NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3.

• Perimeter and Area Class 7 Ex 11.1
• Perimeter and Area Class 7 Ex 11.2
• Perimeter and Area Class 7 Ex 11.4
• Perimeter and Area Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

Question 1.
Find the circumference of the circles with the following radus (Take π = \(\frac { 22 }{ 7 } \))
(a) 14 cm
(b) 28 mm
(c) 21 cm
Solution:

Question 2.
Find the area of the following circles, given that: ( Take π = \(\frac { 22 }{ 7 } \) )
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius = 5 cm.
Solution:

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet ( Take π = \(\frac { 22 }{ 7 } \) )
Solution:
Circumference of the circular sheet = 154 m
Let the radius of the circular sheet be r cm
Then, its circumference = 2nr m According to the question,
Circumference = 2πr = 154

Question 4.
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase if he makes 2 rounds of offense. Also find the cost of the rope, if it costs ₹ 4 per meter. ( Take π = \(\frac { 22 }{ 7 } \) )
Solution:
The diameter of the circular garden (r) = 21 m
Radius of the circular garden (r) = \(\frac { 21 }{ 2 } \) m
∴ Circumference of the circular garden = 2πr
= 2 × \(\frac { 22 }{ 7 } \) × \(\frac { 21 }{ 2 } \) m = 66m
⇒ Length of the rope needed to make 1 round of fence = 66 m
⇒ Length of the rope needed to make 2 rounds of fence
= 66 × 2 m = 132 m
Cost of rope per meter = ₹ 4
∴ Cost of the rope = ₹ 132 × 4 = ₹ 528.

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Solution:

Here, Outer radius, r = 4 cm
Inner radius, r = 3 cm
Area of the remaining sheet = Outer area – Inner area
= π (R² – r²) = 3.14 (4² – 3²) cm²
= 3.14 (16 – 9) cm²
= 3.14 × 7 cm² = 21.98 cm²

Question 6.
Saima wants to put lace on the edge of a circular table cover of a diameter of 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15. (Take π = 3.14)
Solution:
Diameter of the table cover = 1.5 m
⇒ Radius of the table cover (r) = \(\frac { 1.5 }{ 2 } \) m
⇒ Circumference of the table cover = 2πr
= 2 × 3.14 × \(\frac { 1.5 }{ 2 } \) m = 4.71 m
⇒ Length of the lace required = 4.71 m
∵ Cost of lace per meter = ₹ 15
∴ Cost of the lace = ₹ 4.71 × 15 = ₹ 70.65

Question 7.
Find the perimeter of the following figure, which is a semicircle including its diameter.
Solution:

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m². (Take π = 3.14)
Solution:
Diameter of the table-top = 1.6 m
⇒ Radius of the table-top (r) = \(\frac { 1.6 }{ 2 } \) m = 0.8 m
∴ Area of the table-top = πr²
= 3.14 × (0.8)2 m²
= 3.14 × 0.64 m²
= 2.0096 m²
∵ Rate of polishing = ₹ 15 per m²
∴ Cost of polishing the table-top = ₹ 2.0096 × 15
= ₹ 30.144
= ₹ 30.14 (approx.).

Question 9.
Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its side? Which figure encloses more area, the circle or the square? ( Take π = \(\frac { 22 }{ 7 } \) )
Solution:

Question 10.
From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the following figure). Find the area of the remaining sheet. ( Take π = \(\frac { 22 }{ 7 } \) )

Solution:

Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the leftover aluminium sheet? (Take π = 3.14)
Solution:

Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)
Solution:

Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Solution:

Question 14.
A circular flower garden has an area of about 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Taken π = 3.14)
Solution:
The circular area of the sprinkler = πr²
= 3.14 × 12 × 12
= 3.14 × 144 = 452.16 m²
Area of the circular flower garden = 314 m²
Since the area of the circular flower garden is smaller than by sprinkler
Therefore, the sprinkler will water the entire garden.

Question 15.
Find the circumference of the inner and the outer circles as shown in the following figure? (Take π = 3.14)

Solution:
Radius of inner circle = 19 – 10 = 9 m
∴ Circumference of the inner circle = 2 πr = 2 × 3.14 × 9 m = 56.52 cm
The radius of the outer circle = 19 m
∴ Circumference of the outer circle = 2πr = 2 × 3.14 × 19 m = 119.32 m.

Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? ( Take π = \(\frac { 22 }{ 7 } \) )
Solution:

Question 17.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)
Solution:
We know that the minute hand describes one complete revolution in one hour.
∴ Distance covered by its tip = Circumference of the circle of radius 15 cm
= (2 × 3.14 × 15) cm
= 94.2 cm

We hope the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2.

• Perimeter and Area Class 7 Ex 11.1
• Perimeter and Area Class 7 Ex 11.3
• Perimeter and Area Class 7 Ex 11.4
• Perimeter and Area Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

Question 1.
Find the area of the following parallelograms:


Solution:

Question 2.
Find the area of each of the following triangles:


Solution:

Question 3.
Find the missing values:

Solution:

Question 4.
Find the missing values:

Solution:

Question 5.
PQRS is a parallelogram (in Figure). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm.

Solution:
(a) Area of the parallelogram PQRS = base × height = SR × QM = 12 × 7.6 cm² = 91.2 cm²
(b) Area of the parallelogram PQRS = base × height = PS × QN
⇒ 91.2 = 8 × QN
⇒ QN = \(\frac { 912 }{ 8 } \) cm
⇒ QN = 11.4 cm.

Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (in Figure). If the area of a parallelogram is 1470 cm2, AB = 35 cm, and AD = 49 cm, find the length of BM and DL.

Solution:

Question 7.
∆ ABC is right-angled at A (in Figure). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of ∆ ABC. Also, find the length of AD.

Solution:

Question 8.
∆ ABC is isosceles with AB = AC = 7.5 cm, and BC = 9 cm (in Figure). The height of AD from A to BC is 6 cm. Find the area of ∆ ABC. What will be the height from C to AB i.e., CE?

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.4
• Practical Geometry Class 7 Ex 10.5

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

Question 1.
Construct A XYZ in which XY = 4.5 cm, YZ = 5 cm and, ZX = 6 cm.
Solution:
Steps of Construction

1. Draw a line segment YZ of length 5 cm.
   
2. With Y as centre, draw an arc of radius 4.5 cm.
3. With Z as centre, draw an arc of radius 6 cm,
4. Mark the point of intersection of arcs as X.
5. Join XY and XZ. ∆ XYZ is now ready.

Question 2.
Construct an equilateral triangle of side 5.5 cm.
Solution:
Steps of Construction:

1. Draw a line segment BC of length 5.5 cm.
2. With B as centre, draw an arc of radius 5.5 cm.
   
3. With C as centre, draw an arc of radius 5.5 cm.
4. Mark the point of intersection of arcs as A.
5. Join AB and AC. Equilateral ∆ ABC is now ready.

Question 3.
Draw ∆ PQR with PQ = 4 cm, QR =3.5 cm and PR = 4 cm. What type of triangle is this?
Solution:
Steps of Construction:

1. Draw a line segment QR of length 3.5 cm.
2. With Q as centre, draw an arc of radius 4 cm.
3. With R as centre, draw an arc of radius 4 cm.
   
4. Mark the point of intersection of arcs as P.
5. Join PQ and PR.

∆ PQR is now ready,
∵ PQ = PR
∴ ∆ PQR is isosceles.

Question 4.
Construct ∆ ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Solution:
Steps of Construction

1. Draw a line segment BC of length 6 cm.
   
2. With B as centre, draw an arc of radius 2.5 cm.
3. With C as centre, draw an arc of a radius of 6.5 cm.
4. Mark the point of intersection of arcs as A.
5. Join AB and AC.
6. ∆ ABC is now ready. On measurement, ∠B = 90°.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.4
• Practical Geometry Class 7 Ex 10.5

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Question 1.
Construct ADEF such that DE = 5 cm, DF 3 cm, and m ∠EDF = 90°.
Solution:
Steps of Construction:

1. Draw a line segment DE = 5cm.
2. Draw ∠EDX = 90°.
3. With centre D and radius = 3 cm, draw an arc to intersect DX at F.
4. Join EF to obtain the required triangle DBF.

Question 2.
Construct an isosceles triangle in which the length of each of its equal sides is 6.5 cm and the angle between them is 110°.
Solution:
Steps of Construction

1. Draw a line segment QR of length 6.5 cm.
2. At Q, draw QX making 110° with QR, using a protractor.
   
3. With Q as centre, draw an arc of a radius of 6.5 cm. It cuts QX at P.
4. Join PR. ∆ PQR is now obtained.

Question 3.
Construct ∆ ABC with BC = 7.5 cm, AC = 5 cm and m ∠C = 60°.
Solution:
Steps of Construction:

1. Draw a line segment BC = 7.5 cm.
2. Draw ∠BCX = 60°.
3. With C as centre and radius = 5 cm, draw an arc intersecting CX at A.
4. Join AB to obtain the required ∆ABC.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.5

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

Question 1.
Construct ∆ ABC, given m ∠A = 60°, m ∠B = 30° and AB = 5.8 cm.
Solution:
Steps of Construction

1. Draw AB of length 5.8 cm.
2. At A, draw a ray AP making an angle of 60° with AB.
3. At B, draw a ray BQ making an angle of 30° with BA.
4. Mark the point of intersection of two rays as C.
5. ∆ ABC is now completed.

Question 2.
Construct ∆ PQR if PQ = 5 cm, m ∠PQR = 105° and m ∠QRP = 40°.
(Hint: Recall angle-sum property of a triangle).
Solution:
By angle-sum property of a triangle
m ∠RPQ + m ∠PQR + m ∠QRP = 180°
⇒ m ∠RPQ + 105° + 40° = 180°
⇒ m ∠RPQ + 145° = 180°
⇒ m ∠RPQ = 35°
Steps of Construction

1. Draw PQ of length 5 cm.
2. At Q, draw a ray QX making an angle of 105° with QP.
   
3. At P draw a ray PY making an angle of 35° with PQ.
4. Mark the point of intersection of two rays as R.

∆ PQR is now completed.

Question 3.
Examine whether you can construct ∆DEF such that EF = 7.2 cm, m ∠E = 110° and m ∠F = 80°. Justify your answer.
Solution:
m ∠E + m ∠F = 110° + 80° = 190° > 180°
This is not possible since the sum of the measures of the three angles of a triangle is 180°. As such, the sum of two angles of a triangle cannot exceed 180°.
Hence, ∆ DEF cannot be constructed.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.4, drop a comment below and we will get back to you at the earliest.