CBSE Class 7

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.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5.

• 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 10.4

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

Question 1.
Construct the right-angled ∆ PQR where m ∠Q = 90°, QR = 8 cm and PR = 10 cm.
Solution:
Steps of Construction:

1. Draw QR of length 8 cm.
   
2. At Q, draw QX ⊥ QR.
3. With R as centre, draw an arc of radius 10 cm.
4. Mark the meeting point of these two as P.

∆ PQR is now obtained.

Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Solution:
Steps of Construction

1. Draw QR of length 4 cm.
2. At Q, draw QX ⊥ QR.
   
3. With R as centre, draw an arc of radius 6 cm.
4. Mark the meeting point of arc and QX as P.

∆ PQR is now obtained.

Question 3.
Construct an isosceles right-angled triangle ABC where m ∠ACB = 90° and AC = 6 cm.
Solution:
Steps of Construction

1. Draw AC of length 6 cm.
2. At C, draw CX ⊥ CA.
   
3. With C as centre, draw an arc of radius 6 cm to intersect CX at B.
4. Join AB. ∆ACB is now obtained.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.5, drop a comment below and we will get back to you at the earliest.