NCERT Solutions for Class 7 Maths Exercise 11.2 Chapter 11 Perimeter and Area in simple PDF are available here. Triangles as parts of rectangles, generalising for other congruent parts of rectangles, area of a parallelogram and area of a triangle are the topics covered in this exercise of NCERT Solutions for Class 7 Chapter 11. Our expert professors formulate these NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area to support you with your exam preparation to score good marks in Maths, with the help of solutions provided here.
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1. Find the area of each of the following parallelograms:
(a)
Solution:-
From the figure,
Height of parallelogram = 4 cm
Base of parallelogram = 7 cm
Then,
Area of parallelogram = base × height
= 7 × 4
= 28 cm2
(b)
Solution:-
From the figure,
Height of parallelogram = 3 cm
Base of parallelogram = 5 cm
Then,
Area of parallelogram = base × height
= 5 × 3
= 15 cm2
(c)
Solution:-
From the figure,
Height of parallelogram = 3.5 cm
Base of parallelogram = 2.5 cm
Then,
Area of parallelogram = base × height
= 2.5 × 3.5
= 8.75 cm2
(d)
Solution:-
From the figure,
Height of parallelogram = 4.8 cm
Base of parallelogram = 5 cm
Then,
Area of parallelogram = base × height
= 5 × 4.8
= 24 cm2
(e)
Solution:-
From the figure,
Height of parallelogram = 4.4 cm
Base of parallelogram = 2 cm
Then,
Area of parallelogram = base × height
= 2 × 4.4
= 8.8 cm2
2. Find the area of each of the following triangles:
(a)
Solution:-
From the figure,
Base of triangle = 4 cm
Height of height = 3 cm
Then,
Area of triangle = ½ × base × height
= ½ × 4 × 3
= 1 × 2 × 3
= 6 cm2
(b)
Solution:-
From the figure,
Base of triangle = 3.2 cm
Height of height = 5 cm
Then,
Area of triangle = ½ × base × height
= ½ × 3.2 × 5
= 1 × 1.6 × 5
= 8 cm2
(c)
Solution:-
From the figure,
Base of triangle = 3 cm
Height of height = 4 cm
Then,
Area of triangle = ½ × base × height
= ½ × 3 × 4
= 1 × 3 × 2
= 6 cm2
(d)
Solution:-
From the figure,
Base of triangle = 3 cm
Height of height = 2 cm
Then,
Area of triangle = ½ × base × height
= ½ × 3 × 2
= 1 × 3 × 1
= 3 cm2
3. Find the missing values:
|
S.No. |
Base |
Height |
Area of the 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 |
Solution:-
(a)
From the table,
Base of parallelogram = 20 cm
Height of parallelogram =?
Area of the parallelogram = 246 cm2
Then,
Area of parallelogram = base × height
246 = 20 × height
Height = 246/20
Height = 12.3 cm
∴Height of the parallelogram is 12.3 cm.
(b)
From the table,
Base of parallelogram =?
Height of parallelogram =15 cm
Area of the parallelogram = 154.5 cm2
Then,
Area of parallelogram = base × height
154.5 = base × 15
Base = 154.5/15
Height = 10.3 cm
∴Base of the parallelogram is 10.3 cm.
(c)
From the table,
Base of parallelogram =?
Height of parallelogram =8.4 cm
Area of the parallelogram = 48.72 cm2
Then,
Area of parallelogram = base × height
48.72 = base × 8.4
Base = 48.72/8.4
Height = 5.8 cm
∴Base of the parallelogram is 5.8 cm.
(d)
From the table,
Base of parallelogram = 15.6 cm
Height of parallelogram =?
Area of the parallelogram = 16.38 cm2
Then,
Area of parallelogram = base × height
16.38 = 15.6 × height
Height = 16.38/15.6
Height = 1.05 cm
∴Height of the parallelogram is 1.05 cm.
|
S.No. |
Base |
Height |
Area of the Parallelogram |
|
a. |
20 cm |
12.3 cm |
246 cm2 |
|
b. |
10.3 cm |
15 cm |
154.5 cm2 |
|
c. |
5.8 cm |
8.4 cm |
48.72 cm2 |
|
d. |
15.6 cm |
1.05 |
16.38 cm2 |
4. Find the missing values:
|
Base |
Height |
Area of Triangle |
|
15 cm |
87 cm2 |
|
|
31.4 mm |
1256 mm2 |
|
|
22 cm |
170.5 cm2 |
Solution:-
(a)
From the table,
Height of triangle =?
Base of triangle = 15 cm
Area of the triangle = 16.38 cm2
Then,
Area of triangle = ½ × base × height
87 = ½ × 15 × height
Height = (87 × 2)/15
Height = 174/15
Height = 11.6 cm
∴Height of the triangle is 11.6 cm.
(b)
From the table,
Height of triangle =31.4 mm
Base of triangle =?
Area of the triangle = 1256 mm2
Then,
Area of triangle = ½ × base × height
1256 = ½ × base × 31.4
Base = (1256 × 2)/31.4
Base = 2512/31.4
Base = 80 mm = 8 cm
∴Base of the triangle is 80 mm or 8 cm.
(c)
From the table,
Height of triangle =?
Base of triangle = 22 cm
Area of the triangle = 170.5 cm2
Then,
Area of triangle = ½ × base × height
170.5 = ½ × 22 × height
170.5 = 1 × 11 × height
Height = 170.5/11
Height = 15.5 cm
∴Height of the triangle is 15.5 cm.
5. PQRS is a parallelogram (Fig 11.23). 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
Fig 11.23
Solution:-
From the question it is given that,
SR = 12 cm, QM = 7.6 cm
(a) We know that,
Area of the parallelogram = base × height
= SR × QM
= 12 × 7.6
= 91.2 cm2
(b) Area of the parallelogram = base × height
91.2 = PS × QN
91.2 = 8 × QN
QN = 91.2/8
QN = 11.4 cm
6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Fig 11.24
Solution:-
From the question it is given that,
Area of the parallelogram = 1470 cm2
AB = 35 cm
AD = 49 cm
Then,
We know that,
Area of the parallelogram = base × height
1470 = AB × BM
1470 = 35 × DL
DL = 1470/35
DL = 42 cm
And,
Area of the parallelogram = base × height
1470 = AD × BM
1470 = 49 × BM
BM = 1470/49
BM = 30 cm
7. ΔABC is right angled at A (Fig 11.25). 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.
Fig 11.25
Solution:-
From the question it is given that,
AB = 5 cm, BC = 13 cm, AC = 12 cm
Then,
We know that,
Area of the ΔABC = ½ × base × height
= ½ × AB × AC
= ½ × 5 × 12
= 1 × 5 × 6
= 30 cm2
Now,
Area of ΔABC = ½ × base × height
30 = ½ × AD × BC
30 = ½ × AD × 13
(30 × 2)/13 = AD
AD = 60/13
AD = 4.6 cm
8. ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). 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?
Solution:-
From the question it is given that,
AB = AC = 7.5 cm, BC = 9 cm, AD = 6cm
Then,
Area of ΔABC = ½ × base × height
= ½ × BC × AD
= ½ × 9 × 6
= 1 × 9 × 3
= 27 cm2
Now,
Area of ΔABC = ½ × base × height
27 = ½ × AB × CE
27 = ½ × 7.5 × CE
(27 × 2)/7.5 = CE
CE = 54/7.5
CE = 7.2 cm