- NCERT Solutions for Class 7 Maths Chapter 9 Perimeter and Area Ex 9.1
- NCERT Solutions for Class 7 Maths Chapter 9 Perimeter and Area Ex 9.2
NCERT Solutions for Class 7 Maths Chapter 9 Perimeter and Area Ex 9.1 (Rationalised Contents)
Ex 9.1 Class 7 Maths Question 1.
Find the area of each of the following parallelograms:
Solution:
(a) Area of a parallelogram = base × altitude = 7 cm × 4 cm
= 28 cm^{2}
(b) Area of a parallelogram = base × altitude = 5 cm × 3 cm
= 15 cm^{2}
(c) Area of a parallelogram = base × altitude = 2.5 cm × 3.5 cm
= 8.75 cm^{2}
(d) Area of a parallelogram = base × altitude = 5 cm × 4.8 cm
= 24.0 cm^{2}
(e) Area of a parallelogram = base × altitude = 2 cm × 4.4 cm
= 8.8 cm^{2}
(a) Area of a parallelogram = base × altitude = 7 cm × 4 cm
= 28 cm^{2}
(b) Area of a parallelogram = base × altitude = 5 cm × 3 cm
= 15 cm^{2}
(c) Area of a parallelogram = base × altitude = 2.5 cm × 3.5 cm
= 8.75 cm^{2}
(d) Area of a parallelogram = base × altitude = 5 cm × 4.8 cm
= 24.0 cm^{2}
(e) Area of a parallelogram = base × altitude = 2 cm × 4.4 cm
= 8.8 cm^{2}
Ex 9.1 Class 7 Maths Question 2.
Find the area of each of the following triangles:
Solution:
(a) Area of a triangle = ½ × b × h
= ½ × 4 cm × 3 cm
= 6 m^{2}
(b) Area of a triangle = ½ × b × h
= ½ × 5 cm × 3.2 cm
= 8.0 cm^{2}
(c) Area of a triangle = ½ × b × h
= ½ × 3 cm × 4 cm
= 6 cm^{2}
(d) Area of a triangle = ½ × b × h
= ½ × 3 cm × 2 cm
= 3 cm^{2}
Solution:
(a) Area of a triangle = ½ × b × h
= ½ × 4 cm × 3 cm
= 6 m^{2}
(b) Area of a triangle = ½ × b × h
= ½ × 5 cm × 3.2 cm
= 8.0 cm^{2}
(c) Area of a triangle = ½ × b × h
= ½ × 3 cm × 4 cm
= 6 cm^{2}
(d) Area of a triangle = ½ × b × h
= ½ × 3 cm × 2 cm
= 3 cm^{2}
Ex 9.1 Class 7 Maths Question 3.
Find the missing values:S. No.
Base
Height
Area of the parallelogram
(a)
20 cm
246 cm^{2}
(b)
15 cm
154.5 cm^{2}
(c)
8.4 cm
48.72 cm^{2}
(d)
15.6
16.38 cm^{2}
Solution:
(a) Area of a parallelogram = b × h
246 = 20 × h
(b) Area of a parallelogram = b × h
154.5 = b × 15
(c) Area of a parallelogram = b × h
48.72 = b × 8.4
(d) Area of a parallelogram = b × h
16.38 = 15.6 × h
S. No. | Base | Height | Area of the parallelogram |
(a) | 20 cm |
| 246 cm^{2} |
(b) |
| 15 cm | 154.5 cm^{2} |
(c) |
| 8.4 cm | 48.72 cm^{2} |
(d) | 15.6 |
| 16.38 cm^{2} |
Solution:
(a) Area of a parallelogram = b × h
246 = 20 × h
(b) Area of a parallelogram = b × h
154.5 = b × 15
(c) Area of a parallelogram = b × h
48.72 = b × 8.4
(d) Area of a parallelogram = b × h
16.38 = 15.6 × h
Ex 9.1 Class 7 Maths Question 4.
Find the missing values:Base
Height
Area of the triangle
15 cm
—
87 cm^{2}
—
31.4 mm
1256 mm^{2}
22 cm
—
170.5 cm^{2}
Solution:
(i) Area of a triangle = ½ × b × h
So, the height =11.6 cm(ii) Area of a triangle = ½ × b × h
So, the required base = 80 mm.
(iii) Area of a triangle = ½ × b × h
So, the required height = 15.5 cm
Base | Height | Area of the triangle |
15 cm | — | 87 cm^{2} |
— | 31.4 mm | 1256 mm^{2} |
22 cm | — | 170.5 cm^{2} |
Solution:
(i) Area of a triangle = ½ × b × h
(ii) Area of a triangle = ½ × b × h
So, the required base = 80 mm.
(iii) Area of a triangle = ½ × b × h
Ex 9.1 Class 7 Maths Question 5.
PQRS is a parallelogram. 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
= SR × QM (∵ Area of a parallelogram = Base × Height)
= 12 cm × 7.6 cm
= 91.2 cm^{2}(b) Area of the parallelogram PQRS = PS × QN
91.2 = 8 × QN
QN = 91.2/8 = 11.4 cm
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm
Solution:
(a) Area of the parallelogram PQRS
= SR × QM (∵ Area of a parallelogram = Base × Height)
= 12 cm × 7.6 cm
= 91.2 cm^{2}
(b) Area of the parallelogram PQRS = PS × QN
91.2 = 8 × QN
QN = 91.2/8 = 11.4 cm
Ex 9.1 Class 7 Maths Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm^{2}, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Solution:
Area of the parallelogram ABCD = AB × DL
⇒ 1470 cm^{2} = 35 cm × DL
⇒ 1470/35 = DL
∴ DL = 42 cm
Area of the parallelogram ABCD = AD × BM
1470 cm^{2} = 49 cm × BM
⇒ 1470/49 = BM
∴ BM = 30 cm
Hence, BM = 30 cm and DL = 42 cm.
Solution:
Area of the parallelogram ABCD = AB × DL
⇒ 1470 cm^{2} = 35 cm × DL
⇒ 1470/35 = DL
∴ DL = 42 cm
Area of the parallelogram ABCD = AD × BM
1470 cm^{2} = 49 cm × BM
⇒ 1470/49 = BM
∴ BM = 30 cm
Hence, BM = 30 cm and DL = 42 cm.
Ex 9.1 Class 7 Maths Question 7.
Ex 9.1 Class 7 Maths Question 8.
∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height 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:
Area of ∆ABC = ½ × base × height
Related Links:
NCERT Solutions for Maths Class 8
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11
Solution:
Area of ∆ABC = ½ × base × height
Related Links:
NCERT Solutions for Maths Class 8
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11