NCERT solution class 7 chapter 11 Perimeter and Area exercise 11.2 mathematics

EXERCISE 11.2

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Q 1, Ex 11.2 - Perimeter and Area - Chapter 11 - Maths Class 7th - NCERT

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Page No 216:

Question 1:

Find the area of each of the following parallelograms:

Answer:

Area of parallelogram = Base × Height
(a) Height= 4 cm
Base = 7 cm
Area of parallelogram = 7 × 4 = 28 cm²
(b) Height= 3 cm
Base = 5 cm
Area of parallelogram = 5 × 3 = 15 cm²
(c) Height= 3.5 cm
Base = 2.5 cm
Area of parallelogram = 2.5 × 3.5 = 8.75 cm²
(d) Height= 4.8 cm
Base = 5 cm
Area of parallelogram = 5 × 4.8 = 24 cm²
(e) Height= 4.4 cm
Base = 2 cm
Area of parallelogram = 2 × 4.4 = 8.8 cm²

Q 2, Ex 11.2 - Perimeter and Area - Chapter 11 - Maths Class 7th - NCERT

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Question 2:

Find the area of each of the following triangles:

Answer:

(a) Base = 4 cm, height= 3 cm
Area  = 6 cm²
(b) Base = 5 cm, height= 3.2 cm
Area  = 8 cm²
(c) Base = 4 cm, height= 3 cm
Area  = 6 cm²
(d)Base = 3 cm, height= 2 cm
Area = = 3 cm²

Q 3, Ex 11.2 - Perimeter and Area - Chapter 11 - Maths Class 7th - NCERT

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Question 3:

Find the missing values:

So No	Base	Height	Area of parallelogram
a.	20 cm	–	246 cm2
b.	–	15 cm	154.5 cm2
c.	–	8.4 cm	48.72 cm2
d.	15.6 cm	–	16.38 cm2

Answer:

Area of parallelogram = Base × Height
(a) b = 20 cm
h = ?
Area = 246 cm²
20 × h = 246

Therefore, the height of such parallelogram is 12.3 cm.
(b) b = ?
h = 15 cm
Area = 154.5 cm²
b × 15 = 154.5
b = 10.3 cm
Therefore, the base of such parallelogram is 10.3 cm.
(c) b = ?
h = 8.4 cm
Area = 48.72 cm²
b × 8.4 = 48.72

Therefore, the base of such parallelogram is 5.8 cm.
(d) b = 15.6 cm
h = ?
Area = 16.38 cm²
15.6 × h = 16.38

Therefore, the height of such parallelogram is 1.05 cm.

Q 4, Ex 11.2 - Perimeter and Area - Chapter 11 - Maths Class 7th - NCERT

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Page No 217:

Question 4:

Find the missing values:

Base	Height	Area of triangle
15 cm	_______	87 cm2
_______	31.4 mm	1256 mm2
22 cm	_______	170.5 cm2

Answer:

(a) b = 15 cm
h = ?
Area =

Therefore, the height of such triangle is 11.6 cm.
(b) b = ?
h = 31.4 mm
Area =

Therefore, the base of such triangle is 80 mm.
(c) b = 22 cm
h = ?
Area =

Therefore, the height of such triangle is 15.5 cm.

Q 5, Ex 11.2 - Perimeter and Area - Chapter 11 - Maths Class 7th - NCERT

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Question 5:

PQRS is a parallelogram (see the given 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

Answer:

(a) Area of parallelogram = Base × Height = SR × QM
= 7.6 × 12 = 91.2 cm²
(b) Area of parallelogram = Base × Height = PS × QN = 91.2 cm²
QN × 8 = 91.2

Q 6, Ex 11.2 - Perimeter and Area - Chapter 11 - Maths Class 7th - NCERT

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Question 6:

DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (see the given figure). If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Answer:

Area of parallelogram = Base × Height = AB × DL
1470 = 35 × DL

Also, 1470 = AD × BM
1470 = 49 × BM

Q 7, Ex 11.2 - Perimeter and Area - Chapter 11 - Maths Class 7th - NCERT

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Question 7:

ΔABC is right angled at A (see the given figure). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ΔABC. Also find the length of AD.

Answer:

Area == 30 cm²

Q 8, Ex 11.2 - Perimeter and Area - Chapter 11 - Maths Class 7th - NCERT

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Question 8:

ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (see the given figure). The height AD from A to BC, is 6 cm. Find the area of ΔABC. What will be the height from C to AB i.e., CE?

Answer: