Rs Aggarwal 2017 Solutions for Class 7 Math Chapter 20 Mensuration are provided here with simple step-by-step explanations. These solutions for Mensuration are extremely popular among Class 7 students for Math Mensuration Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 7 Math Chapter 20 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

#### Page No 229:

(i) Length = 24.5 m

∴ Area of the rectangle = Length $×$ Breadth
= 24.5 m $×$ 18 m
= 441 m2

(ii) Length = 12.5 m
Breadth = 8 dm = (8 $×$ 10) = 80 cm = 0.8 m     [since 1 dm = 10 cm and 1 m = 100 cm]

∴ Area of the rectangle = Length $×$ Breadth
= 12.5 m $×$ 0.8 m
= 10 m2

#### Page No 229:

We know that all the angles of a rectangle are 90° and the diagonal divides the rectangle into two right angled triangles.
So, 48 m will be one side of the triangle and the diagonal, which is 50 m, will be the hypotenuse.
According to the Pythagoras theorem:
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
Perpendicular = $\sqrt{{\left(\mathrm{Hypotenuse}\right)}^{2}-\left(\mathrm{Base}{\right)}^{2}}$
Perpendicular = $\sqrt{{\left(50\right)}^{2}-{\left(48\right)}^{2}}=\sqrt{2500-2304}=\sqrt{196}=14$ m
∴ Other side of the rectangular plot = 14 m
Length = 48m

∴ Area of the rectangular plot = 48 m $×$ 14 m = 672 m2
Hence, the area of a rectangular plot is 672 m2.

#### Page No 229:

Let the length of the field be 4x m.
∴ Area of the field = (4x $×$ 3x) m2 = 12x2 m2
But it is given that the area is 1728 m2.
∴ 12x2 = 1728
x2 = $\left(\frac{1728}{12}\right)$ = 144
x = $\sqrt{144}$ = 12
∴ Length = (4 $×$ 12) m = 48 m
Breadth = (3 $×$ 12) m =36 m
∴ Perimeter of the field = 2(l + b) units
= 2(48 + 36) m = (2 $×$ 84) m = 168 m
∴ Cost of fencing = Rs (168 $×$ 30) = Rs 5040

#### Page No 229:

Area of the rectangular field = 3584 m2
Length of the rectangular field = 64 m
Breadth of the rectangular field = $\left(\frac{\mathrm{Area}}{\mathrm{Length}}\right)$ = $\left(\frac{3584}{64}\right)$ m = 56 m
Perimeter of the rectangular field = 2 (length + breadth)
= 2(64+ 56) m = (2 $×$ 120) m = 240 m
Distance covered by the boy = 5 $×$ Perimeter of the rectangular field
= 5 $×$ 240 = 1200 m

The boy walks at the rate of 6 km/hr.
or
Rate = $\left(\frac{6×1000}{60}\right)$ m/min = 100 m/min.

∴ Required time to cover a distance of 1200 m = $\left(\frac{1200}{100}\right)$ min = 12 min
Hence, the boy will take 12 minutes to go five times around the field.

#### Page No 229:

Given:
Length of the verandah = 40 m = 400 dm    [since 1 m = 10 dm ]
Breadth of the verandah = 15 m = 150 dm
∴ Area of the verandah= (400 $×$ 150) dm2 = 60000 dm2

Length of a stone = 6 dm
Breadth of a stone = 5 dm
∴ Area of a stone = (6 $×$ 5) dm2 = 30 dm2

∴ Total number of stones needed to pave the verandah =

= $\left(\frac{60000}{30}\right)$ = 2000

#### Page No 229:

Area of the carpet = Area of the room
= (13 m $×$ 9 m) = 117 m2

Now, width of the carpet = 75 cm    (given)
= 0.75 m      [since 1 m = 100 cm]

Length of the carpet = $\left(\frac{117}{0.75}\right)$ m = 156 m
Rate of carpeting = Rs 105 per m
∴ Total cost of carpeting = Rs (156 $×$105) = Rs 16380
Hence, the total cost of carpeting the room is Rs 16380.

#### Page No 229:

Given:
Length of the room = 15 m
Width of the carpet = 75 cm = 0.75 m          (since 1 m = 100 cm)

Let the length of the carpet required for carpeting the room be x m.
Cost of the carpet = Rs. 80 per m
∴ Cost of x m carpet = Rs. (80 $×$ x) = Rs. (80x)
Cost of carpeting the room = Rs. 19200
∴ 80x = 19200 ⇒ x = $\left(\frac{19200}{80}\right)$ = 240
Thus, the length of the carpet required for carpeting the room is 240 m.
Area of the carpet required for carpeting the room = Length of the carpet $×$ Width of the carpet
= ( 240 $×$ 0.75) m2 = 180 m2
Let the width of the room be b m.
Area to be carpeted = 15 m $×$ b m = 15b m2
∴ 15b m2 = 180 m2
b = $\left(\frac{180}{15}\right)$ m = 12 m
Hence, the width of the room is 12 m.

#### Page No 229:

Total cost of fencing a rectangular piece = Rs. 9600
Rate of fencing = Rs. 24
∴ Perimeter of the rectangular field =  m = $\left(\frac{9600}{24}\right)$ m = 400 m

Let the length and breadth of the rectangular field be 5x and 3x, respectively.
Perimeter of the rectangular land =  2(5x + 3x) = 16x
But the perimeter of the given field is 400 m.
∴ 16x = 400
x$\left(\frac{400}{16}\right)$ = 25
Length of the field = (5 $×$ 25) m = 125 m
Breadth of the field = (3 $×$ 25) m = 75 m

#### Page No 230:

Length of the diagonal of the room = $\sqrt{{\mathrm{l}}^{2}+{\mathrm{b}}^{2}+{\mathrm{h}}^{2}}$
= $\sqrt{{\left(10\right)}^{2}+{\left(10\right)}^{2}+{\left(5\right)}^{2}}$ m
= $\sqrt{100+100+25}$m
= $\sqrt{225}$m = 15 m

Hence, length of the largest pole that can be placed in the given hall is 15 m.

#### Page No 230:

Side of the square = 8.5 m
∴ Area of the square = (Side)2
= (8.5 m)2
= 72.25 m2

#### Page No 230:

(i) Diagonal of the square = 72 cm
∴ Area of the square = $\left[\frac{1}{2}×{\left(D\mathrm{iagonal}\right)}^{2}\right]$ sq. unit
= $\left[\frac{1}{2}×{\left(72\right)}^{2}\right]$ cm2
= 2592 cm2

(ii)Diagonal of the square = 2.4 m
∴ Area of the square = $\left[\frac{1}{2}×{\left(D\mathrm{iagonal}\right)}^{2}\right]$ sq. unit
= $\left[\frac{1}{2}×{\left(2.4\right)}^{2}\right]$ m2
= 2.88 m2

#### Page No 230:

We know:
Area of a square = $\left\{\frac{1}{2}×{\left(D\mathrm{iagonal}\right)}^{2}\right\}$ sq. units
Diagonal of the square = units
= $\left(\sqrt{2×16200}\right)$m = 180 m
∴ Length of the diagonal of the square = 180 m

#### Page No 230:

Area of the square = $\left\{\frac{1}{2}×{\left(D\mathrm{iagonal}\right)}^{2}\right\}$ sq. units
Given:
Area of the square field = $\frac{1}{2}$ hectare
= $\left(\frac{1}{2}×10000\right)$ m2 = 5000 m2             [since 1 hectare = 10000 m2 ]

Diagonal of the square =

= $\left(\sqrt{2×5000}\right)$m = 100 m

∴ Length of the diagonal of the square field = 100 m

#### Page No 230:

Area of the square plot = 6084 m2
Side of the square plot = $\left(\sqrt{\mathrm{Area}}\right)$
= $\left(\sqrt{6084}\right)$ m
= $\left(\sqrt{78×78}\right)$m = 78 m

∴ Perimeter of the square plot = 4 $×$ side = (4 $×$ 78) m = 312 m
312 m wire is needed to go along the boundary of the square plot once.

Required length of the wire that can go four times along the boundary = 4 $×$ Perimeter of the square plot
= (4 $×$ 312) m = 1248 m

#### Page No 230:

Side of the square = 10 cm
Length of the wire = Perimeter of the square = 4 $×$ Side = 4 $×$ 10 cm = 40 cm
Length of the rectangle (l) = 12 cm
Let b be the breadth of the rectangle.

Perimeter of the rectangle = Perimeter of the square

⇒ 2(l + b) = 40
⇒ 2(12 + b) = 40
⇒ 24 + 2b = 40
⇒ 2b = 40 - 24 = 16
⇒ b = $\left(\frac{16}{2}\right)$ cm = 8 cm
∴ Breadth of the rectangle = 8 cm
Now, Area of the square = (Side)2 = (10 cm $×$ 10 cm) = 100 cm2
Area of the rectangle = $×$ b = (12 cm $×$ 8 cm) = 96 cm2

Hence, the square encloses more area.
It encloses 4 cm2 more area.

#### Page No 230:

Given:
Length = 50 m
Height = 10 m

Area of the four walls  = {2h(l + b)} sq. unit
= {2 $×$ 10 $×$ (50 + 40)}m2
= {20 $×$ 90} m2 = 1800 m2
Area of the ceiling = $×$ b = (50 m $×$ 40 m) = 2000 m2
∴ Total area to be white washed = (1800 + 2000) m2 = 3800 m2
Rate of white washing = Rs 20/sq. metre
∴ Total cost of white washing = Rs (3800 $×$ 20) = Rs 76000

#### Page No 230:

Let the length of the room be l m.
Given:
Breadth of the room = 10 m
Height of the room = 4 m
Area of the four walls = [2(l + b)h] sq units.
= 168 m2
∴ 168 = [2(l + 10) $×$ 4]
⇒ 168 = [8l + 80]
⇒ 168 - 80 = 8l
⇒  88 = 8l
l = $\left(\frac{88}{8}\right)$ m = 11 m
∴ Length of the room = 11 m

#### Page No 230:

Given:
Length of the room = 7.5 m
Breadth of the room = 3.5 m
Area of the four walls = [2(l + b)h] sq. units.
= 77 m2
∴ 77 = [2(7.5 + 3.5)h]
⇒ 77 = [(2 $×$ 11)h]
⇒ 77 = 22h
h$\left(\frac{77}{22}\right)$ m = $\left(\frac{7}{2}\right)$ m = 3.5 m
∴ Height of the room = 3.5 m

#### Page No 230:

Let the breadth of the room be x m.
Length of the room = 2x m
Area of the four walls = {2(l + b$×$ h} sq. units
120 m2  = {2(2x + x) $×$ 4} m2
⇒ 120 = {8 $×$ 3x }
⇒ 120 = 24x
x =$\left(\frac{120}{24}\right)$ = 5
∴ Length of the room = 2x = (2 $×$ 5) m = 10 m
Breadth of the room = x = 5 m
∴ Area of the floor = l $×$ b = (10 m $×$ 5 m) = 50 m2

#### Page No 230:

Length = 8.5 m
Height = 3.4 m

Area of the four walls = {2(l + b$×$ h} sq. units
= {2(8.5 + 6.5) $×$ 3.4}m2 = {30 $×$ 3.4} m2 = 102 m2
Area of one door = (1.5 $×$ 1) m2 = 1.5 m2
∴ Area of two doors = (2 $×$ 1.5) m2 = 3 m2
Area of one window = (2 $×$ 1) m2 = 2 m2
∴ Area of two windows = (2 $×$ 2) m2 = 4 m2
Total area of  two doors and two windows = (3 + 4) m2
= 7 m2
Area to be painted = (102 - 7) m2 = 95 m2
Rate of painting = Rs 160 per m2
Total cost of painting = Rs (95 $×$ 160) = Rs 15200

#### Page No 232:

Let PQRS be the given grassy plot and ABCD be the inside boundary of the path.

Length = 75 m
Area of the plot = (75 $×$ 60) m2 = 4500 m2
Width of the path = 2 m
∴ AB = (75 - 2 $×$ 2) m = (75 - 4) m =71 m
AD = (60 - 2 $×$ 2) m = (60 - 4) m = 56 m
Area of rectangle ABCD = (71 x 56) m2 = 3976 m2
Area of the path = (Area of PQRS - Area of ABCD)
= (4500 - 3976) m2 = 524 m2
Rate of constructing the path = Rs 125 per m2
∴ Total cost of constructing the path = Rs (524 $×$ 125) = Rs 65,500

#### Page No 232:

Let PQRS be the given rectangular plot and ABCD be the inside boundary of the path.

Length = 95 m
Area of the plot = (95 $×$ 72) m2 = 6,840 m2
Width of the path = 3.5 m
∴ AB = (95 - 2 $×$ 3.5) m = (95 - 7) m = 88 m
AD = (72 - 2 $×$ 3.5) m = (72 - 7) m = 65 m
Area of the grassy rectangle plot ABCD = (88 $×$ 65) m2 = 5,720 m2
Area of the path = (Area PQRS - Area ABCD)
= (6840 - 5720) m2 = 1,120 m2
Rate of constructing the path = Rs. 80 per m2
∴ Total cost of  constructing the path = Rs. (1,120 $×$ 80) = Rs. 89,600
Rate of laying the grass on the plot ABCD = Rs. 40 per m2
∴ Total cost of laying the grass on the plot = Rs. (5,720 $×$ 40) = Rs. 2,28,800
∴ Total expenses involved = Rs. ( 89,600 + 2,28,800) = Rs. 3,18,400

#### Page No 232:

Let ABCD be the saree and EFGH be the part of saree without border.

Length, AB= 5 m
Width of the border of the saree = 25 cm = 0.25 m

∴ Area of ABCD = 5 m $×$ 1.3 m = 6.5 m2

Length, GH = {5 -( 0.25 + 0.25} m = 4.5 m
Breadth, FG = {1.3 - 0.25 + 0.25} m = 0.8 m
∴ Area of EFGH = 4.5 m $×$ .8 m = 3.6 m2

Area of the border = Area of ABCD − Area of EFGH
=  6.5 m2  − 3.6 m2
= 2.9 m2 = 29000 cm2     [since 1 m2 = 10000 cm2]
Rate of printing the border = Rs 1 per 10 cm2
∴ Total cost of printing the border = Rs $\left(\frac{1×29000}{10}\right)$
= Rs 2900

#### Page No 233:

Length, EF = 38 m

∴ Area of EFGH = 38 m $×$  25 m = 950 m2

Length, AB = (38  + 2.5  + 2.5 ) m = 43 m
Breadth, BC = ( 25 + 2.5 + 2.5 ) m = 30 m
∴ Area of ABCD = 43 m $×$ 30 m = 1290 m2

Area of the path = Area of ABCD − Area of PQRS
= 1290 m2 − 950 m2
= 340 m2
Rate of gravelling the path = Rs 120 per m2

∴ Total cost of gravelling the path = Rs (120 $×$ 340)
= Rs 40800

#### Page No 233:

Let EFGH denote the floor of the room.
The white region represents the floor of the 1.25 m verandah.

Length, EF = 9.5 m

∴ Area of EFGH = 9.5 m  $×$  6 m = 57 m2

Length, AB = (9.5  + 1.25  + 1.25 ) m = 12 m
Breadth, BC = ( 6 + 1.25 + 1.25 ) m = 8.5 m
∴ Area of ABCD = 12 m $×$ 8.5 m = 102 m2

Area of the verandah = Area of ABCD − Area of EFGH
= 102 m2 − 57 m2
= 45 m2
Rate of cementing the verandah = Rs 80 per m2

∴ Total cost of cementing the verandah = Rs ( 80 $×$ 45)
= Rs 3600

#### Page No 233:

Side of the flower bed = 2 m 80 cm = 2.80 m      [since 100 cm = 1 m]

∴ Area of the square flower bed = (Side)2 = (2.80 m )2 = 7.84 m2
Side of the flower bed with the digging strip = 2.80 m + 30 cm + 30 cm
= (2.80 + 0.3 + 0.3) m = 3.4 m
Area of the enlarged flower bed with the digging strip = (Side )2 = (3.4 )2 = 11.56 m2

∴ Increase in the area of the flower bed = 11.56 m2 − 7.84 m2
= 3.72 m2

#### Page No 233:

Let the length and the breadth of the park be 2x m and x m, respectively.
Perimeter of the park = 2(2x + x) = 240 m
⇒ 2(2x + x) = 240
⇒ 6x = 240
x = $\left(\frac{240}{6}\right)$ m =40 m
∴ Length of the park = 2x = (2 $×$ 40) = 80 m
Breadth = x = 40 m
Let PQRS be the given park and ABCD be the inside boundary of the path.

Length = 80 m
Area of the park = (80 $×$ 40) m2 = 3200 m2
Width of the path = 2 m
∴ AB = (80 - 2 $×$ 2) m = (80 - 4) m =76 m
AD = (40 - 2 $×$ 2) m = (40 - 4) m = 36 m
Area of the rectangle ABCD = (76 $×$ 36) m2 = 2736 m2
Area of the path = (Area of PQRS - Area of ABCD)
= (3200 - 2736) m2 = 464 m2
Rate of paving the path = Rs. 80 per m2
∴ Total cost of  paving the path = Rs. (464 $×$ 80) = Rs. 37,120

#### Page No 233:

Length of the hall, PQ = 22 m
Breadth of the hall, QR = 15.5 m

∴ Area of the school hall PQRS = 22 m $×$ 15.5 m = 341 m2
Length of the carpet, AB = 22 m − ( 0.75 m + 0.75 m) = 20.5 m         [since 100 cm = 1 m]
Breadth of the carpet, BC = 15.5 m − ( 0.75 m + 0.75 m) = 14 m
∴ Area of the carpet ABCD = 20.5 m $×$ 14 m = 287 m2
Area of the strip = Area of the school hall (PQRS) − Area of the carpet (ABCD)
= 341 m2 − 287 m2
= 54 m2

Area of 1 m length of the carpet = 1 m $×$ 0.82 m = 0.82 m2

∴ Length of the carpet whose area is 287 m2 = 287 m2 ÷ 0.82 m2 = 350 m
Cost of the 350 m long carpet = Rs 60 $×$ 350 = Rs 21000

#### Page No 233:

Let ABCD be the square lawn and PQRS be the outer boundary of the square path.

Let a side of the lawn (AB) be x m.
Area of the square lawn = x2
Length, PQ = (x m + 2.5 m + 2.5 m) = (x + 5) m
∴ Area of PQRS = (x + 5)2 = (x2 + 10x + 25) m2

Area of the path = Area of PQRS − Area of the square lawn (ABCD)
⇒ 165 = x2 + 10x + 25 x2
⇒ 165 = 10x + 25
⇒ 165 − 25 = 10x
⇒ 140 = 10x
x = 140 ÷ 10 = 14
∴ Side of the lawn = 14 m

∴ Area of the lawn = (Side)2 = (14 m)2 = 196 m2

#### Page No 233:

Area of the path = 305 m2

Let the length of the park be 5x m and the breadth of the park be 2x m.

∴ Area of the rectangular park = 5x $×$ 2x = 10x2 m2
Width of the path = 2.5 m
Outer length, PQ = 5x m + 2.5 m + 2.5 m = (5x + 5) m
Outer breadth, QR = 2x + 2.5 m + 2.5 m = (2x + 5) m
Area of PQRS = (5x + 5) $×$ (2x + 5)  = (10x2 + 25x + 10x + 25) = (10x2 + 35x + 25) m2
∴ Area of the path = [(10x2 + 35x + 25) − 10x2 ] m2
⇒  305 = 35x + 25
⇒ 305 − 25 = 35x
⇒ 280 = 35x
x = 280 ÷ 35 = 8

∴ Length of the park = 5x = $×$ 8 = 40 m
Breadth of the park = 2x = 2 $×$ 8 = 16 m

#### Page No 233:

Let ABCD be the rectangular park.
Let EFGH and IJKL be the two rectangular roads with width 5 m.

Length of the rectangular park, AD = 70 m
Breadth of the rectangular park, CD = 50 m
∴ Area of the rectangular park = Length $×$ Breadth = 70 m $×$ 50 m = 3500 m2
Area of road EFGH = 70 m $×$ 5 m = 350 m2
Area of road IJKL = 50 m $×$ 5 m = 250 m2

Clearly, area of MNOP is common to both the two roads.

∴ Area of MNOP = 5 m $×$ 5 m = 25 m2

Area of the roads = Area (EFGH) + Area (IJKL) − Area (MNOP)
= (350  + 250 ) m2− 25 m2 = 575 m2
It is given that the cost of constructing the roads is Rs. 120/m2.

Cost of constructing 575 m2 area of the roads = Rs. (120 × 575)
= Rs. 69000

#### Page No 233:

Let ABCD be the rectangular field and PQRS and KLMN be the two rectangular roads with width 2 m and 2.5 m, respectively.

Length of the rectangular field, CD = 115 cm
Breadth of the rectangular field, BC = 64 m
∴ Area of the rectangular lawn ABCD = 115 m $×$ 64 m = 7360 m2
Area of the road PQRS = 115 m $×$ 2 m = 230 m2
Area of the road KLMN = 64 m $×$ 2.5 m = 160 m2

Clearly, the area of EFGH is common to both the two roads.

∴ Area of EFGH = 2 m $×$ 2.5 m = 5 m2

∴ Area of the roads = Area (KLMN) + Area (PQRS) − Area (EFGH)
= (230 m2 + 160 m2) − 5 m2 = 385 m2

Rate of gravelling the roads = Rs 60 per m2
∴ Total cost of gravelling the roads = Rs (385 $×$ 60)
= Rs 23,100

#### Page No 233:

Let ABCD be the rectangular field and KLMN and PQRS be the two rectangular roads with width 2.5 m and 2 m, respectively.

Length of the rectangular field CD = 50 cm
Breadth of the rectangular field BC = 40 m
∴ Area of the rectangular field ABCD = 50 m $×$ 40 m = 2000 m2
Area of road KLMN = 40 m $×$ 2.5 m = 100 m2
Area of road PQRS = 50 m $×$ 2 m = 100 m2

Clearly, area of EFGH is common to both the two roads.

∴ Area of EFGH = 2.5 m $×$ 2 m = 5 m2

∴ Area of the roads = Area (KLMN) + Area (PQRS) − Area (EFGH)
= (100 m2 + 100 m2) − 5 m2 = 195 m2

Area of the remaining portion of the field = Area of the rectangular field (ABCD) − Area of the roads
= (2000 − 195) m2
= 1805 m2

#### Page No 233:

(i) Complete the rectangle as shown below:

Area of the shaded region = [Area of rectangle ABCD - Area of rectangle EFGH] sq. units
= [(43 m $×$ 27 m) - {(43 - 2 $×$ 1.5) m x (27 - 1 $×$ 2) m}]
= [(43 m $×$ 27 m) - {40 m $×$ 25 m}]
= 1161 m2 - 1000 m2
= 161 m2

(ii) Complete the rectangle as shown below:

Area of the shaded region = [Area of square ABCD - {(Area of EFGH) + (Area of IJKL) - (Area of MNOP)}] sq. units
= [(40 $×$ 40) - {(40 $×$ 2) + (40 $×$ 3) - (2 $×$ 3)}] m2
= [1600 - {(80 + 120 - 6)] m2
= [1600 - 194] m2
= 1406 m2

#### Page No 234:

(i) Complete the rectangle as shown below:

Area of the shaded region = [Area of rectangle ABCD - Area of rectangle EFGD] sq. units
= [(AB $×$ BC) - (DG $×$ GF)] m2
= [(24 m $×$ 19 m) - {(24 - 4) m $×$ 16.5 m} ]
= [(24 m $×$ 19 m) - (20 m $×$ 16.5) m]
= (456 - 330) m2 = 126 m2

(ii) Complete the rectangle by drawing lines as shown below:

Area of the shaded region ={(12 $×$ 3) + (12 $×$ 3) + (5$×$ 3) + {(15 - 3 - 3) $×$3)} cm2
= { 36 + 36 + 15 + 27} cm2
= 114 cm2

#### Page No 234:

Divide the given figure in four parts shown below:

Given:
Width of each part = 0.5 m

Now, we have to find the length of each part.

Length of part I = 3.5 m
Length of part II = (3.5 - 0.5 - 0.5) m = 2.5 m
Length of part III = (2.5 - 0.5 - 0.5) = 1.5 m
Length of part IV = (1.5 - 0.5 - 0.5) = 0.5 m
∴ Area of the shaded region = [Area of part (I) + Area of part (II) + Area of part (III) + Area of part (IV)] sq. units
= [(3.5 $×$ 0.5) + (2.5 $×$ 0.5) + ( 1.5 $×$ 0.5) + (0.5 $×$ 0.5)] m2
= [1.75 + 1.25 + 0.75 + 0.25] m2
= 4 m2

#### Page No 237:

Base = 32 cm
Height = 16.5 cm

∴ Area of the parallelogram = Base $×$ Height
=  32 cm $×$ 16.5 cm
= 528 cm2

#### Page No 237:

Base = 1 m 60 cm = 1.6 m             [since 100 cm = 1 m]
Height = 75 cm = 0.75 m

∴ Area of the parallelogram = Base $×$ Height
= 1.6 m $×$ 0.75 m
= 1.2 m2

#### Page No 237:

(i) Base = 14 dm = (14 $×$ 10) cm = 140 cm                  [since 1 dm  = 10 cm]
Height = 6.5 dm = (6.5 $×$ 10) cm = 65 cm

Area of the parallelogram  = Base $×$ Height
= 140 cm $×$ 65 cm
= 9100 cm2

(ii) Base = 14 dm = (14 $×$ 10) cm                            [since 1 dm = 10 cm and 100 cm  = 1 m]
= 140 cm = 1.4 m
Height = 6.5 dm = (6.5 $×$ 10) cm
= 65 cm = 0.65 m

∴ Area of the parallelogram = Base $×$ Height
= 1.4 m $×$ 0.65 m
= 0.91 m2

#### Page No 237:

Area of the given parallelogram = 54 cm2
Base of the given parallelogram = 15 cm
∴ Height of the given parallelogram = $\frac{\mathrm{Area}}{\mathrm{Base}}$$\left(\frac{54}{15}\right)$ cm = 3.6 cm

#### Page No 237:

Base of the parallelogram = 18 cm
Area of the parallelogram = 153 cm2
∴ Area of the parallelogram = Base $×$ Height
⇒ Height = = $\left(\frac{153}{18}\right)$ cm = 8.5 cm
Hence, the distance of the given side from its opposite side is 8.5 cm.

#### Page No 237:

Base, AB = 18 cm
Height, AL = 6.4 cm
∴ Area of the parallelogram ABCD = Base $×$ Height
= (18 cm $×$ 6.4 cm) = 115.2 cm2                    ... (i)

Now, taking BC as the base:
Area of the parallelogram ABCD = Base $×$ Height
= (12 cm $×$ AM)                          ... (ii)
From equation (i) and (ii):
12 cm $×$ AM = 115.2 cm2
⇒ AM = $\left(\frac{115.2}{12}\right)$cm
= 9.6 cm

#### Page No 237:

ABCD is a parallelogram with side AB of length 15 cm and the corresponding altitude AE of length 4 cm.
The adjacent side AD is of length 8 cm and the corresponding altitude is CF.

Area of a parallelogram = Base × Height

We have two altitudes and two corresponding bases.

AD $×$ CF = AB $×$ AE
⇒ 8 cm $×$ CF = 15 cm $×$4 cm

CF =$\left(\frac{15×4}{8}\right)$ cm = $\left(\frac{15}{2}\right)$ cm = 7.5 cm
Hence, the distance between the shorter sides is 7.5 cm.

#### Page No 237:

Let the base of the parallelogram be x cm.
Then, the height of the parallelogram will be $\frac{1}{3}$x cm.
It is given that the area of the parallelogram is 108 cm2.

Area of a parallelogram =  Base $×$ Height
∴ 108 cm2 = x $×$ $\frac{1}{3}$x
108 cm2 = $\frac{1}{3}$x2
x2 = (108 $×$ 3) cm2 = 324 cm2
x2 = (18 cm)2
x = 18 cm

∴ Base = x = 18 cm
Height = $\frac{1}{3}$x $\left(\frac{1}{3}×18\right)$ cm
= 6 cm

#### Page No 237:

Let the height of the parallelogram be x cm.
Then, the base of the parallelogram will be 2x cm.
It is given that the area of the parallelogram is 512 cm2.

Area of a parallelogram =  Base $×$ Height
∴ 512 cm2 = 2x $×$ x
512 cm2 = 2x2
x2 =$\left(\frac{512}{2}\right)$ cm2 = 256 cm2
x2 = (16 cm)2
x = 16 cm

∴ Base = 2x = $×$ 16
= 32 cm
Height = x = 16 cm

#### Page No 237:

A rhombus is a special type of a parallelogram.

The area of a parallelogram is given by the product of its base and height.
∴ Area of the given rhombus = Base × Height
(i) Area of the rhombus = 12 cm $×$ 7.5 cm = 90 cm2

(ii) Base = 2 dm = (2 $×$ 10) = 20 cm    [since 1 dm = 10 cm]
Height = 12.6 cm
∴ Area of the rhombus = 20 cm $×$ 12.6 cm = 252 cm2

#### Page No 237:

(i)
Length of one diagonal = 16 cm
Length of the other diagonal = 28 cm
∴ Area of the rhombus = $\frac{1}{2}$ $×$ (Product of the diagonals)
= $\left(\frac{1}{2}×16×28\right)$ cm2 = 224 cm2
(ii)
Length of one diagonal = 8 dm 5 cm = (8 $×$ 10 + 5) cm = 85 cm              [since 1 dm = 10 cm]
Length of the other diagonal = 5 dm 6 cm = (5 $×$ 10 + 6) cm = 56 cm
∴ Area of the rhombus = $\frac{1}{2}$ $×$ (Product of the diagonals)
= $\left(\frac{1}{2}×85×56\right)$ cm2
= 2380 cm2

#### Page No 237:

Let ABCD be the rhombus, whose diagonals intersect at O.

AB = 20 cm and AC = 24 cm
The diagonals of a rhombus bisect each other at right angles.

Therefore, ΔAOB is a right angled triangle, right angled at O.

Here, OA =$\frac{1}{2}\mathrm{AC}$ = 12 cm
AB = 20 cm

By Pythagoras theorem:
(AB)2 = (OA)2 + (OB)2
⇒ (20)2 = (12)2 + (OB)2
⇒ (OB)2 = (20)2 − (12)2
⇒ (OB)2 = 400 − 144 = 256
⇒ (OB)2 = (16)2
⇒ OB = 16 cm
∴ BD = 2 $×$ OB = 2 $×$ 16 cm = 32 cm

∴  Area of the rhombus ABCD = $\left(\frac{1}{2}×\mathrm{AC}×\mathrm{BD}\right)$ cm2
= $\left(\frac{1}{2}×24×32\right)$ cm2
= 384 cm2

#### Page No 237:

Area of a rhombus =  $\frac{1}{2}$ $×$ (Product of the diagonals)
Given:
Length of one diagonal = 19.2 cm
Area of the rhombus = 148.8 cm2

∴ Length of the other diagonal = $\left(\frac{148.8×2}{19.2}\right)$ cm = 15.5 cm

#### Page No 237:

Perimeter of the rhombus = 56 cm
Area of the rhombus = 119 cm2
Side of the rhombus = $\frac{\mathrm{Perimeter}}{4}$ = $\left(\frac{56}{4}\right)$ cm = 14 cm
Area of a rhombus = Base $×$ Height

∴  Height of the rhombus = $\frac{\mathrm{Area}}{\mathrm{Base}}$= $\left(\frac{119}{14}\right)$ cm
= 8.5 cm

#### Page No 237:

Given:
Height of the rhombus = 17.5 cm
Area of the rhombus = 441 cm2

We know:
Area of a rhombus = Base $×$ Height

∴  Base of the rhombus =$\frac{\mathrm{Area}}{\mathrm{Height}}$$\left(\frac{441}{17.5}\right)$ cm = 25.2 cm
Hence, each side of a rhombus is 25.2 cm.

#### Page No 237:

Area of a triangle = $\frac{1}{2}$ $×$ Base $×$ Height
= $\left(\frac{1}{2}×24.8×16.5\right)$ cm2 = 204.6 cm2
Given:
Area of the rhombus = Area of the triangle
Area of the rhombus = 204.6 cm2

Area of the rhombus = $\frac{1}{2}$ $×$ (Product of the diagonals)
Given:
Length of one diagonal = 22 cm
∴ Length of the other diagonal = $\left(\frac{204.6×2}{22}\right)$ cm
= 18.6 cm

#### Page No 242:

We know:
Area of a triangle = $\frac{1}{2}×\mathrm{Base}×\mathrm{Height}$
(i) Base = 42 cm
Height = 25 cm
∴ Area of the triangle = $\left(\frac{1}{2}×42×25\right)$ cm2 = 525 cm2
(ii) Base = 16.8 m
Height = 75 cm = 0.75 m      [since 100 cm = 1 m]
∴ Area of the triangle = $\left(\frac{1}{2}×16.8×0.75\right)$ m2 = 6.3 m2
(iii) Base = 8 dm = (8 $×$ 10) cm = 80 cm     [since 1 dm = 10 cm]
Height = 35 cm
∴ Area of the triangle = $\left(\frac{1}{2}×80×35\right)$ cm2 = 1400 cm2

#### Page No 242:

Height of a triangle = $\frac{2×\mathrm{Area}}{\mathrm{Base}}$
Here, base = 16 cm and area = 72 cm2

∴ Height = $\left(\frac{2×72}{16}\right)$ cm = 9 cm

#### Page No 242:

Height of a triangle = $\frac{2×\mathrm{Area}}{\mathrm{Base}}$
Here, base = 28 m and area = 224 m2

∴ Height = $\left(\frac{2×224}{28}\right)$ m = 16 m

#### Page No 242:

Base of a triangle = $\frac{2×\mathrm{Area}}{\mathrm{Height}}$
Here, height = 12 cm and area = 90 cm2

∴ Base = $\left(\frac{2×90}{12}\right)$ cm = 15 cm

#### Page No 242:

Total cost of cultivating the field = Rs. 14580

Rate of cultivating the field = Rs. 1080 per hectare

Area of the field = hectare
= $\left(\frac{14580}{1080}\right)$ hectare
=  13.5 hectare
= (13.5 $×$ 10000) m2 = 135000 m2       [since 1 hectare = 10000 m2 ]
Let the height of the field be x m.
Then, its base will be 3x m.
Area of the field = $\left(\frac{1}{2}×3x×x\right)$ m2 = $\left(\frac{3{x}^{2}}{2}\right)$ m2
$\left(\frac{3{x}^{2}}{2}\right)$ = 135000

x = $\sqrt{90000}$ = 300
∴ Base = (3 $×$ 300) = 900 m
Height = 300 m

#### Page No 242:

Let the length of the other leg be h cm.
Then, area of the triangle = $\left(\frac{1}{2}×14.8×h\right)$ cm2 = (7.4 h) cm2
But it is given that the area of the triangle is 129.5 cm2.
∴ 7.4h = 129.5
h = $\left(\frac{129.5}{7.4}\right)$ = 17.5 cm
∴ Length of the other leg = 17.5 cm

#### Page No 242:

Here, base = 1.2 m and hypotenuse = 3.7 m

In the right angled triangle:

Perpendicular = $\sqrt{\left(H\mathrm{ypotenuse}{\right)}^{2}-\left(B\mathrm{ase}{\right)}^{2}}$

$=\sqrt{{\left(3.7\right)}^{2}-{\left(1.2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{13.69-1.44}\phantom{\rule{0ex}{0ex}}=\sqrt{12.25}\phantom{\rule{0ex}{0ex}}=3.5$
Area = $\left(\frac{1}{2}×\mathrm{base}×\mathrm{perpendicular}\right)$ sq. units
= $\left(\frac{1}{2}×1.2×3.5\right)$ m2
∴ Area of the right angled triangle = 2.1 m2

#### Page No 242:

In a right angled triangle, if one leg is the base, then the other leg is the height.
Let the given legs be 3x and 4x, respectively.
Area of the triangle = $\left(\frac{1}{2}×3x×4x\right)$ cm2
⇒ 1014 = (6x2)
⇒ 1014 = 6x2
x2 = $\left(\frac{1014}{6}\right)$ = 169
x = $\sqrt{169}$ = 13
∴ Base = (3 $×$ 13) = 39 cm
Height = (4 $×$ 13) = 52 cm

#### Page No 243:

Consider a right-angled triangular scarf (ABC).
Here, ∠B= 90°
BC = 80 cm
AC = 1 m = 100 cm

Now, AB2 + BC2 = AC2
⇒ AB2 = AC2 - BC2 = (100)2 - (80)2
= (10000 - 6400) = 3600
⇒ AB = $\sqrt{3600}$  = 60 cm
Area of the scarf ABC = $\left(\frac{1}{2}×BC×AB\right)$ sq. units
= $\left(\frac{1}{2}×80×60\right)$ cm2
= 2400 cm2 = 0.24 m2     [since 1 m2 = 10000 cm2]
Rate of the cloth = Rs 250 per m2
∴ Total cost of the scarf = Rs (250 $×$ 0.24) = Rs 60
Hence, cost of the right angled scarf is Rs 60.

#### Page No 243:

(i) Side of the equilateral triangle = 18 cm
Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^{2}$  sq. units
=  $\frac{\sqrt{3}}{4}{\left(18\right)}^{2}$ cm2 = $\left(\sqrt{3}×81\right)$ cm2
= (1.73 $×$ 81) cm2 = 140.13 cm2

(ii) Side of the equilateral triangle = 20 cm
Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^{2}$  sq. units
=  $\frac{\sqrt{3}}{4}{\left(20\right)}^{2}$ cm2 = $\left(\sqrt{3}×100\right)$ cm2
= (1.73 $×$ 100) cm2 = 173 cm2

#### Page No 243:

It is given that the area of an equilateral triangle is $16\sqrt{3}$ cm2.

We know:
Area of an equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^{2}$ sq. units

∴ Side of the equilateral triangle = $\left[\sqrt{\left(\frac{4×\mathrm{Area}}{\sqrt{3}}\right)}\right]$ cm
=   $\left[\sqrt{\left(\frac{4×16\sqrt{3}}{\sqrt{3}}\right)}\right]$cm = $\left(\sqrt{4×16}\right)$cm = $\left(\sqrt{64}\right)$cm = 8 cm

Hence, the length of the equilateral triangle is 8 cm.

#### Page No 243:

Let the height of the triangle be h cm.
Area of the triangle = sq. units
= $\left(\frac{1}{2}×24×h\right)$ cm2

Let the side of the equilateral triangle be a cm.
Area of the equilateral triangle = $\left(\frac{\sqrt{3}}{4}{a}^{2}\right)$ sq. units
= $\left(\frac{\sqrt{3}}{4}×24×24\right)$ cm2 = $\left(\sqrt{3}×144\right)$ cm2
$\left(\frac{1}{2}×24×h\right)$ = $\left(\sqrt{3}×144\right)$
⇒ 12 h = $\left(\sqrt{3}×144\right)$
h = $\left(\frac{\sqrt{3}×144}{12}\right)=\left(\sqrt{3}×12\right)=\left(1.73×12\right)=20.76$ cm
∴ Height of the equilateral triangle = 20.76 cm

#### Page No 243:

(i) Let a = 13 m, b = 14 m and c = 15 m
s = $\left(\frac{a+b+c}{2}\right)$ = $\left(\frac{13+14+15}{2}\right)=\left(\frac{42}{2}\right)$m = 21 m
∴  Area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ sq. units
= $\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}$m2
=   $\sqrt{21×8×7×6}$ m2
= $\sqrt{3×7×2×2×2×7×2×3}$ m2
= (2 $×$$×$$×$ 7) m2
= 84 m2

(ii) Let a = 52 cm, b = 56 cm and c = 60 cm
s = $\left(\frac{a+b+c}{2}\right)$ = $\left(\frac{52+56+60}{2}\right)=\left(\frac{168}{2}\right)$ cm = 84 cm
∴  Area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ sq. units
= $\sqrt{84\left(84-52\right)\left(84-56\right)\left(84-60\right)}$cm2
=   $\sqrt{84×32×28×24}$ cm2
= $\sqrt{12×7×4×8×4×7×3×8}$ cm2
= $\sqrt{2×2×3×7×2×2×2×2×2×2×2×7×3×2×2×2}$ cm2
= (2 $×$$×$$×$$×$$×$$×$$×$$×$ 7) m2
= 1344 cm2

(iii) Let a = 91 m, b = 98 m and c = 105 m
s = $\left(\frac{a+b+c}{2}\right)$ = $\left(\frac{91+98+105}{2}\right)=\left(\frac{294}{2}\right)$ m = 147 m
∴  Area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ sq. units
= $\sqrt{147\left(147-91\right)\left(147-98\right)\left(147-105\right)}$m2
=  $\sqrt{147×56×49×42}$ m2
= $\sqrt{3×49×8×7×49×6×7}$ m2
= $\sqrt{3×7×7×2×2×2×7×7×7×2×3×7}$ m2
= ( 2 $×$$×$$×$$×$$×$ 7) m2
= 4116 m2

#### Page No 243:

Let a = 33 cm, b = 44 cm and c = 55 cm
Then, s = $\frac{a+b+c}{2}$ = $\left(\frac{33+44+55}{2}\right)$ cm = $\left(\frac{132}{2}\right)$ cm = 66 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
= $\sqrt{66\left(66-33\right)\left(66-44\right)\left(66-55\right)}$ cm2
=  $\sqrt{66×33×22×11}$ cm2
= $\sqrt{6×11×3×11×2×11×11}$ cm2
= (6 $×$ 11 $×$ 11) cm2 = 726 cm2

Let the height on the side measuring 44 cm be h cm.
Then, Area  = $\frac{1}{2}×b×h$
⇒ 726 cm2 = $\frac{1}{2}×44×h$
h = $\left(\frac{2×726}{44}\right)$ cm = 33 cm.
∴  Area of the triangle = 726 cm2
Height corresponding to the side measuring 44 cm = 33 cm

#### Page No 243:

Let a = 13x cm, b = 14x cm and c = 15x cm
Perimeter of the triangle = 13x + 14x + 15x = 84 (given)
⇒ 42x = 84
x = $\frac{84}{42}=2$
a = 26 cm , b = 28 cm and c = 30 cm

s = $\frac{a+b+c}{2}$ = $\left(\frac{26+28+30}{2}\right)$cm = $\left(\frac{84}{2}\right)$cm = 42 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
= $\sqrt{42\left(42-26\right)\left(42-28\right)\left(42-30\right)}$ cm2
=  $\sqrt{42×16×14×12}$ cm2
= $\sqrt{6×7×4×4×2×7×6×2}$ cm2
= (2 $×$$×$$×$ 7) cm2 = 336 cm2
Hence, area of the given triangle is 336 cm2.

#### Page No 243:

Let a = 42 cm, b = 34 cm and c = 20 cm
Then, s = $\frac{a+b+c}{2}$ = $\left(\frac{42+34+20}{2}\right)$ cm = $\left(\frac{96}{2}\right)$ cm = 48 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
= $\sqrt{48\left(48-42\right)\left(48-34\right)\left(48-20\right)}$ cm2
=  $\sqrt{48×6×14×28}$ cm2
= $\sqrt{6×2×2×2×6×14×2×14}$ cm2
= (2 $×$$×$$×$ 14) cm2 = 336 cm2

Let the height on the side measuring 42 cm be h cm.
Then, Area  = $\frac{1}{2}×b×h$
⇒ 336 cm2 = $\frac{1}{2}×42×h$
h = $\left(\frac{2×336}{42}\right)$ cm = 16 cm
∴ Area of the triangle = 336 cm2
Height corresponding to the side measuring 42 cm = 16 cm

#### Page No 243:

Let each of the equal sides be a cm.
b = 48 cm
a = 30 cm
Area of the triangle = $\left\{\frac{1}{2}×b×\sqrt{{a}^{2}-\frac{{b}^{2}}{4}}\right\}$ sq. units
= $\left\{\frac{1}{2}×48×\sqrt{{\left(30\right)}^{2}-\frac{{\left(48\right)}^{2}}{4}}\right\}$ cm2 = $\left(24×\sqrt{900-\frac{2304}{4}}\right)$ cm2
= $\left(24×\sqrt{900-576}\right)$ cm2 = $\left(24×\sqrt{324}\right)$ cm2 = (24 $×$ 18) cm2 = 432 cm2
∴ Area of the triangle = 432 cm2

#### Page No 243:

Let each of the equal sides be a cm.
a + a + 12 = 32 ⇒ 2a = 20  ⇒ a = 10
b = 12 cm and a = 10 cm
Area of the triangle = $\left\{\frac{1}{2}×b×\sqrt{{a}^{2}-\frac{{b}^{2}}{4}}\right\}$ sq. units
= $\left\{\frac{1}{2}×12×\sqrt{100-\frac{144}{4}}\right\}$ cm2 = $\left(6-\sqrt{100-36}\right)$ cm2
= $\left(6×\sqrt{64}\right)$ cm2 = (6 $×$ 8) cm2
= 48 cm2

#### Page No 243:

We have:
AC = 26 cm, DL = 12.8 cm and BM = 11.2 cm

Area of ΔADC $\frac{1}{2}$ $×$ AC $×$ DL
= $\frac{1}{2}$ $×$ 26 cm $×$ 12.8 cm = 166.4 cm2
Area of ΔABC = $\frac{1}{2}$ $×$ AC $×$ BM
= $\frac{1}{2}$ $×$ 26 cm $×$ 11.2 cm = 145.6 cm2

∴ Area of the quadrilateral ABCD = Area of ΔADC  + Area of ΔABC
= (166.4 + 145.6) cm2
= 312 cm2

#### Page No 243:

First, we have to find the area of ΔABC and ΔACD.
For ΔACD:
Let a = 30 cm, b = 40 cm and c = 50 cm
s = $\left(\frac{a+b+c}{2}\right)=\left(\frac{30+40+50}{2}\right)=\left(\frac{120}{2}\right)=60$ cm
∴  Area of triangle ACD = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$  sq. units
= $\sqrt{60\left(60-30\right)\left(60-40\right)\left(60-50\right)}$ cm2
=  $\sqrt{60×30×20×10}$ cm2
= $\sqrt{360000}$ cm2
= 600 cm2
For ΔABC:
Let a = 26 cm, b = 28 cm and c = 30 cm
s = $\left(\frac{a+b+c}{2}\right)=\left(\frac{26+28+30}{2}\right)=\left(\frac{84}{2}\right)=42$ cm
∴  Area of triangle ABC = $\sqrt{s\left(s-a\right)\left(s-b\left(s-c\right)\right)}$  sq. units
= $\sqrt{42\left(42-26\right)\left(42-28\right)\left(42-30\right)}$ cm2
=  $\sqrt{42×16×14×12}$ cm2
= $\sqrt{2×3×7×2×2×2×2×2×7×3×2×2}$ cm2
= (2 $×$$×$$×$$×$$×$ 7) cm2
= 336 cm2
∴ Area of the given quadrilateral ABCD =  Area of ΔACD + Area of ΔABC
= (600 + 336) cm2 = 936 cm2

#### Page No 243:

Area of the rectangle = AB $×$ BC
= 36 m $×$ 24 m
= 864 m2
Area of the triangle  = $\frac{1}{2}$ $×$ AD $×$ FE
= $\frac{1}{2}$ $×$ BC $×$ FE       [since AD = BC]
= $\frac{1}{2}$ $×$ 24 m $×$ 15 m
= 12 m $×$15 m = 180 m2
∴ Area of the shaded region = Area of the rectangle − Area of the triangle
= (864 − 180) m2
=  684 m2

#### Page No 244:

Join points PR and SQ.
These two lines bisect each other at point O.

Here, AB = DC = SQ = 40 cm
AD = BC =RP = 25 cm

Also, OP = OR $\frac{RP}{2}=\frac{25}{2}$ = 12.5 cm
From the figure we observe:
Area of ΔSPQ = Area of ΔSRQ
∴ Area of the shaded region   = 2 $×$ (Area of ΔSPQ)
= 2 $×$ ($\frac{1}{2}$$×$ SQ $×$OP)
= 2 $×$ ($\frac{1}{2}$ $×$ 40 cm $×$ 12.5 cm)
= 500 cm2

#### Page No 244:

(i) Area of rectangle ABCD = (10 cm x 18 cm) = 180 cm2

Area of triangle I = $\left(\frac{1}{2}×6×10\right)$ cm2 = 30 cm2
Area of triangle II = $\left(\frac{1}{2}×8×10\right)$ cm2 = 40 cm2
∴ Area of the shaded region = {180 - ( 30 + 40)} cm2 = { 180 - 70}cm2 = 110 cm2

(ii) Area of square ABCD = (Side)2 = (20 cm)2 = 400 cm2

Area of triangle I = $\left(\frac{1}{2}×10×20\right)$ cm2 = 100 cm2
Area of triangle II = $\left(\frac{1}{2}×10×10\right)$ cm2 = 50 cm2
Area of triangle III = $\left(\frac{1}{2}×10×20\right)$ cm2 = 100 cm2
∴ Area of the shaded region = {400 - ( 100 + 50 + 100)} cm2 = { 400 - 250}cm2 = 150 cm2

#### Page No 244:

Let ABCD be the given quadrilateral and let BD be the diagonal such that BD is of the length 24 cm.
Let AL ⊥ BD and CM ⊥ BD
Then, AL = 5 cm and CM = 8 cm
Area of the quadrilateral ABCD = (Area of ΔABD + Area of ΔCBD)
=  $\left[\left(\frac{1}{2}×BD×AL\right)+\left(\frac{1}{2}×BD×CM\right)\right]$ sq. units
=  $\left[\left(\frac{1}{2}×24×5\right)+\left(\frac{1}{2}×24×8\right)\right]$ cm2
= ( 60 + 96) cm2 = 156 cm2

∴ Area of the given quadrilateral = 156 cm

#### Page No 248:

Here, r = 15 cm
∴ Circumference = $2\mathrm{\pi r}$
= ( 2 $×$ 3.14 $×$ 15) cm
= 94.2 cm
Hence, the circumference of the given circle is 94.2 cm

#### Page No 248:

(i) Here, r = 28 cm
∴ Circumference = 2π r
= $\left(2×\frac{22}{7}×28\right)$cm
=  176 cm
Hence, the circumference of the given circle is 176 cm.

(ii) Here, r = 1.4 m
∴ Circumference = 2π r
= $\left(2×\frac{22}{7}×1.4\right)$ m
=  $\left(2×22×0.2\right)$ m = 8.8 m
Hence, the circumference of the given circle is 8.8 m.

#### Page No 248:

(i) Here, d = 35 cm
Circumference = 2π r
= $\left(\mathrm{\pi }d\right)$    [since 2r = d]
=  $\left(\frac{22}{7}×35\right)$ cm = (22 $×$ 5) = 110 cm
Hence, the circumference of the given circle is 110 cm.

(ii) Here, d = 4.9 m
Circumference =2π r
= $\left(\mathrm{\pi }d\right)$    [since 2r = d]
=  $\left(\frac{22}{7}×4.9\right)$ m = (22 $×$ 0.7) = 15.4 m
Hence, the circumference of the given circle is 15.4 m.

#### Page No 248:

Circumference of the given circle = 57.2 cm
∴ C = 57.2 cm
Let the radius of the given circle be r cm.
C = $2\mathrm{\pi r}$
r = $\frac{\mathrm{C}}{2\mathrm{\pi }}$ cm
r = $\left(\frac{57.2}{2}×\frac{7}{22}\right)$ cm = 9.1 cm
Thus, radius of the given circle is 9.1 cm.

#### Page No 248:

Circumference of the given circle = 63.8 m
∴ C = 63.8 m
Let the radius of the given circle be r cm.
C = $2\mathrm{\pi r}$
r = $\frac{\mathrm{C}}{2\mathrm{\pi }}$
r = $\left(\frac{63.8}{2}×\frac{7}{22}\right)$m =10.15 m
∴ Diameter of the given circle = 2r = (2 $×$ 10.15) m = 20.3 m

#### Page No 248:

Let the radius of the given circle be r cm.
Then, its circumference = $2\mathrm{\pi r}$

Given:
(Circumference) - (Diameter) = 30 cm
∴ ($2\mathrm{\pi r}$ - 2r ) =  30
$2r\left(\mathrm{\pi }-1\right)=30$
$2r\left(\frac{22}{7}-1\right)=30$
$2r×\frac{15}{7}=30$
$r=\left(30×\frac{7}{30}\right)=7$
∴ Radius of the given circle = 7 cm

#### Page No 248:

Let the radii of the given circles be 5x and 3x, respectively.
Let their circumferences be C1 and C2, respectively.

C1 = $2×\pi ×5x=10\mathrm{\pi }x$

C2 = $2×\pi ×3x=6\mathrm{\pi }x$
$\frac{{C}_{1}}{{C}_{2}}=\frac{10\pi x}{6\pi x}=\frac{5}{3}$
⇒ C1:C2 = 5:3
Hence, the ratio of the circumference of the given circle is 5:3.

#### Page No 248:

Radius of the circular field, r = 21 m.
Distance covered by the cyclist = Circumference of the circular field
= $2\mathrm{\pi r}$
= $\left(2×\frac{22}{7}×21\right)$ m = 132 m
Speed of the cyclist = 8 km per hour = = $\left(\frac{20}{9}\right)\mathrm{m}/\mathrm{s}$

Time taken by the cyclist to cover the field =
= $\left[\frac{132}{\left(\frac{20}{9}\right)}\right]\mathrm{s}$
= $\left(\frac{132×9}{20}\right)\mathrm{s}$
= 59.4 s

#### Page No 248:

Let the inner and outer radii of the track be r metres and R metres, respectively.

Then, $2\mathrm{\pi r}$ = 528
$2\mathrm{\pi R}$ = 616
$2×\frac{22}{7}×r=528$
$2×\frac{22}{7}×R=616$
r = $\left(528×\frac{7}{44}\right)=84$
R = $\left(616×\frac{7}{44}\right)=98$
⇒ (R - r) = (98 - 84) m = 14 m
Hence, the width of the track is 14 m.

#### Page No 248:

Let the inner and outer radii of the track be r metres and (r + 10.5) metres, respectively.

Inner circumference = 330 m
$2\mathrm{\pi r}=330$$2×\frac{22}{7}×r=330$
⇒ r =
Inner radius of the track = 52.5 m
∴ Outer radii of the track = (52.5 + 10.5) m = 63 m

∴ Circumference of the outer circle =
Rate of fencing = Rs. 20 per metre
∴ Total cost of fencing the outer circle = Rs. (396 $×$ 20) = Rs. 7920

#### Page No 248:

We know that the concentric circles are circles that form within each other, around a common centre point.
Radius of the inner circle, r = 98 cm
∴  Circumference of the inner circle = $2\mathrm{\pi r}$
= $\left(2×\frac{22}{7}×98\right)$ cm = 616 cm

Radius of the outer circle, R = 1 m 26 cm = 126 cm              [since 1 m = 100 cm]
∴  Circumference of the outer circle = $2\mathrm{\pi R}$
= $\left(2×\frac{22}{7}×126\right)$ cm = 792 cm
∴ Difference in the lengths of the circumference of the circles = (792 - 616) cm = 176 cm
Hence, the circumference of the second circle is 176 cm larger than that of the first circle.

#### Page No 248:

Length of the wire = Perimeter of the equilateral triangle
= 3 $×$ Side of the equilateral triangle = (3 $×$ 8.8) cm = 26.4 cm
Let the wire be bent into the form of a circle of radius r cm.
Circumference of the circle = 26.4 cm
$2\mathrm{\pi r}=26.4$
$2×\frac{22}{7}×r=26.4$
r = $\left(\frac{26.4×7}{2×22}\right)$ cm = 4.2 cm

∴ Diameter = 2r = (2 × 4.2) cm = 8.4 cm
Hence, the diameter of the ring is 8.4 cm.

#### Page No 248:

Circumference of the circle = Perimeter of the rhombus
= 4  × Side of the rhombus = (4  ×  33) cm = 132 cm

∴ Circumference of the circle = 132 cm
⇒ $2\mathrm{\pi r}=132$
$2×\frac{22}{7}×r=132$
r = $\left(\frac{132×7}{2×22}\right)$cm = 21 cm
Hence, the radius of the circle is 21 cm.

#### Page No 248:

Length of the wire = Perimeter of the rectangle
= 2(l + b) = 2  × (18.7 + 14.3) cm = 66 cm

Let the wire be bent into the form of a circle of radius r cm.

Circumference of the circle = 66 cm

$2\mathrm{\pi r}=66$
$\left(2×\frac{22}{7}×r\right)=66$
r = $\left(\frac{66×7}{2×22}\right)$ cm = 10.5 cm

Hence, the radius of the circle formed is 10.5 cm.

#### Page No 248:

It is given that the radius of the circle is 35 cm.
Length of the wire = Circumference of the circle
⇒ Circumference of the circle  = $2\mathrm{\pi r}$ = $\left(2×\frac{22}{7}×35\right)$ cm = 220 cm
Let the wire be bent into the form of a square of side a cm.
Perimeter of the square = 220 cm
⇒ 4a = 220
a = $\left(\frac{220}{4}\right)$cm = 55 cm
Hence, each side of the square will be 55 cm.

#### Page No 248:

Length of the hour hand (r)= 4.2 cm.
Distance covered by the hour hand in 12 hours = $2\mathrm{\pi r}$ = $\left(2×\frac{22}{7}×4.2\right)$ cm = 26.4 cm

∴ Distance covered by the hour hand in 24 hours = (2  × 26.4) = 52.8 cm
Length of the minute hand (R)= 7 cm
Distance covered by the minute hand in 1 hour = $2\mathrm{\pi R}$ = $\left(2×\frac{22}{7}×7\right)$ cm = 44 cm

∴ Distance covered by the minute hand in 24 hours = (44  × 24) cm = 1056 cm

∴ Sum of the distances covered by the tips of both the hands in 1 day = (52.8 + 1056) cm
= 1108.8 cm

#### Page No 248:

Given:
Diameter of the well (d) = 140 cm.
Radius of the well (r) = $\left(\frac{140}{2}\right)$cm = 70 cm

Let the radius of the outer circle (including the stone parapet) be R cm.
Length of the outer edge of the parapet = 616 cm
$2\mathrm{\pi R}=616$
$\left(2×\frac{22}{7}×R\right)=616$
R = $\left(\frac{616×7}{2×22}\right)$ cm = 98 cm

Now, width of the parapet = {Radius of the outer circle (including the stone parapet) - Radius of the well}
= {98 -70} cm = 28 cm
Hence, the width of the parapet is 28 cm.

#### Page No 249:

It may be noted that in one rotation, the bus covers a distance equal to the circumference of the wheel.
Now, diameter of the wheel = 98 cm
∴ Circumference of the wheel = $\mathrm{\pi d}$ = $\left(\frac{22}{7}×98\right)$ cm = 308 cm
Thus, the bus travels 308 cm in one rotation.

∴ Distance covered by the bus in 2000 rotations = (308  × 2000) cm
= 616000 cm
= 6160 m          [since 1 m = 100 cm]

#### Page No 249:

It may be noted that in one revolution, the cycle covers a distance equal to the circumference of the wheel.
Diameter of the wheel = 70 cm
∴ Circumference of the wheel = $\mathrm{\pi d}$ = $\left(\frac{22}{7}×70\right)$ cm = 220 cm
Thus, the cycle covers 220 cm in one revolution.

∴ Distance covered by the cycle in 250 revolutions = (220  × 250) cm
= 55000 cm
= 550 m              [since 1 m = 100 cm]

Hence, the cycle will cover 550 m in 250 revolutions.

#### Page No 249:

Diameter of the wheel = 77 cm
⇒ Radius of the wheel = $\left(\frac{77}{2}\right)$ cm
Circumference of the wheel = $2\mathrm{\pi r}$
= $\left(2×\frac{22}{7}×\frac{77}{2}\right)$cm = (22  × 11) cm = 242 cm
= $\left(\frac{242}{100}\right)$m = $\left(\frac{121}{50}\right)$m
Distance covered by the wheel in 1 revolution = $\left(\frac{121}{50}\right)$ m
Now, $\left(\frac{121}{50}\right)$ m is covered by the car in 1 revolution.
(121  × 1000) m will be covered by the car in $\left(1×\frac{50}{121}×121×1000\right)$ revolutions, i.e. 50000 revolutions.
∴ Required number of revolutions = 50000

#### Page No 249:

It may be noted that in one revolution, the bicycle covers a distance equal to the circumference of the wheel.
Total distance covered by the bicycle in 5000 revolutions = 11 km

⇒ 5000 × Circumference of the wheel = 11000 m                [since 1 km = 1000 m]

Circumference of the wheel = $\left(\frac{11000}{5000}\right)$ m =2.2 m = 220 cm               [since 1 m = 100 cm]

Circumference of the wheel =
⇒ 220 cm =
⇒ Diameter of the wheel = $\left(\frac{220×7}{22}\right)$ cm = 70 cm
Hence, the circumference of the wheel is 220 cm and its diameter is 70 cm.

#### Page No 252:

(i) Given:
r = 21 cm

∴ Area of the circle = $\left({\mathrm{\pi r}}^{2}\right)$ sq. units
= $\left(\frac{22}{7}×21×21\right)$ cm2 = $\left(22×3×21\right)$ cm2 = 1386 cm2

(ii) Given:
r = 3.5 m

Area of the circle = $\left({\mathrm{\pi r}}^{2}\right)$ sq. units
= $\left(\frac{22}{7}×3.5×3.5\right)$ m2 = $\left(22×0.5×3.5\right)$ m2 = 38.5 m2

#### Page No 252:

(i) Given:
d = 28 cm ⇒ r = $\left(\frac{d}{2}\right)$ = $\left(\frac{28}{2}\right)$ cm = 14 cm
Area of the circle = $\left({\mathrm{\pi r}}^{2}\right)$ sq. units
= $\left(\frac{22}{7}×14×14\right)$ cm2 = $\left(22×2×14\right)$ cm2 = 616 cm2

(ii) Given:
r = 1.4 m ⇒ r = $\left(\frac{d}{2}\right)$ = $\left(\frac{1.4}{2}\right)$m = 0.7 m
Area of the circle = $\left({\mathrm{\pi r}}^{2}\right)$ sq. units
= $\left(\frac{22}{7}×0.7×0.7\right)$ m2 = $\left(22×0.1×0.7\right)$ m2 = 1.54 m2

#### Page No 252:

Let the radius of the circle be r cm.
Circumference = $\left(2\mathrm{\pi r}\right)$cm
$\left(2\mathrm{\pi r}\right)$ = 264
$\left(2×\frac{22}{7}×r\right)=264$
r = $\left(\frac{264×7}{2×22}\right)$ = 42
∴ Area of the circle = ${\mathrm{\pi r}}^{2}$
= $\left(\frac{22}{7}×42×42\right)$ cm2
= 5544 cm2

#### Page No 252:

Let the radius of the circle be r m.
Then, its circumference will be $\left(2\mathrm{\pi r}\right)$m.
$\left(2\mathrm{\pi r}\right)$ = 35.2
$\left(2×\frac{22}{7}×r\right)=35.2$
r = $\left(\frac{35.2×7}{2×22}\right)$ = 5.6
∴ Area of the circle = ${\mathrm{\pi r}}^{2}$
= $\left(\frac{22}{7}×5.6×5.6\right)$ m2 = 98.56 m2

#### Page No 252:

Let the radius of the circle be r cm.
Then, its area will be ${\mathrm{\pi r}}^{2}$ cm2.
∴ ${\mathrm{\pi r}}^{2}$ = 616
$\left(\frac{22}{7}×r×r\right)$ = 616
r2 = $\left(\frac{616×7}{22}\right)$ = 196
r = $\sqrt{196}$ = 14
⇒  Circumference of the circle = $\left(2\mathrm{\pi r}\right)$ cm
= $\left(2×\frac{22}{7}×14\right)$ cm = 88 cm

#### Page No 252:

Let the radius of the circle be r m.
Then, area = ${\mathrm{\pi r}}^{2}$ m2
∴ ${\mathrm{\pi r}}^{2}$ = 1386
$\left(\frac{22}{7}×r×r\right)$ = 1386
r2 = $\left(\frac{1386×7}{22}\right)$ = 441
r = $\sqrt{441}$ = 21
⇒  Circumference of the circle = $\left(2\mathrm{\pi r}\right)$ m
= $\left(2×\frac{22}{7}×21\right)$ m = 132 m

#### Page No 252:

Let r1 and r2 be the radii of the two given circles and A1 and A2 be their respective areas.

$\frac{{r}_{1}}{{r}_{2}}=\frac{4}{5}$
$\frac{{A}_{1}}{{A}_{2}}=\frac{\mathrm{\pi }{{r}_{1}}^{2}}{\mathrm{\pi }{{r}_{2}}^{2}}=\frac{{{r}_{1}}^{2}}{{{r}_{2}}^{2}}={\left(\frac{{r}_{1}}{{r}_{2}}\right)}^{2}={\left(\frac{4}{5}\right)}^{2}=\frac{16}{25}$
Hence, the ratio of the areas of the given circles is 16:25.

#### Page No 252:

If the horse is tied to a pole, then the pole will be the central point and the area over which the horse will graze will be a circle. The string by which the horse is tied will be the radius of the circle.
Thus,
Radius of the circle (r) = Length of the string = 21 m

Now, area of the circle = ${\mathrm{\pi r}}^{2}$ = $\left(\frac{22}{7}×21×21\right)$ m2 = 1386 m2
∴ Required area = 1386 m2

#### Page No 252:

Let a be one side of the square.
Area of the square = 121 cm2                (given)
⇒ a2 = 121
⇒ a = 11 cm   (since 11  × 11 = 121)
Perimeter of the square = 4  × side = 4a = (4  × 11) cm = 44 cm
Length of the wire = Perimeter of the square
= 44 cm
The wire is bent in the form of a circle.
Circumference of a circle = Length of the wire
∴ Circumference of a circle = 44 cm
$2\mathrm{\pi r}=44$
$\left(2×\frac{22}{7}×r\right)=44$
r = $\left(\frac{44×7}{2×22}\right)$= 7 cm
∴ Area of the circle = ${\mathrm{\pi r}}^{2}$
= $\left(\frac{22}{7}×7×7\right)$ cm2
= 154 cm2

#### Page No 252:

It is given that the radius of the circle is 28 cm.

Length of the wire = Circumference of the circle
⇒ Circumference of the circle  = $2\mathrm{\pi r}=\left(2×\frac{22}{7}×28\right)$ cm = 176 cm
Let the wire be bent into the form of a square of side a cm.

Perimeter of the square = 176 cm

⇒ 4a = 176
a = $\left(\frac{176}{4}\right)$cm = 44 cm
Thus, each side of the square is 44 cm.

Area of the square = (Side)2 = (a)2 = (44 cm)2
= 1936 cm2
∴ Required area of the square formed = 1936 cm2

#### Page No 252:

Area of the acrylic sheet = 34 cm  × 24 cm = 816 cm2
Given that the diameter of a circular button is 3.5 cm.
∴ Radius of the circular button (r)= $\left(\frac{3.5}{2}\right)$ cm = 1.75 cm
∴ Area of 1 circular button = ${\mathrm{\pi r}}^{2}$
= $\left(\frac{22}{7}×1.75×1.75\right)$ cm2
= 9.625 cm2
∴ Area of 64 such buttons = (64  × 9.625) cm2 = 616 cm2
Area of the remaining acrylic sheet = (Area of the acrylic sheet - Area of 64 circular buttons)
= (816 - 616) cm2 = 200 cm2

#### Page No 253:

Area of the rectangular ground = 90 m  × 32 m = (90  × 32) m2 = 2880 m2
Given:
Radius of the circular tank (r) = 14 m
∴ Area covered by the circular tank = = $\left(\frac{22}{7}×14×14\right)$ m2
= 616 m2

∴ Remaining portion of the rectangular ground for turfing = (Area of the rectangular ground - Area covered by the circular tank)
= (2880 - 616) m2 = 2264 m2
Rate of turfing = Rs 50 per sq. metre
∴ Total cost of turfing the remaining ground = Rs (50  × 2264) = Rs 1,13,200

#### Page No 253:

Area of each of the four quadrants is equal to each other with radius 7 cm.

Area of the square ABCD = (Side)2 = (14 cm)2 = 196 cm2
Sum of the areas of the four quadrants = $\left(4×\frac{1}{4}×\frac{22}{7}×7×7\right)$ cm2
= 154 cm2
∴ Area of the shaded portion  = Area of square ABCD - Areas of the four quadrants
= (196 - 154) cm2
= 42 cm2

#### Page No 253:

Let ABCD be the rectangular field.

Here, AB = 60 m
BC = 40 m

Let the horse be tethered to corner A by a 14 m long rope.

Then, it can graze through a quadrant of a circle of radius 14 m.
∴ Required area of the field = $\left(\frac{1}{4}×\frac{22}{7}×14×14\right)$ m2 = 154 m2
Hence, horse can graze 154 m2 area of the rectangular field.

#### Page No 253:

Diameter of the big circle = 21 cm
Radius = $\left(\frac{21}{2}\right)$ cm = 10.5 cm
∴ Area of the bigger circle = ${\mathrm{\pi r}}^{2}$ = $\left(\frac{22}{7}×10.5×10.5\right)$ cm2
= 346.5 cm2

Diameter of circle I  = $\frac{2}{3}$ of the diameter of the bigger circle
= $\frac{2}{3}$ of 21 cm = $\left(\frac{2}{3}×21\right)$ cm = 14 cm
Radius of circle I (r1) = $\left(\frac{14}{2}\right)$ cm = 7 cm
∴ Area of circle I = ${{\mathrm{\pi r}}_{1}}^{2}$ = $\left(\frac{22}{7}×7×7\right)$ cm2
= 154 cm2

Diameter of circle II  = $\frac{1}{3}$ of the diameter of the bigger circle
= $\frac{1}{3}$ of 21 cm = $\left(\frac{1}{3}×21\right)$ cm = 7 cm
Radius of circle II (r2) = $\left(\frac{7}{2}\right)$ cm = 3.5 cm
∴ Area of circle II = ${{\mathrm{\pi r}}_{2}}^{2}$ = $\left(\frac{22}{7}×3.5×3.5\right)$ cm2
= 38.5 cm2

∴ Area of the shaded portion = {Area of the bigger circle - (Sum of the areas of circle I and II)}
= {346.5 - (154 + 38.5)} cm2
= {346.5 - 192.5} cm2

= 154 cm2
Hence, the area of the shaded portion is 154 cm2

#### Page No 253:

Let ABCD be the rectangular plot of land that measures 8 m by 6 m.
∴ Area of the plot = (8 m  × 6 m) = 48 m2
Area of the four flower beds = $\left(4×\frac{1}{4}×\frac{22}{7}×2×2\right)$ m2 = $\left(\frac{88}{7}\right)$ m2
Area of the circular flower bed in the middle of the plot = ${\mathrm{\pi r}}^{2}$
= $\left(\frac{22}{7}×2×2\right)$ m2 = $\left(\frac{88}{7}\right)$ m2

Area of the remaining part = $\left\{48-\left(\frac{88}{7}+\frac{88}{7}\right)\right\}$ m2
= $\left\{48-\frac{176}{7}\right\}$ m2
= $\left\{\frac{336-176}{7}\right\}$ m2 = $\left(\frac{160}{7}\right)$ m2 = 22.86 m2
∴ Required area of the remaining plot = 22.86 m2

#### Page No 253:

(c) 192 cm2

Let ABCD be the rectangular plot.
Then, AB = 16 cm
AC = 20 cm

Let BC = x cm
From right triangle ABC:
AC2 = AB2 + BC2
⇒ (20)2 = (16)2 + x2
x2 = (20)2 - (16)2 ⇒ {400 - 256} = 144
x = $\sqrt{144}$ = 12
∴ BC = 12 cm
∴ Area of the plot = (16 × 12) cm2 = 192 cm2

#### Page No 253:

(b) 72 cm2

Given:
Diagonal of the square = 12 cm
∴ Area of the square = $\left\{\frac{1}{2}×{\left(\mathrm{Diagonal}\right)}^{2}\right\}$ sq. units.
= $\left\{\frac{1}{2}×{\left(12\right)}^{2}\right\}$ cm2
= 72 cm2

#### Page No 254:

(b) 20 cm

Area of the square = $\left\{\frac{1}{2}×{\left(D\mathrm{iagonal}\right)}^{2}\right\}$ sq. units.
Area of the square field = 200 cm2

Diagonal of a square =
= $\left(\sqrt{2×200}\right)$ cm = $\left(\sqrt{400}\right)$ cm = 20 cm
∴ Length of the diagonal of the square = 20 cm

#### Page No 254:

(a) 100 m

Area of the square = $\left\{\frac{1}{2}×{\left(D\mathrm{iagonal}\right)}^{2}\right\}$sq. units.
Given:
Area of square field = 0.5 hectare
= $\left(0.5×10000\right)$m2                     [since 1 hectare = 10000 m2]
= 5000 m2

Diagonal of a square =
= $\left(\sqrt{2×5000}\right)$m = 100 m
Hence, the length of the diagonal of a square field is 100 m.

#### Page No 254:

(c) 90 m

Let the breadth of the rectangular field be x m.
Length = 3x m
Perimeter of the rectangular field = 2(l + b)
⇒ 240 = 2( x + 3x)
⇒ 240 = 2(4x)
⇒ 240 = 8x     ⇒ x = $\left(\frac{240}{8}\right)=30$
∴ Length of the field = 3x = (3 × 30) m = 90 m

#### Page No 254:

(d) 56.25%

Let the side of the square be a cm.
Area of the square = (a)2 cm2
Increased side = (a + 25% of a) cm
= $\left(a+\frac{25}{100}a\right)$ cm = $\left(a+\frac{1}{4}a\right)\mathrm{cm}=\left(\frac{5}{4}a\right)$ cm
Area of the square = ${\left(\frac{5}{4}a\right)}^{2}c{\mathrm{m}}^{2}=\left(\frac{25}{16}{a}^{2}\right)$ cm2
Increase in the area = $\left[\left(\frac{25}{16}{a}^{2}\right)-{a}^{2}\right]$ cm2= $\left(\frac{25{a}^{2}-16{a}^{2}}{16}\right)$ cm2 = $\left(\frac{9{a}^{2}}{16}\right)$ cm2
% increase in the area =
= $\left[\frac{\left(\frac{9}{16}{a}^{2}\right)}{{a}^{2}}×100\right]$ = $\left(\frac{9×100}{16}\right)=56.25$

#### Page No 254:

(b) 1:2

Let the side of the square be a.
Length of its diagonal = $\sqrt{2}a$
∴ Required ratio = $\frac{{a}^{2}}{{\left(\sqrt{2}a\right)}^{2}}=\frac{{a}^{2}}{2{a}^{2}}=\frac{1}{2}=1:2$

#### Page No 254:

(c) A > B

We know that a square encloses more area even though its perimeter is the same as that of the rectangle.

∴ Area of a square  > Area of a rectangle

#### Page No 254:

(b) 13500 m2

Let the length of the rectangular field be 5x.
Perimeter of the field = 2(l + b) = 480 m          (given)
⇒ 480 = 2(5x + 3x)  ⇒ 480 = 16x
x$\frac{480}{16}$ = 30
∴ Length = 5x = (5 × 30) = 150 m
Breadth = 3x = (3 × 30) = 90 m
∴ Area of the rectangular park = 150 m × 90 m = 13500 m2

#### Page No 254:

(a) 6 m

Total cost of carpeting = Rs 6000
Rate of carpeting = Rs 50 per m
∴ Length of the carpet = $\left(\frac{6000}{50}\right)$ m = 120 m
∴ Area of the carpet = $\left(120×\frac{75}{100}\right)$ m2 =  90 m2     [since 75 cm = ]
Area of the floor = Area of the carpet = 90 m2
∴ Width of the room =

#### Page No 254:

(a) 84 cm2

Let a = 13 cm, b = 14 cm and c = 15 cm
Then, s = $\frac{a+b+c}{2}$ = $\left(\frac{13+14+15}{2}\right)$ cm = 21 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
= $\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}$ cm2
=  $\sqrt{21×8×7×6}$ cm2
= $\sqrt{3×7×2×2×2×7×2×3}$ cm2
= (2 × 2 × 3 × 7) cm2
= 84 cm2

#### Page No 254:

(b) 48 m2

Base = 12 m
Height = 8 m
Area of the triangle = sq. units
= $\left(\frac{1}{2}×12×8\right)$ m2
= 48 m2

#### Page No 254:

(b) 4 cm

Area of the equilateral triangle = $4\sqrt{3}$ cm2

We know:
Area of an equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^{2}$ sq. units
∴ Side of the equilateral triangle = $\left[\sqrt{\left(\frac{4×\mathrm{Area}}{\sqrt{3}}\right)}\right]$ cm
=   $\left[\sqrt{\left(\frac{4×4\sqrt{3}}{\sqrt{3}}\right)}\right]$cm = $\left(\sqrt{4×4}\right)$ cm = $\left(\sqrt{16}\right)$cm = 4 cm

#### Page No 254:

(c) $16\sqrt{3}$ cm2

It is given that one side of an equilateral triangle is 8 cm.
∴ Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^{2}$ sq. units
=  $\frac{\sqrt{3}}{4}{\left(8\right)}^{2}$ cm2
= $\left(\frac{\sqrt{3}}{4}×64\right)$ cm2 = $16\sqrt{3}$ cm2

#### Page No 254:

(b) $2\sqrt{3}$ cm2

Let ΔABC be an equilateral triangle with one side of the length a cm.
Diagonal of an equilateral triangle = $\frac{\sqrt{3}}{2}a$ cm
$\frac{\sqrt{3}}{2}a=\sqrt{6}$
a = $\frac{\sqrt{6}×2}{\sqrt{3}}=\frac{\sqrt{3}×\sqrt{2}×2}{\sqrt{3}}=2\sqrt{2}$ cm
Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{a}^{2}$
= $\frac{\sqrt{3}}{4}{\left(2\sqrt{2}\right)}^{2}$ cm2 = $\left(\frac{\sqrt{3}}{4}×8\right)$ cm2 = $2\sqrt{3}$ cm2

#### Page No 254:

(b) 72 cm2

Base of the parallelogram = 16 cm
Height of the parallelogram = 4.5 cm
∴ Area of the parallelogram = Base × Height
= (16 × 4.5) cm2  = 72 cm2

#### Page No 255:

(b) 216 cm2

Length of one diagonal = 24 cm
Length of the other diagonal = 18 cm
∴ Area of the rhombus = $\frac{1}{2}$ × (Product of the diagonals)
= $\left(\frac{1}{2}×24×18\right)$ cm2 = 216 cm2

#### Page No 255:

(c) 154 cm2

Let the radius of the circle be r cm.
Circumference = $2\mathrm{\pi r}$

$r\left(2\mathrm{\pi }-1\right)=37$
r = $\frac{37}{\left(2\mathrm{\pi }-1\right)}$ = $\frac{37}{\left(2×\frac{22}{7}-1\right)}=\frac{37}{\left(\frac{44}{7}-1\right)}=\frac{37}{\left(\frac{44-7}{7}\right)}=\left(\frac{37×7}{37}\right)=7$
∴ Radius of the given circle is 7 cm.
∴ Area = ${\mathrm{\pi r}}^{2}$ = $\left(\frac{22}{7}×7×7\right)$ cm2 = 154 cm2

#### Page No 255:

(c) 54 m2

Given:
Perimeter of the floor = 2(l + b) = 18 m
Height of the room = 3 m

∴ Area of the four walls = {2(l + b) × h}
= Perimeter × Height
= 18 m × 3 m = 54 m2

#### Page No 255:

(a) 200 m

Area of the floor of a room = 14 m × 9 m = 126 m2

Width of the carpet = 63 cm = 0.63 m                (since 100 cm = 1 m)

∴ Required length of the carpet =
=

#### Page No 255:

(c) 120 cm2

Let the length of the rectangle be x cm and the breadth be y cm.
Area of the rectangle = xy cm2
Perimeter of the rectangle = 2( x + y) = 46 cm          (given)
⇒ 2( x + y) = 46
⇒ ( x + y) = $\left(\frac{46}{2}\right)$ cm = 23 cm

Diagonal of the rectangle = $\sqrt{{x}^{2}+{y}^{2}}$ = 17 cm
⇒  $\sqrt{{x}^{2}+{y}^{2}}$ = 17

Squaring both the sides, we get:
x2 + y2 = (17)2
x2 + y2 = 289

Now, (x2 + y2) = ( x + y)2 - 2xy
⇒ 2xy = ( x + y)2 - (x2 + y2)
= (23)2 - 289
= 529 - 289 = 240
xy = $\left(\frac{240}{2}\right)$ cm2 = 120 cm2

#### Page No 255:

(b) 3:1

Let a side of the first square be a cm and that of the second square be b cm.
Then, their areas will be a2 and b2, respectively.
Their perimeters will be 4a and 4b, respectively.

According to the question:
$\frac{{a}^{2}}{{b}^{2}}=\frac{9}{1}$${\left(\frac{a}{b}\right)}^{2}=\frac{9}{1}={\left(\frac{3}{1}\right)}^{2}$$\frac{a}{b}=\frac{3}{1}$

∴ Required ratio of the perimeters = $\frac{4a}{4b}=\frac{4×3}{4×1}=\frac{3}{1}$= 3:1

#### Page No 255:

(d) 4:1

Let the diagonals be 2d and d.
Area of the square = $\left[\frac{1}{2}×{\left(\mathrm{Diagonal}\right)}^{2}\right]$ sq. units
Required ratio = $\frac{{A}_{1}}{{A}_{2}}=\frac{\frac{1}{2}{\left(2d\right)}^{2}}{\frac{1}{2}{\left(d\right)}^{2}}=\frac{4{d}^{2}}{{d}^{2}}=\frac{4}{1}=4:1$

#### Page No 255:

(c) 49 m

Let the width of the rectangle be x m.

Given:
Area of the rectangle = Area of the square
⇒ Length × Width = Side × Side
⇒ (144 × x) = 84 × 84
∴ Width (x) = $\left(\frac{84×84}{144}\right)$ m = 49 m
Hence, width of the rectangle is 49 m.

#### Page No 255:

(d) $4:\sqrt{3}$

Let one side of the square and that of an equilateral triangle be the same, i.e. a units.
Then, Area of the square = (Side)2 = (a)2
Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^{2}$ = $\frac{\sqrt{3}}{4}{\left(\mathrm{a}\right)}^{2}$
∴ Required ratio = $\frac{{a}^{2}}{\frac{\sqrt{3}}{4}{a}^{2}}=\frac{4}{\sqrt{3}}=4:\sqrt{3}$

#### Page No 255:

(a) $\sqrt{\mathrm{\pi }}:1$

Let the side of the square be x cm and the radius of the circle be r cm.
Area of the square = Area of the circle
⇒ (x)2 = ${\mathrm{\pi r}}^{2}$
∴ Side of the square (x) = $\sqrt{\mathrm{\pi }}r$
Required ratio =
= $\frac{x}{r}=\frac{\sqrt{\mathrm{\pi }}r}{r}=\frac{\sqrt{\mathrm{\pi }}}{1}=\sqrt{\mathrm{\pi }}:1$

#### Page No 255:

(b)

Let the radius of the circle be r cm.
Then, its area = ${\mathrm{\pi r}}^{2}$ cm2
∴  ${\mathrm{\pi r}}^{2}$ = 154
$\frac{22}{7}×r×r=154$
r2$\left(\frac{154×7}{22}\right)$ = 49
r$\sqrt{49}$cm = 7 cm

Side of the equilateral triangle =  Radius of the circle
= 7 cm
∴ Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^{2}$ sq. units

= $\frac{\sqrt{3}}{4}{\left(7\right)}^{2}$ cm2

= $\frac{49\sqrt{3}}{4}$ cm2

#### Page No 255:

(c) 12 cm

Area of the rhombus = $\frac{1}{2}$ × (Product of the diagonals)
Given:
Length of one diagonal = 6 cm
Area of the rhombus = 36 cm2

∴ Length of the other diagonal = $\left(\frac{36×2}{6}\right)$ cm = 12 cm

#### Page No 256:

(d) 24 cm

Let the length of the shorter diagonal of the rhombus be x cm.
∴ Longer diagonal = 2x

Area of the rhombus = $\frac{1}{2}$× (Product of its diagonals)
⇒ 144 = $\left(\frac{1}{2}×x×2x\right)$

⇒ 144 = $\left(\frac{2{x}^{2}}{2}\right)$ = $\left({x}^{2}\right)$

x = $\left(\sqrt{144}\right)$cm = 12 cm

∴ Length of the longer diagonal = 2x
= (2 × 12) cm
= 24 cm

#### Page No 256:

(c) 17.60 m

Let the radius of the circle be r m.
Area = ${\mathrm{\pi r}}^{2}$ m2
∴ ${\mathrm{\pi r}}^{2}$ = 24.64
$\left(\frac{22}{7}×r×r\right)$ = 24.64
r2 = $\left(\frac{24.64×7}{22}\right)$ = 7.84
r = $\sqrt{7.84}$ = 2.8 m
⇒  Circumference of the circle = $\left(2\mathrm{\pi r}\right)$ m
= $\left(2×\frac{22}{7}×2.8\right)$ m = 17.60 m

#### Page No 256:

(c) 3 cm

Suppose the radius of the original circle is r cm.
Area of the original circle = ${\mathrm{\pi r}}^{2}$

Radius of the circle = (r +1) cm
According to the question:

$\mathrm{\pi }\left({\mathrm{r}}^{2}+1+2\mathrm{r}\right)={\mathrm{\pi r}}^{2}+22$
${\mathrm{\pi r}}^{2}+\mathrm{\pi }+2\mathrm{\pi r}={\mathrm{\pi r}}^{2}+22$
$\mathrm{\pi }+2\mathrm{\pi r}=22$      [cancel ${\mathrm{\pi r}}^{2}$ from both the sides of the equation]
$\mathrm{\pi }\left(1+2\mathrm{r}\right)=22$
$\left(1+2r\right)=\frac{22}{\mathrm{\pi }}=\left(\frac{22×7}{22}\right)=7$
⇒ 2r = 7 -1 = 6
r = $\left(\frac{6}{2}\right)$ cm = 3 cm
∴ Original radius of the circle = 3 cm

#### Page No 256:

(c) 1000

Radius of the wheel = 1.75 m
Circumference of the wheel = $2\mathrm{\pi r}$
= $\left(2×\frac{22}{7}×1.75\right)$cm = (2 × 22 × 0.25) m = 11 m

Distance covered by the wheel in 1 revolution is 11 m.
Now, 11 m is covered by the car in 1 revolution.
(11 × 1000) m will be covered by the car in $\left(1×\frac{1}{11}×11×1000\right)$ revolutions, i.e. 1000 revolutions.
∴ Required number of revolutions = 1000

#### Page No 257:

We know that all the angles of a rectangle are 90° and the diagonal divides the rectangle into two right angled triangles.
So, one side of the triangle will be 48 m and the diagonal, which is 50 m, will be the hypotenuse.
According to Pythagoras theorem:
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
Perpendicular = $\sqrt{{\left(\mathrm{Hypotenuse}\right)}^{2}-\left(\mathrm{Base}{\right)}^{2}}$
Perpendicular = $\sqrt{{\left(50\right)}^{2}-{\left(48\right)}^{2}}=\sqrt{2500-2304}=\sqrt{196}=14$ m
∴ Other side of the rectangular plot = 14 m

∴ Area of the rectangular plot = 48 m × 14 m = 672 m2
Hence, the area of a rectangular plot is 672 m2.

#### Page No 257:

Length = 9 m
Height = 6.5 m

Area of the four walls = {2(l + b) × h} sq. units
= {2(9 + 8) × 6.5} m2 = {34 × 6.5} m2 = 221 m2
Area of one door = (2 × 1.5) m2 = 3 m2
Area of one window = (1.5 × 1) m2 = 1.5 m2
∴ Area of four windows = (4 × 1.5) m2 = 6 m2
Total area of one door and four windows = (3 + 6) m2
= 9 m2
Area to be painted = (221 - 9) m2 = 212 m2
Rate of painting = Rs 50 per m2
Total cost of painting = Rs ( 212 × 50) = Rs 10600

#### Page No 257:

Given that the diagonal of a square is 64 cm.

Area of the square = $\left\{\frac{1}{2}×{\left(\mathrm{diagonal}\right)}^{2}\right\}$ sq. units.
= $\left\{\frac{1}{2}×{\left(64\right)}^{2}\right\}$ cm2 =$\left\{\frac{1}{2}×4096\right\}$ cm2 = 2048 cm2
∴ Area of the square = 2048 cm2

#### Page No 257:

Let ABCD be the square lawn and PQRS be the outer boundary of the square path.

Let one side of the lawn (AB) be x m.
Area of the square lawn = x2
Length PQ = (x m + 2 m + 2 m) = (x + 4) m
∴ Area of PQRS = (x + 4)2 = (x2 + 8x + 16) m2

Now, Area of the path = Area of PQRS − Area of the square lawn
⇒ 136 = x2 + 8x + 16 x2
⇒ 136 = 8x + 16
⇒ 136 16 = 8x
⇒ 120 = 8x
x = 120 ÷ 8 = 15
∴ Side of the lawn = 15 m

∴ Area of the lawn = (Side)2 = (15 m)2 = 225 m2

#### Page No 257:

Let ABCD be the rectangular park.
EFGH and IJKL are the two rectangular roads with width 2 m.

Length of the rectangular park AD = 30 cm
Breadth of the rectangular park CD = 20 cm
Area of the road EFGH = 30 m × 2 m = 60 m2
Area of the road IJKL = 20 m × 2 m = 40 m2
Clearly, area of MNOP is common to the two roads.
∴ Area of MNOP = 2 m × 2 m = 4 m2

∴ Area of the roads = Area (EFGH) + Area (IJKL) − Area (MNOP)
= (60  + 40 ) m2 − 4 m2 = 96 m2

#### Page No 257:

Let ABCD be the rhombus whose diagonals intersect at O.

Then, AB = 13 cm
AC = 24 cm
The diagonals of a rhombus bisect each other at right angles.
Therefore, ΔAOB is a right-angled triangle, right angled at O, such that:
OA = $\frac{1}{2}$AC = 12 cm
AB = 13 cm

By Pythagoras theorem:
(AB)2 = (OA)2 + (OB)2
⇒ (13)2 = (12)2 + (OB)2
⇒ (OB)2 = (13)2 − (12)2
⇒ (OB)2 = 169 − 144 = 25
⇒ (OB)2 = (25)2
⇒ OB = 5 cm
∴ BD = 2 × OB = 2 × 5 cm = 10 cm

∴ Area of the rhombus ABCD = $\frac{1}{2}$ × AC × BD cm2
= $\left(\frac{1}{2}×24×10\right)$ cm2 = 120 cm2

#### Page No 257:

Let the base of the parallelogram be x m.
Then, the altitude of the parallelogram will be 2x m.
It is given that the area of the parallelogram is 338 m2.

Area of a parallelogram =  Base × Altitude
∴ 338 m2x × 2x
338 m2 = 2x2
x2 = $\left(\frac{338}{2}\right)$ m2 = 169 m2
x2 = 169 m2
x = 13 m
∴ Base = x m = 13 m
Altitude = 2x m = (2 × 13)m = 26 m

#### Page No 257:

Consider ΔABC.
Here, ∠B = 90°
AB = 24 cm
AC = 25 cm

Now, AB2 + BC2 = AC2
⇒ BC2 = AC2 - AB2
= (252 - 242)
= (625 - 576)
= 49
⇒ BC = $\left(\sqrt{49}\right)$cm = 7 cm
Area of ΔABC = $\left(\frac{1}{2}×\mathrm{BC}×\mathrm{AB}\right)$ sq. units
= $\left(\frac{1}{2}×7×24\right)$ cm2 = 84 cm2
Hence, area of the right angled triangle is 84 cm2.

#### Page No 257:

Radius of the wheel = 35 cm
Circumference of the wheel = $2\mathrm{\pi r}$
= $\left(2×\frac{22}{7}×35\right)$cm = (44 × 5) cm = 220 cm
= $\left(\frac{220}{100}\right)$ m = $\left(\frac{11}{5}\right)$ m
Distance covered by the wheel in 1 revolution = $\left(\frac{11}{5}\right)$ m
Now, $\left(\frac{11}{5}\right)$ m is covered by the car in 1 revolution.
Thus, (33 × 1000) m will be covered by the car in $\left(1×\frac{5}{11}×33×1000\right)$ revolutions, i.e. 15000 revolutions.
∴ Required number of revolutions = 15000

#### Page No 257:

Let the radius of the circle be r cm.
∴ Area = $\left({\mathrm{\pi r}}^{2}\right)$ cm2
∴ ${\mathrm{\pi r}}^{2}$ = 616
$\left(\frac{22}{7}×r×r\right)$ = 616
r2 = $\left(\frac{616×7}{22}\right)$ = 196
r = $\sqrt{196}$ = 14 cm
Hence, the radius of the given circle is 14 cm.

#### Page No 257:

(a) 14 cm

Let the radius of the circle be r cm.
Then, its area will be $\left({\mathrm{\pi r}}^{2}\right)$ cm2.
∴ ${\mathrm{\pi r}}^{2}$ = 154
$\left(\frac{22}{7}×r×r\right)$ = 154
r2 = $\left(\frac{154×7}{22}\right)$ = 49
r = $\sqrt{49}$ = 7 cm
∴ Diameter of the circle = 2r = (2 × 7) cm = 14 cm

#### Page No 257:

(b) 154 cm2

Let the radius of the circle be r cm.
Circumference = $\left(2\mathrm{\pi r}\right)$cm
$\left(2\mathrm{\pi r}\right)$ = 44
$\left(2×\frac{22}{7}×r\right)=44$
r = $\left(\frac{44×7}{2×22}\right)$ = 7 cm
∴ Area of the circle = ${\mathrm{\pi r}}^{2}$
= $\left(\frac{22}{7}×7×7\right)$ cm2 = 154 cm2

#### Page No 257:

(c) 98 cm2

Given that the diagonal of a square is 14 cm.

Area of a square = $\left\{\frac{1}{2}×{\left(D\mathrm{iagonal}\right)}^{2}\right\}$sq. units.
= $\left\{\frac{1}{2}×{\left(14\right)}^{2}\right\}$ cm2 =$\left\{\frac{1}{2}×196\right\}$ cm2 = 98 cm2
Hence, area of the square is 98 cm2.

#### Page No 257:

(b) 10 cm

Given that the area of the square is 50 cm2.
We know:
Area of a square = $\left\{\frac{1}{2}×{\left(\mathrm{Diagonal}\right)}^{2}\right\}$sq. units
∴ Diagonal of the square = = $\left(\sqrt{2×50}\right)$cm = $\left(\sqrt{100}\right)$cm = 10 cm
Hence, the diagonal of the square is 10 cm.

#### Page No 257:

(a) 192 m2

Let the length of the rectangular park be 4x.
Perimeter of the park = 2(l + b) = 56 m (given)
⇒ 56 = 2(4x + 3x)
⇒ 56 = 14x
x = $\frac{56}{14}$= 4
Length = 4x = (4 × 4) = 16 m
Breadth = 3x = (3 × 4) = 12 m
∴ Area of the rectangular park = 16 m × 12 m = 192 m2

#### Page No 257:

(a) 84 cm2

Let a = 13 cm, b = 14 cm and c = 15 cm
s = $\frac{a+b+c}{2}$ = $\left(\frac{13+14+15}{2}\right)$cm = 21 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
= $\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}$ cm2
=  $\sqrt{21×8×7×6}$ cm2
= $\sqrt{3×7×2×2×2×7×2×3}$ cm2
= (2 × 2 × 3 × 7) cm2 = 84 cm2

#### Page No 257:

(a)

Given that each side of an equilateral triangle is 8 cm.
∴ Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^{2}$ sq. units

=  $\frac{\sqrt{3}}{4}{\left(8\right)}^{2}$ cm2

= $\left(\frac{\sqrt{3}}{4}×64\right)$ cm2 = $16\sqrt{3}$ cm2

#### Page No 257:

(b) 91 cm2

Base  = 14 cm
Height = 6.5 cm
∴ Area of the parallelogram = Base × Height
= (14 × 6.5) cm2
=  91 cm2

#### Page No 257:

(b) 135 cm2

Area of the rhombus = $\frac{1}{2}$ ×  (Product of the diagonals)
= $\left(\frac{1}{2}×18×15\right)$ cm2 = 135 cm2
Hence, the area of the rhombus is 135 cm2.

#### Page No 258:

(i) If d1 and d2 be the diagonals of a rhombus, then its area is $\left(\frac{1}{2}{d}_{1}{d}_{2}\right)$ sq. units.
Area of a rhombus = $\frac{1}{2}$× (Product of its diagonals)

(ii) If l, b and h are the length, breadth and height respectively of a room, then area of its 4 walls = 2h(l + b) sq. units.

(iii) 1 hectare = (10000) m2
(since 1 hectometre = 100 m)
∴1 hectare = (100 × 100) m2
(iv) 1 acre = 100 m2
(v) If each side of a triangle is a cm, then its area = $\frac{\sqrt{3}}{4}{a}^{2}$ cm2.
Area of equilateral triangle with side a = $\left(\frac{\sqrt{3}}{4}{a}^{2}\right)$sq. units.

#### Page No 258:

(i) F
Area of a triangle = $\frac{1}{2}×\mathrm{Base}×\mathrm{Height}$

(ii) T
Area of a parallelogram = Base × Height

(iii) F
Area of a circle = ${\mathrm{\pi r}}^{2}$

(iv) T
Circumference of a circle = $2\mathrm{\pi r}$2πr2

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