NCERT Solution For Mensuration

Sol. Let x be the breadth of the rectangle. It is given that the perimeter of a rectangle = perimeter of the square.

∴ 2(80 + x) = 4 × 60

⇒80 + x = 120

⇒x = 120 – 80 = 40

i.e. breadth of the rectangle = 40 m

Now, Area of the square = (60 × 60) m^{2}

= 3600 m^{2}

and the area of the rectangle = (40 × 80) m^{2}

= 3200 m^{2}

Hence, the square field has a larger area.

Sol. Area of the garden = Area of the outer square

� Area of the inner rectangle

= 25 × 25 m^{2} � 20 × 15 m^{2}

(625 � 300) m^{2} = 325 m^{2}

Cost of developing a garden @ Rs. 55 per sq. meter

= Rs. (55 × 325) = Rs. 17875

Sol. Total area of the garden

= Area of the rectangular portion + The sum of the areas of the pair of semi circles

Perimeter of the garden

= 2 × length of rectangular portion + circumference of circle

Sol. Area of one tile = base × height

= (24 × 10) cm^{2}

= 240 cm^{2}

Number of tiles required to cover the floor

Sol. Let us mark points A, B, C and D in the given figures as shown. Let A be the point in each figure from where the ant start moving on the food pieces. She is to reach the initial point after moving around the boundary of each food piece.

For food piece (a)

Distance moved = Arc AB + BA

Sol. Area of top surface of a table

= Area of the trapezium

Sol. Let the required side be x cm.

Then, area of the trapezium

But, the area of the trapezium = 34 cm^{2} (given) ∴ 2 (10 + x) = 34

⇒10 + × = 17

⇒x = 17 – 10 = 7

Hence, the outer side = 7 cm

Sol. Let ABCD be the given trapezium in which BC 48 m, CD = 17 m and AD = 40 m

Through D, draw DL ⊥ BC.

Now, BL = AD = 40 m

and LC = BC – BL = (48 – 40) m = 8 m

Applying pythagoras theorem in right ΔDLC, we have

DL^{2} = DC^{2} – LC^{2} = 17^{2} – 8^{2} = 289 – 64 = 225 ⇒ DL

Now, area of the trapezium ABCD

= (44 × 15) m^{2} = 660 m^{2}.

Sol. Let ABCD be the given quadrilateral in which BE ⊥ AC and DF ⊥ AC.

It is given that

AC = 24 m, BE = 8 m and DF= 13 m.

Now, area of quad. ABCD

= area of ΔABC + area of ΔACD

= (12 × 8 + 12 × 13) m^{2}

= (96 + 156) m^{2} = 252 m^{2}

Sol.

Sol. Let ABCD be a rhombus of side 6 cm and whose altitude DE = 4 cm. Also, one of its diagonals, BD = 8 cm.

Area of the rhombus ABCD

Hence, the other diagonal is 6 cm.

Sol. Area of the floor = 3000 � Area of one tile

Sol. Let the parallel sides of the trapezium shaped field be x m and 2x m. Then, its area.

∴The length of the side along the river is 2 × 70, i.e., 140 metres.

Sol. Area of the octagonal surface ABCDEFGH = Area (trap. ARCH) + Area (rect. HCDG) + Area (trap. GDEF)

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Sol. Taking Jyoti’s diagram:

Area of the pentagonal shaped park

= 2 × Area of trapezium ABEF

Taking Kavita’s diagram:

Area of the pentagonal shaped park

Sol. Total surface area of first box

= 2(lb + bh + lh)

= 2(60 × 40 + 40 × 50 + 60 × 50)cm^{2}

= 200(24 + 20 + 30) cm^{2}

= 200 × 74 cm^{2} = 14800 cm^{2}

Total surface area of second box

= 6 (Edge)2 = 6 × 50 × 50 cm^{2}

= 15000 cm^{2}

Since the total surface area of first box is less than that of the second, therefore the first box i.e., (a) requires the least amount of material to make.

Sol. Total surface area of suitcase

= 2[(80) (48) + (48) (24) + (24) (80))

= 2[3840 + 1152 + 1920]

= 13824 cm^{2}

Total surface area of 100 suitcase

= (13824 × 100) cm^{2} =1382400 cm^{2}

Required tarpaulin = Length × Breadth

1382400 cm^{2} = Length x96 cm

Thus 144 m of tarpaulin is required to cover 100 suitcases.

Sol. Let a be the side of the cube having surface area 600 cm^{2}.

∴ 6a^{2} = 600 ⇒ a2 = 100 ⇒ a = 10

Hence, the side of the cube = 10 cm.

Sol. Here l = 2m, b = 1m and h = 1.5 m

Area to be painted

= 2bh + 2lh + lb

(2 × 1 × 1.5 + 2 × 2 × 1.5 + 2 × 1)m^{2}

= (3 + 6 + 2) m^{2} = 11m^{2}

How many cans of paint will she need to paint the room?

Sol. Here l = 15m, b = 10 m and h = 7 m

Area to be painted

= 2bh + 2lh + lb

= 2 × 10 × 7 + 2 × 15 × 7 + 15 × 10)m^{2}

= (140 + 210 150) m^{2} = 500m^{2}

Since each can of paint covers 100 m^{2}, therefore number of cans required

Sol. Try yourself

Sol. Here, r = 7m and h = 3 m.

Sheet of metal required to make a closed cylinder = Total surface area of the cylinder. = (2πrh + 2πr^{2}) sq. units.

Sol. A hollow cylind :r is cut alcng its height to form a rectangular sheet.

Area of cylinder = Area of rectangular sheet 4224 cm^{2} = 33 cm × Length

Thus, the length of the rectangular sheet is 128 cm. Perimeter of the rectangular sheet

= 2 (Length + Width)

= [2(128 + 33)] cm

=(2 × 161)cm

= 322 cm

Sol. In one revolution, the roller will cover an area equal to its lateral surface area. Thus, in 1 revolution, area of the road covered = 27πrh

Sol. Since the company places a label around the surface of the cylindrical container of radius 7 cm and height 20 cm such that it is placed 2 cm from top and bottom.

We have to find the curved surface of a cylinder of radius 7 cm and height (20 � 4) cm i.e., 16 cm.

This curved surface area

(a) To find how much it can hold.

(b) Number of cement bags required to plaster it.

(c) To find the number of smaller tanks that can be filled with water from it.

Sol. (a) volume (b) Surface area (c) Volume

Sol. The heights and diameters of these cylinders A and B are interchanged.

We know that,

Volume of cylinder = πr^{2}h

If measures of r and h are same, then the cylinder with greater radius will have greater area.

As the radius of cylinder B is greater, therefore, the volume of cylinder B will be greater.

Let us verify by calculating the volume of both the cylinders.

Thus, the surface area of cylinder B is also greater than the surface area of cylinder A.

Sol. Volume of the cuboid = 900 cm^{3}

⇒ (Area of the base) × Height = 900 cm^{3} 180 × Height = 900

Hence, the height of the cuboid is 5 cm.

Sol. Volume of cuboid = (60 × 54 × 30)cm^{3}

= 97200 cm^{3}

Volume of cube = (6 × 6 × 6) cm^{3}

= 216 cm^{3}

Sol. Let the h be the height of cylinder whose radius,

Hence, the height of cylinder is 1 metre.

Sol. Quantity of milk that can be stored in the tank

= Volume of the tank

= πr^{2}h, where r = 1.5 and h = 7 m

= (49.5 × 1000) litres [∵ 1 m^{3} = 1000 Itr.]

= 49500 litres.

(i) How many times Al its surface area increase?

(ii) How many times will its volume increase?

Sol. Let x units be the edge of the cube. Then, its surface area = 6x^{2} and its volume = x^{3}.

When its edge is doubled,

(i) Its surf ace area = 6 (2x)^{2} = 6 × 4x^{2} = 24x^{2}

⇒ The surface area of the new cube will be 4 times that of the original cube.

(ii) Its volume = (2x)^{3} = 8x^{3}

⇒ The volume of the new cube will be 8 times that of the original cube.

Sol. Volume of the reservoir = 108 m^{3}

= 108 × 1000 litres

= 108000 litres

Since water is pouring into reservoir @ 60 litres per minute.

∴ Time taken to fill the reservoir