By using Ganita Prakash Book Class 6 Solutions and Chapter 6 Perimeter and Area Class 6 NCERT Solutions Question Answer, students can improve their problem-solving skills.
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 = ?.
(b) Perimeter of a square = 20 cm; length of a side = ?.
(c) Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?.
Solution:
(a) Given, perimeter of rectangle =14 cm, breadth = 2 cm
We know that the perimeter of a rectangle = 2 × (length + breadth)
⇒ 14 cm = 2 × (length + 2 cm)
⇒ 2l = 14 – 4
⇒ l = \(\frac{10}{2}\) = 5 cm
Hence, the length of the rectangle is 5 cm.
(b) Given, perimeter of the square = 20 cm
We know that the perimeter of square = 4 × side
⇒ 20 = 4 × side
⇒ side = \(\frac{20}{4}\) = 5 cm
Hence, side of the square is 5 cm.
(c) Given, perimeter of rectangle =12 m and length = 3 m
Let breadth = b cm
We know that perimeter of rectangle = 2 (length + breadth)
⇒ 12 = 2(3 + b)
⇒ 2b = 12 – 6 = 6
⇒ b = 3 cm
Hence, breadth of the rectangle is 3 cm.
Question 2.
A rectangle having side lengths 5 cm and 3 cm 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 a side of the square?
Solution:
Given, length of rectangle = 5 cm and breadth = 3 cm
We know that
perimeter of rectangle = 2 × (length × breadth)
= 2 × (5 + 3) = 16 cm
Now, if we bend the wire to form a square, the total length of the wire (16 cm) will be divided equally among the four sides of the square.
So, each side of the square = \(\frac{\text { Perimeter }}{4}\)
= \(\frac{16}{4}\) = 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:
Given, length = 150 m and breadth = m
Perimeter of the park = 2 × (Length + Breadth)
= 2 × (150 +120)
= 540m
Given, cost of fencing the rectangular park = ₹ 40 per metre
∴ Cost of fencing the rectangular park = 540 × ₹ 40
= ₹ 21600
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 of equal length?
Solution:
Given, total length of string = 36 cm
(i) For a square,
each side of the square = \(\frac{\text { Total length of string }}{4}\)
= \(\frac{36}{4}\) = 9 cm
(ii) For a triangle with all sides of equal length,
each side of the triangle = \(\frac{\text { Total length of string }}{3}\)
= \(\frac{36}{3}\) = 12 cm
(iii) For a hexagon with all sides of equal length,
each side ot the hexagon = \(\frac{\text { Total length of string }}{6}\)
= \(\frac{36}{6}\) = 6 cm
Question 6.
A farmer has a rectangular field having length 230 m and breadth 160 m. He wants to fence it with 3 rounds of rope as shown. What is the total length of rope needed?
Solution:
We have,
length of rectangular field = 230 m
and breadth of rectangular field = 160 m
Perimeter of rectangular field = 2 × (length × breadth)
= 2 × (230 +160)
= 2 × 390 = 780 m
Since, the farmer wants to fence the field with 3 round of rope.
∴ Total length of rope needed = 3 × perimeter
= 3 × 730 m
= 2340 m
6.1 Perimeter Figure it Out (Page No. 133 – 134)
Matha Pachchi!
Each track is a rectangle. Akshi’s track has length 70 m and breadth 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:
Akshi runs on a rectangular track with a length of 70 metres and a breadth of 40 metres.
∴ Perimeter of track = 2 × (length + breadth)
= 2 × (70 + 40) = 220 m
Since, the distance covered in one round = 220 m
∴ Total distance covered in 5 rounds = 5 × 220 m
= 1100 m
Question 2.
Find out the total distance Toshi has covered in 7 rounds.
Who ran a longer distance?
Solution:
Toshi runs on a rectangular track with a length of 60 m and breadth of 30 m.
∴ Perimeter of track = 2 × (length + breadth)
= 2 × (60 + 30) = 180 m
Since, the distance covered in ope round = 180 m
∴ Total distance covered in 7 rounds = 7 × 180 m
= 1260 m
So, 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.
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:
We have, length of th4 rectangular garden = 25 m
and area of rectangular garden = 300 sq metre,
∵ Area of rectangle = Length × Width
⇒ 300 = 25 × width
⇒ width = \(\frac{300}{25}\) = 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:
We have, length of rectangular plot = 500 m and width = 200 m
∴ Area of the plot = Length × Width
= 500 × 200 = 100000 sq metre
Given, rate of tiling per hundred sq metre = ₹ 8
Rate of 1 m² = ₹ \(\frac{8}{100}\)
∴ Cost of 100000 m² = 100000 × \(\frac{8}{100}\) = ₹ 8000
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:
Given, length of the coconut grove = 100 m
and breadth of the coconut grove = 50 m
∴ Area of the coconut grove = 100 m × 50 m = 5000 m²
∵ Each coconut tree requires 25 sq. m
∴ Number of tree = \(\frac{\text { Area of the coconut grove }}{\text { Area of each coconut tree }}\)
= \(\frac{5000}{25}\) = 200
Area of each coconut tree 5000 = 200
Hence, 200 trees can be planted in this grove.
Question 4.
By splitting the following figures into rectangles, find their areas (all measures are given in metres):
Solution:
Let the given figure be divided into rectangles A, B,C and D and their length and breadth be written on the figure.
For Rectangle A,
Length = 4 m and breadth = 2 m
Now, area of the Rectangle A = Length × Breadth
= 4 × 2 = 8m²
For Rectangle B,
Length = 3 m and breadth = 3 m
Area of the Rectangle B = Length × Breadth
= 3 × 3 = 9m²
For Rectangle C,
Length = 2 m and breadth = 1 m
Area of the Rectangle C = Length × Breadth
= 2 × 1 = 2 m²
For Rectangle D, length = 3 m and breadth = 3 m
∴ Area of the Rectangle D = Length × Breadth
= 3 × 3= 9m²
Now, total area of the given Figure
= Area of the Rectangle A + Area of the Rectangle B + Area of the Rectangle A – Area of the Rectangle D
= 8 + 9 + 2 + 9 = 28 m²
Hence, the required area is 28 m².
(b) Let the given figure is divided into rectangles A, B and C and their length and breadth are written on the figure.
For rectangle A,
length = 2 m and breadth = 1 m
Area of the rectangle A = Length × Breadth
= 2 × 1 = 2 m²
For rectangle B,
length = 5 m and breadth = 1 m
Area of the rectangle B = Length x Breadth
= 5 x 1 = 5m²
For rectangle C, length = 2 m and breadth = 1 m
∴ Area of the Rectangle C = Length × Breadth
= 2 × 1 = 2 m²
Now, total area of the given figure
= Area of the rectangle A + Area of the rectangle B + Area of the rectangle C
= 2 + 5 + 2 = 9m²
Hence, the area of the given figure is 9 m².
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:
Here, we can see that some shapes have identical areas.
Specifically
Shapes A and B These shapes are identical, meaning they cover the same amount of space, so they have the ^ same area.
Shapes C and E These shapes also have the same area because they are identical in size and shape.
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:
Shape D is twice as big as shape C. This means that if you place two shape C pieces together. Then, they exactly cover shape D.
The relationship between these shapes
Shape D can be completely filled by combining shape C and shape E. So, area of shape D is equal to the sum of the area of shape C and E.
Each of shapes C and E has half the area of shape D.
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:
When comparing the area of shape A to shape G, we get shape A has the same area of shape G. Both shapes are identical in size and cover the same amount of space. So, neither is bigger than the other, they are equal.
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:
The big square formed by all seven tangram pieces has an area that is 8 times the 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 area of rectangle is 8 times the area of shape C.
The area remain the same as the large square because the number of pieces and their combined area does not change only their arrangement does.
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 and the rectangle are different. Even though both shapes have the same area, the arrangement of the pieces affects the perimeter.
The square is more compact, while the rectangle is more spread out, resulting in different perimeter.
6.2 Area Figure it Out (Page No. 140)
Question 1.
Find the area of the following figures. (Page 140)
Solution:
(i) In the figure, full small square = 3
Exactly half square = 2
∴ Total area = 3 × 1 + 2 × \(\frac{1}{2}\)
= 3 + 1
= 4 sq. units
(ii) In the figure, full small square = 6
Exactly half square = 6
∴ Total area = 6 × 1 + 6 × \(\frac{1}{2}\)
. = 6 + 31= 9 sq. units
(iii) In the figure, full small square = 7
Exactly half square = 6
∴ Total area = 7 × 1 + 6 × \(\frac{1}{2}\)
= 7 + 3 = 10 sq. units
(iv) In the figure, full small square = 8
Exactly half square = 6
Total area = 8 × 1 + 6 × \(\frac{1}{2}\)
= 8 + 3 = 11 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
6.3 Area of A Triangle Figure it Out (Page 145)
Question 1.
Making it ‘More’ or ‘Less’.
Observe these two figures. Is there any similarity or difference between the two?
Using 9 unit squares (having an area of 9 sq units), we have made figures with two different perimeters—the first figure has a perimeter of 12 units and the second has a perimeter of 20 units. Arrange or draw different figures with 9 sq units to get other perimeters. Each square should align with atleast one other square on atleast one side completely and together all squares should form a single connected figure with no holes.
Using 9 unit squares, solve the following.
(i) What is the smallest perimeter possible?
(ii) What is the largest perimeter possible?
(iii) Make a figure with a perimeter of 18 units.
(iv) 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?
Solution:
(i) The smallest perimeter possible using 9 unit squares is 12 units when arranged in a 3 × 3 square.
(ii) The largest perimeter possible using 9 unit squares is 20 units when arranged in a straight line of 1 by 9 squares.
(iii) A possible figure with a perimeter of 18 units can be arranged in a 2 × 4 rectangle with one additional square attached on any side.
(iv) Yes, different shaped figure can have the same perimeter. The perimeter depends on the arrangement of the squares and not just the number of squares.
6.3 Area of A Triangle Figure it Out (Page 146)
Question 1.
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.
(i) Find the missing measurements.
(ii) Find out the areas of his house.
Solution:
(i) For utility
width = 15 – 12 = 3ft and length = 15 ft
Area = 15 × 3 = 45 sq. ft
For small bedroom 15ft × _ ft = 180
⇒ Missing term = \(\frac{180}{15}\) = 12 ft
For garden
Length = 15 + 5 = 20 ft
and width = 3 ft
∴ Area = 20 × 3 = 60 sq. ft
For parking
Length = 15 ft
and width = 3 ft
Area = 15 × 3 = 45 sq. ft
For hall
Area = (20 × 12) + (5 × 5)
= 240 sq. ft 4- 25 sq. ft
= 265 sq. ft
(ii) The area of house = Length × Width
= 35 × 30
= 1050 sq. ft
6.3 Area of A Triangle Figure it Out (Page 147)
Question 1.
Now, find out the missing dimensions and area of Sharan’s home. Below is the plan.
Some of the measurements are given.
(i) Find the missing measurements.
(ii) Find out the area of his 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:
(i) For utility
Area = 70 sq. ft
∴ Dimensions = 10 × 7
For small bedroom
Area = 12 × 10 = 120 sq. ft
For hall
Dimensions = 23 × 15
∴ Area = 345 sq. ft For toilet
Dimensions = 5 × 10
∴ Area = 50 sq. ft
For entrance Dimensions = 7 × 15
∴ Area = 105 sq. ft
(ii) Length of house = 42 ft and width = 25 ft
∴ Area = Length × Width
= 42 × 25
= 1050 sq. ft
Hence, the area of Sharan’s and Charan’s house is same but perimeter of Sharan’s house is more than the Charan’s house.
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 × 10m and 2m × 7m.
Solution:
First, find the area of each rectangle.
For first rectangle
Length = 10 m and breadth = 5 m
The area of first rectangle = Length × Breadth
= 10 m × 5 m = 50 sq. m
For second rectangle
Length = 7 m and breadth = 2 m
The area of second rectangle = 7 m × 2 m = 14 sq. m
Total area = Area of first rectangle + Area of second rectangle
= 50 + 14
= 64sq. m
The possible dimensions of a new rectangle with this area can be
(i) 8 m × 8 m (since, 8 × 8 = 64)
Or
(ii) 16 m × 4 m (since, 16 × 4 = 64)
Question 2.
The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
Solution:
Given that length of rectangular garden = 50 m
and the area of rectangular garden = 1000 sq. m
Let width of the garden be b.
We know that
area of rectangular garden = Length × Width
∴ 1000 = 50 × b
⇒ b = 1522
⇒ b = 20 m
Hence, the width of rectangular garden = 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:
Given, length of the floor = 5 m
and width of the floor = 4 m .
∴ Area of the floor = Length × Width
= 5 × 4 = 20 sq. m
Also, given that length of the square carpet = 3 m
The area of the square carpet = 3 × 3 = 9 sq m
Hence, the area of the floor laid with carpet is 9 sq m.
Therefore, the area of the floor that is not carpeted = Area of the floor – Area of the floor laid with carpeted.
= 20 – 9
= 11 sq.m
Question 4.
Four flower beds having sides 2 m long and 1 m wide are dug at the four comers of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?
Solution:
We have, length of the garden = 15 m
and width of the garden = 12 m
∴ The area of the garden = Length × Width
= 15 × 12
= 180 sq. m
Now, also given that length of a flower bed = 2 m
and width of a flower bed = 1 m
The area of a flower bed = Length × Width
= 2 × 1
= 2 sq. m
Total area of 4 flower beds = 4 × 2
= 8 sq. m
The area available for laying down a lawn
= Area of the garden – Area of 4 flower beds
= 180 – 8
= 172 sq. m
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:
For shape A, can be arrange it as 9 units by 2 units, giving a perimeter of 22 units.
For shape B, can be arrange it as 5 units by 4 units, giving a perimeter of 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.