Draw Circles That Cover a Given Area

CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Expanse InText Questions and Answers are provided by experts in social club to assist students secure adept marks in exams.

Form seven Maths NCERT Solutions Chapter 11 Perimeter and Area InText Questions

Effort These (Page 205)

Question i.
Ayush and Deeksha made pictures. Ayush made his moving-picture show on a rectangular canvass of length 60 cm and breadth 20 cm while Deeksha fabricated hers on a rectangular sheet of length forty cm and breadth 35 cm. Both of these pictures have to be separately framed and laminated. Who has to pay more for framing, if the toll of framing is ₹ 3.00 per cm?
If the rate of lamination is ₹ ii.00 per cm², who has to pay more for lamination?
Solution:
For finding the cost of framing, nosotros need to find the perimeter and then multiple it by the rate for framing.
In case of Ayush :
Perimeter of the rectangular sheet = 2 (Length + Latitude)
= ii × (60 + 20) cm = two × 80 cm = 160 cm
∴ Cost of framing at the charge per unit of ₹ 3 per cm
= ₹ (3 × 160) = ₹ 480

In example of Deeksha :
Perimeter of the rectangular sheet
= ii × (twoscore + 35) cm
= 2 × 75 cm = 150 cm
∴ Cost of framing at the charge per unit of ₹ iii per cm
= ₹ (three × 150) = ₹ 450
Thus, Ayush paid more than for framing (as 480 > 450).
For finding the cost of lamination, we demand to observe the expanse and and so multiply it past the rate for lamination.

In the example of Ayush :
Area of the rectangular sheet
= (60 × 20) cmⁱⁱ
=1200 cm²
∴ Cost of lamination at the rate of ₹ 2 per cm²
= ₹ (two × 1200)
= ₹ 2400

In the example of Deeksha :
Area of the rectangular canvass
= (40 × 35) cm² = 1400 cm²
∴ Cost of lamination at the charge per unit of ₹ 2 per cm²
= ₹ (2 × 1400)
= ₹ 2800
Thus, Deeksha paid more for lamination (every bit 2800 > 2400).

Try These (Page 205)

Question 1.
What would you demand to find, surface area, or perimeter to respond the following?
(i) How much space does a blackboard occupy?
Solution:
For finding the space occupied past a blackboard, we should detect the surface area of the blackboard.

(2) What is the length of a wire required to debate a rectangular flower bed?
Solution:
For fencing a rectangular flower bed, we demand to notice the perimeter of the flower bed.

(iii) What distance would you encompass by taking ii rounds of a triangular park?
Solution:
For finding the distance covered past taking two rounds of a triangular park, nosotros need to find the perimeter of the park and then multiply it by 2.

(4) How much plastic sheet do yous need to encompass a rectangular swimming pool?
Solution:
For finding the plastic sheet required to cover a rectangular swimming pool, we need to find the surface area of the puddle.

Try These (Page 206)

Question 1.
Tanya needed a foursquare of side four cm for completing higher. She had a rectangular sheet of length 28 cm and breadth 21 cm [See fig. (i)] She cuts off a foursquare of side iv cm from the rectangular canvas. Her friend saw the remaining sail

[Come across fig.(ii)] and asked Tanya, "Has the perimeter of the sheet increased or decreased now?"

(i) Has the total length of side AD increased after cut off the foursquare?
Solution:
There has been a modify in the length of side Advertisement after cutting the square.

(ii) Has the area increased or decreased?
Solution:
The area has decreased.

(iii) Tanya cuts off ane more than foursquare from the opposite side. (Come across fig.) Will (the perimeter of the remaining sheet increase further?

Solution:
Yeah, the perimeter increases.

(iv) Will the area increase or decrease further?
Solution:
The area will decrease.

(5) So, what tin nosotros infer from this?
Solution:
Clearly, if the perimeter increases, it is not necessary that the area also increases.

Try These (Page 206)

Question 1.
Experiment with several such shapes and cut-outs. You might find it useful to depict these shapes on squared sheets and compute their areas and perimeters. You have seen that increase in perimeter does not mean that area will as well increase.
Solution:
Exercise yourself (Based on the experiment).

Question 2.
Give two. examples where area increases as the perimeter increases.
Solution:
(i)

Consider a rectangle ABCD.
Area (ABCD) = A₁ = viii × 3 cm² = 24 cm
Perimeter (ABCD) = P₁ = 2(8 + 3) cm = 22 cm

Let u.s.a. consider another rectangle ABEF.
Expanse (ABEF) = A₂ = x × three cm² = 30 cm^two
Perimeter (ABEF) = P₂ = 2(ten + 3) cm = 26 cm
Thus, A₂ > A₁ => P₂ > P₁
i.e., as surface area increases, perimeter increases.
(ii)

Consider a square ABCD.
Area (ABCD) = three × three cm^two = ix cm^two
Perimeter (ABCD) = three × 3 cm = 9 cm

Allow u.s. consider a rectangle ABEF.
Area (ABEF) = iii × 5 cm² = 15 cm²
Perimeter (ABEF) = two(three + v) cm = xvi cm
Thus, A₂ > A₁ => P_two > P_one
i.e., every bit expanse increases, perimeter increases.

Question 3.
Give 2 examples where the area does not increase when the perimeter increases.
Solution:
(i)

Consider a rectangle ABCD.
Expanse (ABCD) = A₁ =10 × eight cm^two = 80 cm²

Perimeter (ABCD) = P_two = 2(8 +ten) cm = 36 cm
Cut-off a triangle from the rectangle ABCD as shown.
Its area = A₂

Its perimeter = P₂ = (8 + 10 + 8 + 3 + iii + 5 + 3) cm = 40 cm
Clearly, A₁ < A_two, i.eastward., area does non increases, simply P₂ > P_two, i.e., perimeter increases.

(ii)

Let us have another cut equally shown.
Its area = A₃
= (A₂ – 4 × 3) cm²
= (74 – 12) cm²
= 62 cm^two
Its perimeter = P₃
= (ii + 3 + 4 + three + 2 + x + eight + 3 + iii + 5 + iii) cm = 46 cm
Clearly, A₃ < A₂, but P_iii > P_two, i.e., area decreases and perimeter increases.

Try These (Page 209)

Question 1.
Take a rectangle of sides eight cm and 5 cm. Cut the rectangle along its diagonal to get ii triangles. Superpose one triangle on the other.
ing
(i) Are they exactly the same in size?
Solution:
These triangles are exactly the same in size.
(2) Can yous say that both the triangles are equal in area?
Solution:
Yeah, both the triangles are equal in surface area.
(iii) Are the triangles congruent also?
Solution:
Yes, by the SSS criterion of congruence, these triangles are coinciding.
(4) What is the area of each of these triangles.
Solution:
The area of each triangle

Where 50 = length and b = latitude
∴ Surface area of each triangle = \(\frac{1}{2}\) (8 × v) cm² = twenty cm²

Question ii.
Take a foursquare of side 5 cm and carve up information technology into 4 triangles as shown.

(i) Are the four triangles equal in area?
Solution:
Yes, the four triangles are equal in surface area.
(two) Are they congruent to each other?
Solution:
Yeah, by superimposing, nosotros find that the triangles are congruent.
(three) What is the expanse of each triangle?
Solution:
The expanse of each triangle
= \(\frac{one}{four}\) (Area of the square)
= \(\frac{1}{iv}\) (Side)^two = \(\frac{ane}{4}\) (v)² cm² = 6.25 cm²

Try These (Folio 210)

Question one.
Each of the following rectangles of length 6 cm and latitude 4 cm is composed of coinciding polygons. Detect the area of each polygon.
Solution:
Given, length of rectangle = 6 cm
and breadth of rectangle = iv cm
∴ Area of a rectangle = Length × Breadth = 6 × four = 24 cmⁱⁱ

(a)

From the figure, it is articulate that the rectangle is divided into vi coinciding polygons or parts.
∴ Expanse of each polygon = \(\frac{i}{6}\) × (Expanse of a rectangle)
= \(\frac{one}{six}\) × 24 = iv cm^two

(b)

From the figure, it is articulate that the rectangle is divided into iv congruent polygons.
∴ Area of each polygon = \(\frac{1}{four}\) × (Area of a rectangle)
= \(\frac{1}{4}\) × 24 = 6 cm²

(c)

From the figure, it is articulate that the rectangle is divided into ii congruent polygons.
∴ Area of each polygon = \(\frac{1}{2}\) × (Area of a rectangle)
= \(\frac{one}{2}\) × 24 =12 cm²

(d)

From the figure, it is clear that the rectangle is divided into two coinciding polygons.
∴ Area of each polygon = \(\frac{1}{ii}\) × (Area of a rectangle)
= \(\frac{1}{two}\) × 24 = 12 cm²

From the figure, it is clear that the rectangle is divided into 6 congruent polygons.
∴ Expanse of each polygon = \(\frac{ane}{8}\) × (Area of a rectangle)
= \(\frac{1}{8}\) × 24 = 3 cmⁱⁱ

Try These (Page 211)

Question i.
Consider the following parallelograms

Detect the areas of the parallelograms by counting the squares enclosed inside the figures and also find the perimeters by measuring the sides.
Complete the post-obit table :

Solution:
On counting the squares enclosed within the figures of the parallelogram, nosotros discover in each case these are xv in numbers. And so, the surface area of each parallelogram = 15 sq. units. Let us draw a perpendicular DM to the base AB, produced if necessary, equally shown :

Let us observe the other side Ad of the parallelogram ABCD in each case in order to find the perimeter in each case.
From right-angled ΔADM, using Pythagoras theorem,

On filling the to a higher place data in the given tabular array, nosotros have

We observe that all these parallelograms take equal areas but different perimeters.

Try These (Page 212)

Question i.
Consider the post-obit parallelograms with the sides 7 cm and 5 cm.
(a)

(b)

(C)

(d)

Find the perimeter and the area of each of these parallelograms. Analyze your results. What practice you lot infer from this?
Solution:
The perimeter and the area of each parallelogram are equally under :

We observe that all these parallelograms take dissimilar areas but equal perimeters.
Thus, to find the area of a parallelogram, we demand to know just the base and the corresponding height of the parallelogram.

Try These (Folio 212)

Question 1.
Discover the area of the following parallelograms :
(i)

Solution:
Area = base x height
= (viii × three.5) cm² = 28 cm^two
(2)

Solution:
Area = base x peak
= (8 × 2.5) cmⁱⁱ = 20 cmⁱⁱ
(iii) In a parallelogram ABCD, AB = 7.2 cm and the perpendicular from C on AB is 4.v cm.
Solution:
Area of parallelogram ABCD
= AB × length of perpendicular from C on AB
= (7.2 × 4.5) cm^two
= 32.4 cm²

Endeavour These (Page 213)

Question ane.
Try the higher up activeness with different types of triangles.
Solution:
Here, AABC is the given triangle. We have fixed another AACP in such a way that we get a parallelogram ABCP in each instance every bit shown in the effigy below:

Question 2.
Take unlike parallelograms. Dissever each of the parallelograms into 2 triangles by cutting forth whatever of its diagonals. Is the triangle congruent?
Solution:
Here, ABCD is the given parallelogram and each of the parallelograms is divided into two triangles by cutting along its diagonal \(\frac{A C}{B D}\). The triangles thus formed are congruent.

Try These (Page 213)

Question 1.
In the adjacent figures, all the triangles are on the base AB = half-dozen cm. What tin you say about the height of each of the triangles corresponding to the base AB?

Can we say all the triangles are equal in area?
Solution:
Clearly, from the given figures, the heights of each of the triangles respective to base AB are equal.
Since the expanse of a triangle = \(\frac{1}{2}\) × base × height
So, we can say that all the triangles are equal in the area as they have the aforementioned base and equal meridian.

Endeavour These (Page 214)

Question 1.
Consider the birdbrained-angled triangle ABC of base 6 cm. It's height Ad which is perpendicular from the vertex A is outside the triangle. Tin y'all find the area of a triangle?

Solution:
Yes, its surface area can be found out.
Area = \(\frac{1}{two}\) × base × height

Try These (Page 219)

Question i.
In the given effigy,

(a) Which square has the larger perimeter?
Solution:
The outer square has a larger area.

(b) Which is larger, the perimeter of a smaller square or the circumference of the circumvolve.
Solution:
The circumference of the circle is larger than the perimeter of the smaller square.

Try These (Page 219)

Question ane.
Accept one each of the quarter plate and half plate. Roll in one case each of these on a table-meridian. Which plate covers more distance in one consummate revolution? Which plate volition take fewer revolutions to cover the length of the table-acme?

Solution:
Doing as directed, nosotros observe that the half plate covers more than altitude in i complete revolution than the altitude covered by the quarter plate. Too, the one-half plate takes fewer revolutions to cover the length of the table-top.

Try These (Page 221)

Question ane.
Consider the following:
(i) A farmer dug a bloom bed of a radius of 7 k at the centre of a field. He needs to purchase fertilizer. If 1 kg of fertilizer is required for 1 square meter area, how much fertilizer should he purchase?
Solution:
Area of the circular flower bed = π r², where r = 7 k

Fertilizer required for the flower bed at the rate of i kg per one thou² = (154 × 1) kg = 154 kg

(2) What will be the cost of polishing a circular table-top of radius 2 m at the rate of ₹ ten per foursquare meter?
Solution:
Area of the top of round table = π rⁱⁱ, where r = 2 m

Toll of polishing at the rate of ₹ 2 per sq. meter

=₹ 25.fourteen (approx)

Endeavor These (Page 222)

Question 1.
Describe circles of different radii on graph paper. Notice the surface area by counting the number of squares. As well, observe the area by using the formula. Compare the two answers.
Solution:
Let us draw two circles of radius 1 cm and 2 cm on the graph paper of one cm² square.

Past using graph newspaper, we find that their corresponding areas are 4 cm² and 12 cmⁱⁱ respectively.
By using the formula, nosotros. find that their corresponding areas are

= 12.57 cmⁱⁱ (approx.)
Thus, we see that the ii values differ.

Attempt These (Folio 224)

Question 1.
Tin you lot convert one km^two into m^two?
Solution:
We know that 1 km = g m
1 kmⁱⁱ =1 km × 1 km
1 km^two = 1000 grand × g grand
= 1000000 m²

Try These (Page 225)

Question one.
Convert the following:

(i) fifty cm² in mmⁱⁱ
Solution:
ane cm^two = 1 cm x 1 cm
= 10 mm × 10 mm (∵ 1 cm = 10 mm)
= 100 mm^two
50 cm² = (50 × 100) mm^two = 5000 mm²

(ii) 2 ha in k^two
Solution:
We know that a square of side 100 1000 has an area of 1 hectare (ha).
1 hectare -100 m × 100 m
=10000 yard^two

(three) ten grand² in cm²
Solution:
2 hectares =two × 10000 thousand² = 20000 m²
1 m² =ane grand × 1 m
= 100 cm × 100 cm (∵ 1 g = 100 cm)
= 10000 cm²
10 g² = 10 × 10000 cm^two = 100000 cm²

(4) grand cm² in m²
Solution:
∴ k cm^two =

mcphersonsheyesseet.blogspot.com

Source: https://www.cbsetuts.com/ncert-solutions-for-class-7-maths-chapter-11-intext-questions/