Free access of the complete Ganita Prakash Book Class 6 Solutions and Chapter 6 Perimeter and Area Class 6 NCERT Solutions Question Answer are crafted in simple format to align with the latest CBSE syllabus.
Class 6 Maths Chapter 6 Perimeter and Area Solutions
Perimeter and Area Class 6 Solutions Questions and Answers
6.1 Perimeter Figure it Out (Page No. 132)
Question 1.
Find the missing terms:
(a) Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?
Solution:
We know that perimeter of a rectangle = 2(l + b)
Here, the perimeter of the rectangle = 14 cm and breadth b = 2 cm, l =?
Thus 14 = 2(l + 2)
⇒ 14 = 2l + 4
⇒ 2l = 14 – 4 = 10
⇒ l = \(\frac{10}{2}\) = 5 cm
(b) Perimeter of a square = 20 cm; side of a length = ?
Solution:
We know that the perimeter of the square = 4 × a
where a = side of the square
∴ 20 = 4 × a
⇒ a = 5 cm
(c) Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?
Solution:
Perimeter of rectangle = 2(l + b)
⇒ 12 = 2(3 + b)
⇒ 12 = 6 + 2b
⇒ 12 – 6 = 2b
⇒ 2b = 6 m
⇒ b = 3 m
Question 2.
A rectangle with 5 cm and 3 cm side lengths is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of the side of the square?
Solution:
The perimeter of rectangle = perimeter of the square
⇒ 2(l + b) = 4 × side
⇒ 2(5 + 3) = 4 × side
⇒ 16 = 4 × side
⇒ Side = 4 cm
∴ the length of a side of the square = 4 cm
Question 3.
Find the length of the third side of a triangle having a perimeter of 55 cm and having two sides of length 20 cm and 14 cm, respectively.
Solution:
We know that the perimeter of a triangle is the sum of the length of all three sides.
Here, we have
the perimeter of triangle = 55 cm
and two sides are 20 cm and 14 cm.
Perimeter of triangle = 20 + 14 + Third side of triangle
⇒ Third side of triangle = 55 – (20 + 14) = 21 cm
Question 4.
What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m, if the fence costs ?40 per metre?’
Solution:
Length of the rectangular park = 150 m Breadth of the rectangular park = 120 m
∴ Perimeter of rectangular park = 2 (length + breadth)
= 2 (150 m + 120 m) = 2 (270 m) = 540 m
Since, the cost of fencing per metre = ₹ 40
∴ The cost of fencing the rectangular park = ₹ (540 x 40) = ₹ 21,600
Question 5.
A piece of string is 36 cm long. What will be the length of each side, if it is used to form:
(a) A square,
(b) A triangle with all sides of equal length, and
(c) A hexagon (a six-sided closed figure) with sides oil equal length?
Solution:
(a) Given, a piece of string is 36 cm long
∴ length of each side of the square = a
perimeter = 36
⇒ 4a = 36
⇒ a = 9 cm
(b) Length of each side of the triangle = 3a (Given)
perimeter = 36
⇒ 3a = 36
⇒ a = 12 cm
(c) Length of each side of hexagon = a
perimeter = 36
⇒ 36 = 6a
⇒ a = 6 cm
Question 6.
A farmer has a rectangular field with having length of 230 m and a breadth of 160 m. He wants to fence it with 3 rounds of rope as shown. What is the total length of rope needed?
Solution:
Perimeter of the rectangular field = 2(l + b)
Here l = 230 m, b = 160 m
∴ P = 2(230 + 160)
= 2 (390)
= 780 m
Distance covered by a farmer in one round = 780 m
∴ Total length of rope needed = 3 × 780 = 2340 m
6.1 Perimeter Figure it Out (Page No. 133 – 134)
Akshi and Toshi start running along the rectangular tracks as shown in the figure. Akshi runs along the 1 outer track and completes 5 rounds. Toshi runs along the inner track and completes 7 rounds. Now, they are wondering who ran more. Find out who ran the longer distance.
Each track is a rectangle. Akshi’s track has a length of 70 m and a breadth of 40 m. Running one complete round on this track would cover 220 m, i.e., 2 × (70 + 40)m = 220 m. This is the distance covered by Akshi in one round.
Question 1.
Find out the total distance Akshi has covered in 5 rounds.
Solution:
Distance covered by Akshi in 1 complete round = 220m
So, distance covered by Akshi in 5 rounds = 220 × 5 = 1100 m
Question 2.
Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance?
Sol. Distance covered by Toshi in 1 complete round = 2(60 + 30)
= 2 × 90
= 180 m
Distance covered by Toshi in 7 rounds = 7 × 180 = 1260 m.
As Akshi covered 1100 m and Toshi covered 1260 m, thus Toshi ran a longer distance.
Question 3.
Think and mark the positions as directed—
(a) Mark ‘A’ at the point where Akshi will be after she ran 250 m.
(b) Mark ‘B’ at the point where Akshi will be after she ran 500 m.
(c) Now, Akshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘C’
(d) Mark ‘X’ at the point where Toshi will be after she ran 250 m.
(e) Mark ‘Y’ at the point where Toshi will be after she ran 500 m.
(f) Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘Z’.
Solution:
(a) Akshi’s track has a perimeter of220 m. After running 250 m, Akshi will be 30 m into her second round.
(b) 500 m, Akshi will have completed 2 full round (440 m) and will be 60 m into her third round.
(c) running 1000 m, Akshi will have completed 4 full round (880 m) and will be 120 m into her fifth round. So, she will have finished 4 full rounds.
(d) Toshi’s track has a perimeter of 180 m. running 250 m, Toshi will be 70 m into her second round.
(e) 500 m, Toshi will have completed 2 full round (360 m) and will be 140 m into her third round.
(f) running 1000 m. Toshi will have completed 5 full round (900 m) and will be 100 m into her sixth round. So, she will have finished 5 full round.
InText Questions
Question 1.
Deep Dive: In races, usually there is a common fiish line for all the runners. Here are two square running tracks with the inner track of 100 m each side and outer track of 150 m each side. The common fiishing line for both runners is shown by the flgs in the fiure which are in the center of one of the sides of the tracks.
If the total race is of 350 m, then we have to fid out where the starting positions of the two runners should be on these two tracks so that they both have a common fiishing line after they run for 350 m. Mark the starting points of the runner on the inner track as ‘A’ and the runner on the outer track as ‘B’. (Page 134)
Solution:
Deep Dive: Here the total race is of 350 m. The starting points of the runner on the inner track as ‘A’ and the runner on the outer track as ‘B’ are marked in the figure below:
Question 2.
Write the perimeters of the figures below in terms of straight and diagonal units. (Page 135)
Solution:
1st Figure = 8 straight units + 2 diagonal units
= 8s + 2d units
2nd Figure = 4 straight units + 6 diagonal units
= 4s + 6d units
3rd Figure =12 straight units + 6 diagonal units
= 12s + 6d units
4th Figure =18 straight units + 6 diagonal units
= 18s + 6d units
Question 3.
Find various objects from your surroundings that have regular shapes and find their perimeters. Also, generalise your understanding for the perimeter of other regular polygons. (Page 136)
Solution:
Examples of regular shapes
1. Square coasters
Perimeter of coasters
= AB + BC + CD + DA
= 4 × length of one side
2. Octagonal clock (where all eight sides and all eight angles are equal).
Perimeter of octagonal clock
= AB + BC + CD + DE + EF + FG + GH + HA
= 8 × length of a side
Question 4.
Find out the length of the boundary (i.e., the perimeter) of each of the other arrangements below.
Solution:
(a) Required perimeter
= (6 + 2 + 6 + 2 + 4 + 6 + 2) cm
= 28 cm
(c) Required perimeter
= (2 + 6 + 2 + 2 + 6 + 2 + 2 + 6)cm
= 28cm
(d) Required perimeter
= (6 + 2 + 3 + 2 + 6 + 2 + 3 + 2) cm
= 26 cm
Question 5.
Arrange the two pieces to form a figure with a perimeter of 22 cm.
Solution:
Two pieces figure with a perimeter of 22 cm is as follows:
6.2 Area Figure it Out (Page No. 138)
Question 1.
The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden?
Solution:
Given, area of rectangular garden = 300 sq.m
and length = 25 m
area of rectangular field = l × b
⇒ 300 = 25 × b
⇒ b = 12 m
Question 2.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m?
Solution:
Here, length = 500 m and breadth = 200 m
Hence the area of the rectangular plot = length × breadth
= 500 × 200
= 1,00,000 m2
Now cost of tilling a rectangular plot = \(\frac{8}{100}\)
Hence the cost of tilling 1,00,000 sq. m of rectangular plot = \(\frac{8}{100}\) × 100000 = ₹ 8,000
Question 3.
A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree requires 25 sq m, what is the maximum number of trees that can be planted in this grove?
Solution:
Length of rectangular coconut grove = 100 m
Width of rectangular coconut grove = 50 m
Area of rectangular coconut grove = 100 m × 50 m
= 5000 sq m
Each coconut tree requires 25 sq m
Number of trees that can be planted in this grove
= \(\frac{5000}{25}\) = 200
Question 4.
By splitting the following figures into rectangles, find their areas (all measures are given in metres):
Solution:
(a) Rectangle I: Length = 4 m and breadth = 3 m
Area of rectangle I = 4m × 3m = 12sqm
Rectangle II: Length = 4 m and breadth = 1 m
Area of rectangle II = 4m × 1 m = 4 sq m
Rectangle III: Length = 4 + 1 = 5 m and breadth = 2 m
Area of rectangle III = 5 m × 2 m = 10 sq m
Rectangle IV: Length = 2 m and breadth = 1 m
Area of rectangle IV = 2m × 1m = 2sqm
Area of given figure = Area of rectangle I + Area of rectangle II + Area of rectangle in -Area of rectangle IV = 12 + 4 + 10 + 2 = 28 sq m
(b) Length and breadth of rectangle BCDE are 5 m and 1 m respectively.
So, area of rectangle BCDE = 5m × 1 m = 5 sq m
Area of rectangle ABU = 2 m × 1 m = 2 sqm
Area of rectangle GHEF = 2m × 1 m = 2 sqm
Area of the given figure = Area of rectangle BCDE + Area of rectangle ABIJ + Area of rectangle GHEF = (5 + 2 + 2) sq m = 9 sq m
6.2 Area Figure it Out (Page No. 139)
Cut out the tangram pieces given at the end of your textbook.
Question 1.
Explore and figure out how many pieces have the same area.
Solution:
From Figure we can find out that Shapes A and B, shapes G, D, and F, and also shapes C and E have the same area.
Question 2.
How many times, bigger is Shape D as compared to Shape C? What is the relationship between Shapes C, D, and E?
Solution:
As seen from the given figure, Shape D can be exactly covered using two times Shape C so it shows that Shape D is two times bigger than Shape C.
Hence, Shape D has twice the area of Shape C or Shape E.
Question 3.
Which shape has more area: Shape D or F? Give reasons for your answer.
Solution:
On comparing the area of shape D and shape F, we get
Shape D has more area than shape F. Because shape D is the incomplete combination of two smaller shapes (like shape C and shape £), while shape F is equivalent to only one smaller shape (like shape C or shape F).
Smaller shape (like shape C and shape F), while shape F is equivalent to only one smaller shape (like shape C or shape F). Therefore, shape D covers more space than shape F.
Question 4.
Which shape has more area: Shape F or G? Give reasons for your answer.
Solution:
On comparing shape F with shape G, we get shape G has more area than shape F. This is because shape G is i larger and cover more space compared to shape F. Which is equal to only one smaller piece like shape C or shape E.
Question 5.
What is the area of Shape A as compared to Shape G? Is it twice as big? Four times as big?
Hint: In the tangram pieces, by placing the shapes over each other, we can find out that Shapes A and B have the same area, Shapes C and E have the same area. You would- have also figured out that Shape D can be exactly covered using Shapes C and E, which means Shape D has twice the area of Shape C or shape E, etc.
Solution:
Shape G = 2 × shape C
Now, shape A = 2 × shape F
= 2 (shape C + shape E)
= 2 (2 shape C)
= 4 × shape C
= 2 × shape G
So, area of shape A is twice as compared to shape G.
Question 6.
Can you now figure out the area of the big square formed with all seven pieces in terms of the area of Shape C?
Solution:
Here shape D = 2 × shape C
Shape A and shape B have the same area, also shape A = 4 × shape C.
Also, shape C and shape E have the same area.
Total area of big square = 16 × Area of shape C.
Question 7.
Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle in terms of the area of Shape C now? Give reasons for your answer.
Solution:
The tangram rectangle with all 7 pieces is a tangram square with 5 pieces extended with two big triangles. All seven tans fit together to form a rectangle. Hence area of this rectangle in terms of Shape C is 16 small triangles.
Question 8.
Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same? Give an explanation for your answer.
Solution:
The perimeter of the square is equal to the square formed from these 7 pieces because these are the arrangements of pieces.
InText Questions
Question 1.
Find the area of the following figures. (Page 140)
Solution:
The area of one full small square of the square or graph paper is taken as 1 sq unit.
Ignore portions of the area that are less than half a square. If more than half of a square is in a region, just count it as 1 sq unit.
If exactly half the square is counted, take its area as \(\frac{1}{2}\) sq unit.
(a) There are 3 complete squares and 2 half squares in the figure.
So, area of the figure = (3 × 1 + 2 × \(\frac{1}{2}\))sq units.
= 3 + 1 = 4 sq units
(b) There are 6 complete squares and 6 half squares in the figure.
So, area of the figure = (6 × 1 + 6 × \(\frac{1}{2}\)) sq. units
= (6 + 3) sq units
= 9 sq units
(c) There are 7 complete squares and 6 half squares in the figure.
So, area of the figure = (7 × 1 + 6 × \(\frac{1}{2}\))sq units
= (7 + 3) sq units
= 10 sq units
(d) There are 8 complete squares and 6 half squares in the figure.
So, area of the figure = (8 × 1 + 6 × \(\frac{1}{2}\)) sq un’ts
= (8 + 3) = 11 sq units
Question 2.
Try using different use to measure area. (Page 141-142)
Solution:
Triangle can tile a plane perfectly but measuring area with triangles is more complex.
Merits: Triangle is useful, but the calculations and measurements are not easy as compared to squares.
Rectangles: Can tile a surface perfectly, similar to squares but they require more complex calculations if the rectangles are not uniform.
Merits: Good for measuring areas but can be less intuitive than squares.
Merits of using squares:
1. The area of squares is easy to calculate.
2. Squares covers surface without gaps or overlaps.
3. It fits into graph paper grids nearly.
4. It makes measurement easy.
1. Find the area (in square metres) of the floor outside of the corridor.
Solution:
Area of the floor outside of the corridor = Total floor area – corridor area
2. Find the area (in square metres) occupied by your school playground.
Solution:
If the playground is in rectangle, then area of playground = length × width.
Question 3.
On a squared grid paper (1 square = 1 square unit), make as many rectangles as you can whose lengths and widths are a whole number of units such that the area of the rectangle is 24 square units.
(a) Which rectangle has the greatest perimeter?
(b) Which rectangle has the least perimeter?
(c) If you take a rectangle of area 32 sq cm, what will your answers be? Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter as well as the least perimeter? Give examples and reasons for your answer.
Solution:
If the lengths and widths are a whole number, such that length × width = 24 sq units then the possible pairs are:
1. Length = 1 unit,
2. Length = 2 units,
3. Length = 3 units,
4. Length = 4 units,
5. Length = 6 units,
6. Length = 8 units,
7. Length = 12 units,
8. Length = 24 units,
(a) Rectangle with greatest perimeter is 1 × 24 or 24 × 1
Perimeter = 2 (1 + 24) = 2(25) = 50 units
(b) Rectangle with least perimeter is 4 × 6 or 6 × 4
Perimeter = 2 (6 + 4) = 2 (10) = 20 units
(c) Rectangles with an area of 32 sq cm, the possible pair of length and breadth are as follows:
Length = 1 cm, Breadth = 32 cm
Length = 2 cm, Breadth = 16 cm
Length = 4 cm, Breadth = 8 cm
Length = 8 cm, Breadth = 4 cm
Length = 16 cm, Breadth = 2 cm
Length = 32 cm, Breadth = 1 cm
(a) Rectangle with greatest perimeter are 1 cm × 32 cm 32 cm × 1 cm
Perimeter = 2(1 + 32) cm = 66 cm
(b) Rectangle with least perimeter are 4 cm × 8 cm 8 cm × 4 cm
Perimeter = 2 (8 + 4) cm = 24 cm
We can find out the greatest perimeter if the area is
A, then the dimensions are 1 × A or A × 1.
So perimeter = 2(1 + A)
Example: If area is 24 sq units
Then greatest perimeter (1 x 24) = 2 (1 + 24)
= 50 units
We can find out the least perimeter when the dimensions are least to each other (or approaching a square)
Example: if area is 24 units Least perimeter (4 × 6) = 2 (4 + 6) = 20 units
Question 4.
Check! whether the two triangles overlap each other exactly. Do they have the same area? (Page 142)
Solution:
Yes, the two triangles can overlap each other exactly. The two triangles formed by the diagonal of the rectangle will have the same area. Since diagonal divides the rectangle into two triangles each having half the area of the rectangle.
Question 5.
Can you draw any inferences from this exercise? Please
write it here. _____________________________________
Solution:
The area of each triangle is exactly half of the area of the rectangle.
Question 6.
Can you see some relationship between the blue rectangle and the yellow triangle and their areas? Write the relationship here.
Solution:
The yellow triangle might be equal to the blue rectangle. This is the required relationship.
Question 7.
Use your understanding from previous grades to calculate the area of any closed figure using grid paper and —
1. Find the area of blue triangle BAD. _______
2. Find the area of red triangle ABE. _______
Solution:
Here area of rectangle = 5 × 4 = 20 sq units
1. Area of blue triangle BAD
= \(\frac{1}{2}\) × Area of rectangle ABCD
= \(\frac{1}{2}\) × 20 = 10 sq units
2. Area of triangle AFE = \(\frac{1}{2}\)
= \(\frac{1}{2}\) × (3 × 4) sq cm = 6 sq cm
Area of triangle BFE = \(\frac{1}{2}\) × Area of rectangle CEFB
\(\frac{1}{2}\) × (2 × 4) sq cm = 4 sq cm
So, area of triangle ABE = Area of traingle AFE + Area of triangle BFE = 6 sq units + 4 sq units = 10 sq units
Area of rectangle ABCD = 20 sq units
6.3 Area of A Triangle Figure it Out (Page 144)
Question 1.
Find the areas of the figures below by dividing them into rectangles and triangles.
Solution:
(a) The given figure dividing into one rectangle PQRS and two triangles PTQ and SRU.
In rectangle PQRS, full small square = 20
∴ Area of rectangle PQRS = 20 × 1 = 20 sq. units
In ΔPTQ, more than half of a small square = 2
and less than half of a small square = 2
∴ Area of triangle PTQ = 2 × 1 + 2 × 0 = 2sq. units
In ΔSRU, more than half of a small square = 2
and less than half of a small square = 2
∴ Area of ΔSRU = 2 × 1 + 2 × 0 = 2sq. units
Hence, total area = 20 + 2 + 2 = 24 sq. units
(b) The given figure dividing into one rectangle PQRS and two triangles PQT and SRU.
In rectangle PQRS, full small square = 20
Area of rectangle PQRS = 20 sq. units
In ΔPQT, full small square = 3
more than half of a small square = 3
less than half of a small square = 2
Area of ΔPQT = 3 × 1 + 3 × 1 + 2 × 0 = 3 + 3 = 6sq. units
In ΔSRU, full small square = 2
more than half of a small square = 2
less than half of a small square = 2
Area of ΔSRU = 2 × 1 + 2 × 1 + 2 × 0 = 2 + 2 =4 sq. units
Hence, total area = 20 + 6 + 4 = 30 sq. units
(c) The given figure dividing into one rectangle PQRS and three triangles PTQ, SRV and TUV.
In rectangle PQRS, full small square = 24
∴ Area of rectangle PQRS = 24 × 1 = 24 sq. units
In ΔPTQ, full small square = 1
more than half of a small square = 1
less than half of a small square = 3
Area of ΔPTQ = 1 × 1 + 1 × 1 + 3 × 0 = 1 + 1 = 2 sq. units
In ΔSRV, full small square = 1
more than half of a small square = 2
less than half of a small square = 2
∴ Area of ΔSRU = 1 × 1 + 2 × 1 + 2 × 0 = 1 + 2 = 3 sq. units
(d) The given figure dividing into one rectangle PQRS and two triangles STV and VUR.
In ΔPQRS, full small squares = 12
Area of rectangle PQRS = 12 sq. units
In ΔSTV, more than half of a square = 1
less than half of a square = 1
∴ Area of ΔSTV = 1 × 1 +1 × 0 = 1 sq. unit
In ΔVUR, full small square = 1
more than half of a small square = 2
less than half of a small square = 1
Area of ΔVUR= 1 × 1 + 2 × 1 + 1 × 0 = 1 + 2 = 3sq. units
Hence, total area = 12 + 1 + 3 = 16 sq. units
In ΔTUV, full small square = 9
more than half of a small square = 5
Exactly half of a small square = 2
less than half of a small square = 7
∴ Area of ΔTUV =9 × 1 + 5 × 1 + 2 × 1 + 7 × 0
= 9 + 5 + 1 = 15 sq. units
Hence, total area = 24 + 2 + 3 +15 = 44 sq. units
(e) The given figure divides into one rectangle PQRS and four triangles PTQ, PUS, SWR and QVW.
In rectangle PQRS, full small square = 4
Area of rectangle PQRS = 4 ×x 1 = 4 sq. units
In A PTQ, exactly half of a small square = 2
∴ Area of A PTQ = 2 × \(\frac{1}{2}\) = 1 sq. unit
In APUS, exactly half of a small square = 2
Area of APUS = 2 × \(\frac{1}{2}\) = 1 sq. unit 2
In ASWR, full small square = 1
more than half of a small square = 2
less than half of a small square = 1
∴ Area of ASWR = 1 × 1 + 2 × 1 + 1 × 0 = 1 + 2 = 3 sq. units
In A QVW, more than half a small square = 1
exactly half of a small square = 2
less than half of a small square = 2
∴ Area of AQ VW = 1 × 1 + 2 × —1 – 2 × 0
= 1 +1 + 0 = 2 sq. units
Hence, total area = 4+ 1 + 1 + 3 + 2
= 11 sq. units
InText Questions
Question 1.
Using 9 unit squares, solve the following.
1. What is the smallest perimeter possible?
2. What is the largest perimeter possible?
3. Make a figure with a perimeter of 18 units.
4. Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter? What is your reasoning? (Page 145-147)
Solution:
1. To find the smallest perimeter, arrange the 9 units squares into 3×3 squares.
Perimeter = 2 (3 + 3) = 12 units
So, the smallest perimeter with 9 unit squares is 12 units.
2. To find the largest perimeter, arrange 9 units square into a single row as a column.
Perimeter = 2(1 + 9) = 2(10) = 20 units
3. It is the required figure with a perimeter of 18 units.
4. For 12 units, no other shape is possible.
For perimeter 18 units
Question 2.
Let’s do something tricky now! We have a figure below having perimeter 24 units. _________
Solution:
After attaching a new square in a figure having perimeter = 24 units
(a) Here perimeter = 26 units
So, perimeter increases.
(b) Here perimeter = 22 units
So, perimeter decreases.
(c) Here perimeter = 24 units
So, perimeter stays the same.
Question 3.
Below is the house plan of Charan. It is in a rectangular plot. Look at the plan. What do you notice?
Some of the measurements are given.
(a) Find the missing measurements,
(b) Find out the area of his house.
Solution:
(a) Utility = (15 ft) × (3 ft)
⇒ Area = 45 sq ft
Small Bedroom: (15 ft × 12 ft)
⇒ Area = 180 sq ft
Parking: (15 ft × 3 ft)
⇒ Area = 45 sq ft
Hall area: 20 ft × 12ft + 5ft × 5ft
= 240 sq ft + 25 sq ft = 265 sq ft
Garden: (20 ft × 3 ft)
⇒ Area = 60 sq ft
(b) Area of his house = 35 ft × 30 ft
= 1050 sq ft
Question 4.
Now, find out the missing dimensions and area of Sharan’s home. Below is the plan:
Some of the measurements are given.
(a) Find the missing measurements.
(b) Find out the area of hi& house.
What are the dimensions of all the different rooms in Sharan’s house? Compare the areas and perimeters of Sharan’s house and Charan’s house.
Solution:
(a) Toilet: (5 ft × 10 ft)
⇒ Area = 50 sq ft
Utility: (7 ft × 10 ft)
⇒ Area = 70 sq ft
Entrance: (7 ft × 15 ft)
⇒ Area 105 sq ft
Small Bedroom: (12 ft × 10 ft)
⇒ Area = 120 sq ft
Hall: (23 ft × 15 ft)
⇒ Area = 345 sq ft
(b) Area of his house = 42 ft × 25 ft
= 1050 sq ft
Dimension of different rooms in Sharan’s house are as follows
Master Bedroom: 12 ft × 15 ft
Toilet: 5 ft × 10 ft,
Kitchen: 18 ft × 10 ft
Utility: 7 ft × 10 ft,
Entrance: 7 ft × 15 ft
Hall: 23 ft × 15 ft
Small Bedroom: 12 ft × 10 ft
Now, area of Charan’s house = 1050 sq ft area of Sharan’s house = 1050 sq ft
Both areas are same
Perimeter of Charan’s house = 2 (35 + 30) = 130 ft
Perimeter of Sharan’s house = 2 (42 + 25) = 134 ft
So, the perimeter of Sharan’s house is more than the perimeter of Charan’s house.
Question 5.
Area Maze Puzzles (Page 148)
In each figure, find the missing value of either the length of a side or the area of a region.
Solution:
? sq. cm = 2 × 15 = 30 sq cm
Solution:
? sq. cm = 3 × 3 = 9 sq. cm
Solution:
? sq. cm = 4 × 4 = 16 sq cm
Solution:
? cm × (4 + 3.6) cm = 38 sq cm
⇒ ? cm = 5 cm
6.3 Area of A Triangle Figure it Out (Page 149)
Question 1.
Give the dimensions of a rectangle whose area is the sum, of the areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m.
Solution:
Here, Area of rectangle 1 = 5 × 10 = 50 sq m
Area of rectangle 2 = 2 × 7 = 14 sq m
The Sum of the areas of these 2 rectangles = 50 + 14 = 64 sq m
Now, the total area of the rectangle = 64
Let’s say the sides of the rectangle are Length = x and Width = y
Area of rectangle = x × y
Hence x × y = 64
xy = 64
Let’s say x = 1, then y = \(\frac{64}{1}\) = 64
if x = 2, then y = \(\frac{64}{2}\) = 32
Hence the dimensions of the rectangle are (1 × 64), (2 × 32)
Question 2.
The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
Solution:
Area of rectangular garden = length × width
⇒ 1000 sq m = 50 m × width
⇒ Width = \(\frac{1000 \mathrm{sq} \mathrm{~m}}{50 \mathrm{~m}}\) = 20 m
Thus, the width of the garden is 20 m.
Question 3.
The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted.
Solution:
Dimensions of Room: Length = 5 m; width = 4 m
Dimensions of Carpet: Length = 3 m; width = 3 m
Area of room = 5m × 4m = 20sqm
Area of carpet = 3m × 3m = 9sqm
Area that is not carpeted = Area of room – area of square carpet
= 20 sq m – 9 sq m = 11 sq m
Question 4.
Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?
Solution:
Area of one flower bed = 2 × 1 = 2 m2
Area of four flower beds = 4 × 2 = 8 m2
Area of rectangular garden = 15 × 12 = 180 m2
Area available for laying down a lawn = 180 – 8 = 172 m2
Question 5.
Shape A has an area of 18 square units and Shape B has an area of 20 square units. Shape A has a longer perimeter than Shape B. Draw two such shapes satisfying the given conditions.
Solution:
Perimeter of shape A = 2(9 + 2)
= 2 × 11
= 22 units
Perimeter of shape B = 2(5 + 4)
= 2 × 9
= 18 units
Question 6.
On a page in your book, draw a rectangular border that is 1 cm from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border?
Solution:
Do yourself.
Let the dimensions of the page be x and y.
The dimensions of the inner rectangle = (x – 3) x (y – 2)
∴ Perimeter = 2 × (x – 3) + 2 × (y – 2)
Question 7.
Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area.
Solution:
The inner rectangle should have an area of \(\frac{12 × 8}{2}\) = 48 sq units
So, possible dimensions could be 8 units × 6 units.
Question 8.
A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true here?
(a) The area of each rectangle is larger than the area of the square.
(b) The perimeter of the square is greater than the perimeters of both the rectangles added together.
(c) The perimeters of both the rectangles added together is always 1\(\frac{1}{2}\) times the perimeter of the square.
(d) The area of the square is always three times as large as the areas of both rectangles added together.
Solution:
(c) is true i.e., the perimeters of both the rectangles added together is always 1 \(\frac{1}{2}\) times the perimeter of the square.
Area of square = 16 sq units
Perimeter of square =16 units
Area of both the rectangles = 4 × 2 + 4 × 2 = 8 + 8 = 16
sq units
Perimeter of both the rectangles = 2 (4 + 2) + 2 (4 + 2)
= 12 + 12
= 24 units
= 1 \(\frac{1}{2}\) times (16 units).