = OD<\/p>\n\n\n\n

AO = 16\/2 = 8 cm and BO = 12\/2 = 6 cm<\/p>\n\n\n\n

Now, in right \u2206AOB<\/p>\n\n\n\n

AB2<\/sup> = AO2<\/sup> + BO2<\/sup>  [By Pythagoras Theorem]<\/p>\n\n\n\n

= 82<\/sup> + 62<\/sup><\/p>\n\n\n\n

= 64 + 36<\/p>\n\n\n\n

= 100 = (10)2<\/sup><\/p>\n\n\n\n

\u2234 AB = 10 cm<\/p>\n\n\n\n

Therefore, the side of rhombus = 10 cm<\/p>\n\n\n\n


10. The area of a parallelogram is 98 cm2<\/sup>. If one altitude is half the corresponding base, determine the base and the altitude of the parallelogram.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Area of a parallelogram = 98 cm2<\/sup><\/p>\n\n\n\n

One altitude = Half of its corresponding base<\/p>\n\n\n\n

Let\u2019s consider the base as x cm<\/p>\n\n\n\n

Then altitude = x\/2 cm<\/p>\n\n\n\n

So, area = Base \u00d7 Altitude<\/p>\n\n\n\n

\u21d2 98 = x \u00d7 (x\/2<\/p>\n\n\n\n

x2<\/sup> = 98 \u00d7 2<\/p>\n\n\n\n

= 196<\/p>\n\n\n\n

= (14)2<\/sup><\/p>\n\n\n\n

\u2234 x = 14<\/p>\n\n\n\n

Therefore, Base = 14 cm and altitude = = 7 cm<\/p>\n\n\n\n


11. Preeti is painting the walls and ceiling of a hall whose dimensions are 18 m \u00d7 15 m \u00d7 5 m. From each can of paint 120 m2<\/sup> of area is painted. How many cans of paint does she need to paint the hall?
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Length of a hall (l) = 18 m<\/p>\n\n\n\n

Breadth (b) = 15m and<\/p>\n\n\n\n

height (h) = 5 m<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Area of 4-wall and ceiling = 2(l + b)h + lb<\/p>\n\n\n\n

= 2(18 + 15) \u00d7 5 + 18 \u00d7 15 m2<\/sup><\/p>\n\n\n\n

= 2 \u00d7 33 \u00d7 5 + 270<\/p>\n\n\n\n

= 330 + 270<\/p>\n\n\n\n

= 600 m2<\/sup><\/p>\n\n\n\n

From 1 can an area of 120 m2<\/sup> can be painted<\/p>\n\n\n\n

Hence, total number of cans required to paint the area of 600 m2<\/sup> = 600\/120 = 5 cans<\/p>\n\n\n\n


12. A rectangular paper is size 22 cm \u00d7 14 cm is rolled to form a cylinder of height 14 cm, find the volume of the cylinder. (Take \u03c0 = 22\/7)
Solution:<\/strong><\/p>\n\n\n\n

Given,
<\/strong>
Length of a rectangular paper = 22 cm and breadth = 14 cm<\/p>\n\n\n\n

By rolling it a cylinder is formed whose height is 14 cm<\/p>\n\n\n\n

And, circumference of the base = 22 cm<\/p>\n\n\n\n

We know that, circumference = 2\u03c0r<\/p>\n\n\n\n

So, radius (r) = C\/2\u03c0 = (22 \u00d7 7)\/(2 \u00d7 22) = 7\/2 cm<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Volume of the cylinder so formed = \u03c0r2<\/sup>h<\/p>\n\n\n\n

= 22\/7 \u00d7 7\/2 \u00d7 7\/2 \u00d7 14<\/p>\n\n\n\n

= 539cm3<\/sup><\/p>\n\n\n\n


13. A closed rectangular wooden box has inner dimensions 90 cm by 80 cm by 70 cm. Compute its capacity and the area of the tin foil needed to line its inner surface.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Inner length of rectangular box = 90 cm<\/p>\n\n\n\n

Inner breadth of rectangular box = 80 cm<\/p>\n\n\n\n

Inner height of rectangular box = 70 cm<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Capacity of rectangular box = Volume of rectangular box<\/p>\n\n\n\n

= l \u00d7 b \u00d7 h<\/p>\n\n\n\n

= 90 cm \u00d7 80 cm \u00d7 70 cm<\/p>\n\n\n\n

= 504000 cm3<\/sup><\/p>\n\n\n\n

And,<\/p>\n\n\n\n

Required area of tin foil = 2 (lb + bh + lh)<\/p>\n\n\n\n

= 2(90 \u00d7 80 + 80 \u00d7 70 + 90 \u00d7 70) cm2<\/sup><\/p>\n\n\n\n

= 2(7200 + 5600 + 6300) cm2<\/sup><\/p>\n\n\n\n

= 2 \u00d7 19100 cm2<\/sup><\/p>\n\n\n\n

= 38200 cm2<\/sup><\/p>\n\n\n\n


14. The lateral surface area of a cuboid is 224 cm2<\/sup>. Its height is 7 cm and the base is a square. Find
(i) side of the square base
(ii) the volume of the cuboid.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Lateral surface area of a cuboid is 224 cm2<\/sup><\/p>\n\n\n\n

Height (h) = 7 cm<\/p>\n\n\n\n

(i) 2(l + b) \u00d7 h = 224<\/p>\n\n\n\n

\u21d2 2(l + b) \u00d7 7 = 224<\/p>\n\n\n\n

l + b = 224\/14 = 16 cm<\/p>\n\n\n\n

But l = b [Since, the base of cuboid is a square]<\/p>\n\n\n\n

So, 2 \u00d7 side = 16 cm<\/p>\n\n\n\n

\u21d2 Side = 16\/2 = 8 cm<\/p>\n\n\n\n

(ii) Volume of cuboid = lbh = 8 \u00d7 8 \u00d7 7 cm3<\/sup> = 448 cm3<\/sup><\/p>\n\n\n\n


15. The inner dimensions of a closed wooden box are 2 m by 1.2 m by 0.75 m. The thickness of the wood is 2.5 cm. Find the cost of wood required to make the box if 1 m3<\/sup> of wood costs \u20b95400.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Inner dimensions of wooden box are 2 m, 1.2 m, 0.75 m<\/p>\n\n\n\n

Thickness of the wood = 2.5 cm = 2.5\/100 m = 0.025 m
<\/strong>
Now,<\/p>\n\n\n\n

External dimensions of wooden box are<\/p>\n\n\n\n

= (2 + 2 \u00d7 0.025), (1.2 + 2 \u00d7 0.025), (0.75 + 2 \u00d7 0.025)<\/p>\n\n\n\n

= (2 + 0.05), (1.2 + 0.05), (0.75 + 0.5)<\/p>\n\n\n\n

= 2.05, 1.25, 0.80<\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

Volume of solid = External volume of box \u2013 Internal volume of box<\/p>\n\n\n\n

= 2.05 \u00d7 1.25 \u00d7 0.80 m3<\/sup> \u2013 2 \u00d7 1.2 \u00d7 0.75m3<\/sup><\/p>\n\n\n\n

= 2.05 \u2013 1.80 = 0.25 m3<\/sup><\/p>\n\n\n\n

Given, the cost of wood = \u20b95400 for 1 m3<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Total cost = \u20b95400 \u00d7 0.25 = \u20b95400 \u00d7 25\/100<\/p>\n\n\n\n

= \u20b954 \u00d7 25<\/p>\n\n\n\n

= \u20b91350<\/p>\n\n\n\n


16. A car has a petrol tank 40 cm long, 28 cm wide and 25 cm deep. If the fuel consumption of the car averages 13.5 km per litre, how far can the car travel with a full tank of petrol?
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Capacity of car tank = 40 cm \u00d7 28 cm \u00d7 25 cm = (40 \u00d7 28 \u00d7 25) cm3<\/sup><\/p>\n\n\n\n

= (40 \u00d7 28 \u00d7 25)\/1000 litre [\u2235 1000 cm3<\/sup> = 1 litre]<\/p>\n\n\n\n

Average fuel consumption of car = 13.5 km per litres<\/p>\n\n\n\n

Then, the distance travelled by car is given by<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

= 14 \u00d7 27 km<\/p>\n\n\n\n

= 378 km<\/p>\n\n\n\n

Hence, the car can travel 378 km with a full tank of petrol.<\/p>\n\n\n\n


17. The diameter of a garden roller is 1.4 m and it is 2 m long. How much area it will cover in 5 revolutions?
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Diameter of a garden roller = 1.4 m<\/p>\n\n\n\n

So, its radius (r) = 1.4\/2 = 0.7 m = 70 cm<\/p>\n\n\n\n

and length (h) = 2m<\/p>\n\n\n\n

Now, Curved surface area = 2\u03c0rh<\/p>\n\n\n\n

= 2 \u00d7 22\/7 \u00d7 70 \u00d7 200 cm2<\/sup><\/p>\n\n\n\n

= 88000 cm2<\/sup><\/p>\n\n\n\n

Hence, area covered in 5 revolutions = (88000 \u00d7 5)\/10000 m2<\/sup> = 44 m2<\/sup><\/p>\n\n\n\n


18. The capacity of an open cylindrical tank is 2079 m3<\/sup> and the diameter of its base is 21m. Find the cost of plastering its inner surface at \u20b940 per square metre.
Solution:<\/strong><\/p>\n\n\n\n

Given, capacity of an open cylindrical tank = 2079 m3<\/sup><\/p>\n\n\n\n

Diameter of base = 21 m<\/p>\n\n\n\n

So, radius (r) = 21\/2 m<\/p>\n\n\n\n

Let h be the height, then we have<\/p>\n\n\n\n

\u03c0r2<\/sup>h = 2079<\/p>\n\n\n\n

22\/7 \u00d7 21\/2 \u00d7 21\/2 \u00d7 h = 2079<\/p>\n\n\n\n

h = (2079 \u00d7 2)\/ (11 \u00d7 3 \u00d7 21) = 6 m<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Curved Surface Area + Base area = 2\u03c0rh + \u03c0r2<\/sup><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Hence, the cost of plastering the surface = \u20b940 \u00d7 742.5 = \u20b929700<\/p>\n\n\n\n


19. A solid right circular cylinder of height 1.21 m and diameter 28 cm is melted and recast into 7 equal solid cubes. Find the edge of each cube.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Height of solid right circular cylinder = 1.21 m = 121 cm<\/p>\n\n\n\n

and diameter = 28 cm<\/p>\n\n\n\n

So, radius (r) = 28\/2 = 14 cm<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Volume of the metal used = \u03c0r2<\/sup>h<\/p>\n\n\n\n

= 22\/7 \u00d7 14 \u00d7 14 \u00d7 121 cm3<\/sup><\/p>\n\n\n\n

= 74536 cm3<\/sup><\/p>\n\n\n\n

Thus, the volume of 7 solid cubes = 74536 cm3<\/sup><\/p>\n\n\n\n

And, volume of 1 cube = 74536\/7 = 10648 cm3<\/sup><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Hence, the edge of each cube is 22 cm.<\/p>\n\n\n\n

20. (i) How many cubic metres of soil must be dug out to make a well 20 m deep and 2 m in diameter?
(ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of \u20b950 per m2<\/sup>, find the cost of plastering.
Solution:<\/strong><\/p>\n\n\n\n

(i) Given,<\/p>\n\n\n\n

Depth of a well (h) = 20 m<\/p>\n\n\n\n

and diameter = 2 m<\/p>\n\n\n\n

Radius (r) = 2\/2 = 1 m<\/p>\n\n\n\n

Volume of earth dug out = \u03c0r2<\/sup>h<\/p>\n\n\n\n

= 22\/7 \u00d7 1 \u00d7 1 \u00d7 20<\/p>\n\n\n\n

= 440\/7 m3<\/sup><\/p>\n\n\n\n

(ii) Inner curved surface area = 2\u03c0rh<\/p>\n\n\n\n

= 2 \u00d7 22\/7 \u00d7 1 \u00d7 20<\/p>\n\n\n\n

= 880\/7 m2<\/sup><\/p>\n\n\n\n

The cost of plastering at the rate of \u20b950 per m2<\/sup> = \u20b9 880\/7 \u00d7 50<\/p>\n\n\n\n

= \u20b9 44000\/7<\/p>\n\n\n\n

= \u20b9 6285.70<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 18- Mensuration<\/p>\n\n\n\n

Download PDF<\/strong>: ML Aggarwal Solutions for Class 8 Maths Chapter 18- Mensuration PDF<\/a><\/p>\n\n\n\n

Chapterwise ML Aggarwal Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n