Rs.369

Rs.2000

Area of room = 1.4 \(\times\) 9.5 m² = 1.4\(\times\) 9.5\(\times\) 10000 cm²
Area of one tile = 20\(\times\) 10= 200 cm²

Number of tiles\(=\frac { Area\quad of\quad the\quad room }{ Area\quad of\quad tile } =\frac { 74\times 95\times 100 }{ 200 } \)
\(=\frac { 74\times 95 }{ 2 } =95\times 37=3515\)

(i) Perimeter = AB + AC + BC + BD + DE + CE
= 3+ 3+ 3+ 3+ 3+ 3= 18 cm
(ii) Area of square BCED = Side² = 3² = 9 sq cm

24 sq m

Given, length of a rectangular plot = 500 m
and breadth of a rectangular plot = 200 m
∴ Area of the rectangular plot = Length \(\times\) Breadth
= 500 m\(\times\) 200 m
= 100000 sq m
∴ Cost of tiling per hundred square metres = Rs. 8
∴ Cost of 1 sq m = Rs.\(\frac { 8 }{ 100 } \)
Now, cost of 100000 sq metre =Rs.100000\(\times\frac { 8 }{ 100 } \)
=Rs. 8000
Hence, the cost of tiling rectangular plot is Rs. 8000

Given, the figure is an isosceles triangle, whose sides are 30 cm, 30 cm and 40 cm, respectively.
∴ Perimeter of triangle = Sum of all sides of a triangle
= (30 + 30 + 40) cm = 100 cm
Here, we observe that the perimeter of each figure is
100 cm i.e. they have equal perimeters.

Given, the figure is a square, whose side is 25 cm.
∴ Perimeter of square = 4 \(\times\) Length of a side
= 4 \(\times\) 25 cm = 100 cm

Let ABC be the given triangle and its sides AB = 12 cm and BC=14cm.
Also, perimeter of triangle = 36 cm

We know that,
Perimeter of a triangle = Sum of all its sides
⇒ AB + BC + CA = 36cm
⇒ 12 cm + 14 cm + CA = 36 cm
⇒ (12+14)cm+CA=36cm
⇒ 26 cm+ CA = 36 cm
∴ CA = (36 - 26) cm = 10 cm
Hence, the third side of the triangle is 10 cm.

Here, length of string will be the perimeter of square.
∴ Perimeter of square = 30 cm
We know that, a square has 4 equal sides.
∴ Perimeter of a square = 4 \(\times\) Length of a side
\(Now,length\quad of\quad one\quad side=\frac { Perimeter\quad of\quad a\quad square }{ 4 } \)
\(=\frac { 30 }{ 4 } cm=7.5cm\)
Hence, length of each side of a square is 7.5 cm.

Given, length of the wooden strip = 32 cm
and breadth of the wooden strip = 21 cm
Now, wooden strip required = Perimeter of the photograph
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\) (32 cm + 21 cm)
=2\(\times\)53cm =106 cm
Hence, the required length of wooden strip is 106 cm.

The perimeter of the given figure is 28 cm

From the given figure, we have
AB = 40 cm, BC = 10cm,CD = 40cm and DA =10cm
Now, the sum of the lengths of the four sides
= AB + BC + CD + DA
= 40 cm + 10 cm + 40 cm + 10 cm
= (40 + 10 + 40 + 10) cm = 100 cm
∴ Perimeter of the given figure
= Sum of the lengths of four sides = 100 cm
Henee, the perimeter of the given figure is 100 cm.

Let ABCD be a park whose lengths are BC, AD and widths are AB, CD respectively.

Here, AB = CD = 80 m and BC = DA = 150 m
Now, sum of the lengths of four sides
=AB+ BC + CD +DA
= 80 m + 150 m + 80 m + 150 m
= (80 + 150 + 80 + 150) m = 460 m
∴  Perimeter of the park = Sum of the lengths of four sides of the park =460m
Hence, the distanee eovered by Meera is 460 m.

Given, perimeter of a regular hexagon =30 cm
A regular hexagon has 6 sides.
∴ The perimeter of regular hexagon = 6 \(\times\) Length of a side
⇒30 = 6 \(\times\) Length of a side
∴ The length of a side=\(\frac{30}{6}\)=5cm
Hence, length of each side of a regular hexagon is 5 cm.

434 cm

169 cm²

Let width of rectangle (b)= x cm
Then, length of rectangle (I) = 3x cm
∴ Perimeter = 2 (l + b)
⇒ 40 =2 (3x + x)
⇒ 8x =40
⇒ x=\(\frac{40}{8}\)=5cm
Hence, length is 15 cm and width is 5 cm.

∵ Perimeter = 50 cm [given]
Perimeter of an isosceles triangle = Sum of its all sides
⇒ Perimeter = a + b + c
⇒ 50 = 18 + 18 + Third side
⇒50 - 36 = Third side
⇒ Third side = 14 cm

4 times of the original area

Here, side of the square = 10 cm
∴ Area of square = Side \(\times\) Side = 10 cm \(\times\) 10 cm
= 100 sq cm
Hence, the area of the square is 100 sq cm.

Given, area = 48 cm² and breadth (b) = 6 cm
We know that, Area = Length \(\times\)Breadth\(\Rightarrow Length=\frac { Area }{ Breadth } =\frac { 48 }{ 6 } =8cm\)
Hence, length of the cardboard is 8 cm.

75 cm² (approx)

Given, figure is covered by 3 full squares, 4 half squares, 3 more than half squares and 1 less than half square.
∴ Area = \(\left( 3\times 1+4\times \frac { 1 }{ 2 } +3\times 1+1\times 0 \right) sq\quad units\)
= (3 + 2 + 3 + 0) sq unit = 8 sq units

Given, figure is covered by 5 full squares.
∴. Area = 5 \(\times\) 1 sq unit = 5 sq units

2 cm

140 cm

The perimeter of the given figure is 20 cm.

The perimeter of the given figure is 133 cm.

Perimeter of a squared park = 4 \(\times\) Length of a side = 4 \(\times\) 60 = 240 m
Distance covered in one round = 240 m
Distance covered in three rounds = 3 \(\times\) 240 m = 720 m

(i)

(iii)

4

16 cm²

Given, each side of square is of unit length. Figure contains length of 10 squares and width of 6 squares.
Now, length of rectangle, AD = (BC)
= Sum of length of a side of 10 squares
=1+1+1+1+1+1+1+1+1+1
=10\(\times\)1 =10 units
and breadth of rectangle, AB = (DC)
= Width of 6 squares = 6 \(\times\)1 = 6 units
(a) The perimeter of the rectangle ABCD
= AB+ BC+ CD+DA
= 6+10+ 6+10
= 32 units
(b) The area of the rectangle ABCD = Length \(\times\)Breadth
= AD\(\times\)AB=10\(\times\)6
= 60 units
(c) The total area of rectangle = 60 units

Now, we have to divide the rectangle into 10 equal parts i..e \(\frac{60}{10}=6\) square uni.t.s I.e. we have to take a group of 6-6 square blocks, which is shown in the figure
(d) Now, we find the perimeter of part l.

We know that perimeter of a figure is the total length of its boundary.
∴ Perimeter of part I
= 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1= 12 units
Similarly, we can find the perimeters of remaining 9 parts, all the parts have same perimeter i.e. 12 units.
Yes, all the parts have same perimeter.

Given width of the lawn, AB = EF = 4 m
and length of the lawn, BE = 10m
Also, given length of gap, CD = 1m

Total length of fencing = AB + (BC + DE) + EF
= AB+ (BE -CD)+ EF
= 4 m +( 10 -1) m +4 m
=(4+ 9+ 4) m =17 m
Hence, the length of fencing of the lawn is 17 m.

True, let the length and breadth of a rectangle be I and b respectively.
We know that,
Area of the initial rectangle = Length \(\times\) Breadth
= I \(\times\) b sq units
If length of a rectangle is havled and breadth is doubled.
i.e. New length = \(\frac{l}{2}\)units
and new breadth = 2b units
Then, area of the new rectangle
= New length \(\times\) New breadth =\(\frac{l}{2}\times2b=lb\quad sq\quad units\)

Given, size of lawn in front of Molly's house
=12m\(\times\)8m
Perimeter = 2 (12 + 8) m= 40 m ... (i)
Now, size of lawn in front of Dolly's house
=15m\(\times\)5m
Perimeter = 2 (15+ 5) m = 40 m ... (ii)
From Eqs. (i) and (ii), we get = 40+ 40= 80 m
Hence, total length of bamboo fencing is 80 m.

We know that,
Area of the rectangle = Length \(\times\) Breadth
Here, the area of the rectangle is 16 sq cm.
So, for getting the area 16 sq em, there are three cases.
Case I When length of the rectangle = 16 cm
and breadth of the rectangle = 1cm
∴ Area of the rectangle = Length \(\times\) Breadth
=16 cm \(\times\) 1 cm = 16sq cm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\)(16 cm + 1 cm) = ( 2 \(\times\) 17) cm= 34 cm
Case II When length of the rectangle = 4 cm
and breadth of the rectangle = 4 cm
∴ Area of the rectangle = Length\(\times\) Breadth
= 4 cm \(\times\) 4 cm 16 sq cm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\)(4 cm +4 cm) = (2 \(\times\) 8) cm = 16 cm
Case III When length of the rectangle = 8 cm
and breadth of the rectangle = 2 cm
∴ Area of the rectangle = Length\(\times\) Breadth
=8 cm \(\times\)2 cm =16sq cm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth) = 2 \(\times\) (8 cm + 2 cm)
= (2\(\times\)10) cm = 20 cm
(a) The rectangle which made in Case I has the greatest perimeter.
(b) The rectangle which made in Case II has the least perimeter.
Now, given that the area of the rectangle is 24 sq cm.
Again, for getting the area 24 sq cm, there are three cases:
Case I When length of the rectangle = 24 cm
and breadth of the rectangle = 1 cm
∴ Area of the rectangle = Length \(\times\) Breadth
= 24 cm\(\times\) 1 cm = 24 sq cm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\) (24 cm + 1 cm) = (2 \(\times\) 25)cm = 50 cm
Case II When length of the rectangle = 12 cm
and breadth of the rectangle = 2 cm
∴Area of the rectangle = Length \(\times\)Breadth
= 12 cm \(\times\) 2 cm = 24 sqcm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\) (12 cm +2 cm)
= (2 \(\times\)14) cm = 28 cm
Case III When length of the rectangle = 6 cm
and breadth of the rectangle = 4 cm
∴ Area of the rectangle = Length \(\times\) Breadth
= 6 cm \(\times\) 4 cm = 24 sqcm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\) (6 cm +4 cm)
= 2 \(\times\) 10 cm = 20 cm
(a) The rectangle which made in Case I has the greatest perimeter.
(b) The rectangle which made in Case III has the least perimeter.
Yes, it is possible to predict the shape of the rectangle with greatest perimeter and with least perimeter.
e.g. A rectangle having area 16 sq ern and greatest perimeter 34 cm is of the shape 16 cm \(\times\) 1 cm
(i.e. length = 16 cm and breadth = 1 cm), also a rectangle
having area 24 sq cm and least perimeter 20 cm is of the shape 6 cm \(\times\) 4 cm (i.e. length = 6 cm and breadth = 4 cm).
The rectangle with the greatest length has the maximum perimeter and the rectangle with the smallest length has the least perimeter.

Let the given figure is divided into rectangles A, Band C and their length and breadth are written on the figure.
For rectangle A,
Length = 2 cm and breadth = 1 cm
∴ Area of the rectangle A = Length \(\times\) Breadth
=2 \(\times\) 1=2 sq cm
For rectangle B,
Length = 5 ern and breadth = 1 cm

∴  Area of the rectangle B
= Length \(\times\) Breadth = 5 \(\times\)1 = 5 sq cm
For rectangle C,
Length = 2 cm and breadth = 1cm
∴ Area of the rectangle C = Length \(\times\) Breadth
=2 \(\times\) 1=2 sqcm
Now, total area of the given figure = Area of rectangle A + Area of rectangle B + Area of rectangle C
= (2 + 5 + 2) sq cm = 9 sq cm
Hence, the area of the given figure is 9 sq cm

(a) Here, length of the rectangle = 9 m
and breadth of the rectangle = 6 m
∴ Area of the rectangle = Length \(\times\) Breadth = 54 sq m
Hence, the area of the rectangle is 54 sq m.
(b) Here, length of the rectangle = 17 m
and breadth of the rectangle = 3 m
∴ Area of the rectangle = Length \(\times\) Breadth
= 17m \(\times\) 3m = 51 sq m
Hence, the area of the rectangle is 51 sq m.
(c) Here, length of the rectangle = 14 m
and breadth of the rectangle = 4 m
∴ Area of the rectangle = Length \(\times\) Breadth
= 14 m \(\times\) 4 m = 56 sq m
Hence, the area of the rectangle is 56 sq m.
Now, we have 56> 54> 51
Hence, the rectangle having sides 4 m and 14 m has the largest area and the rectangle having sides 17m and 3 m has the smallest area.

Length = \(\frac{6}{2}\) - 1 = 2 m

Third side = 12 - (5 + 4) = 3 cm

Length of side = \(\frac{9}{3}\)=3 m

Perimeter = 2(9 + 1) = 20 cm

Perimeter = 2(25 + 20) = 90 cm

Distance covered = 2(20 + 10) = 60 m