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

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

Ex 11.2 Class 7 Maths Question 1.

Find the area of each of the following parallelograms:

Solution:

(a) Area of the parallelogram

= base × altitude = 7 cm × 4 cm

= 28 cm2

(b) Area of the parallelogram

= base × altitude = 5 cm × 3 cm

= 15 cm2

(c) Area of the parallelogram

= base × altitude = 2.5 cm × 3.5 cm

= 8.75 cm2

(d) Area of the parallelogram

= base × altitude = 5 cm × 4.8 cm

= 24.0 cm2

(e) Area of the parallelogram

= base × altitude = 2 cm × 4.4 cm

= 8.8 cm2

Ex 11.2 Class 7 Maths Question 2.

Find the area of each of the following triangles:

Solution:

Area of the triangle = (frac{1}{2}) × b × h

= (frac{1}{2}) × 4 cm × 3 cm

= 6m2

(b) Area of the triangle = (frac{1}{2}) × b × h

= (frac{1}{2}) × 5 cm × 3.2 cm

= 8.0 cm2

(c) Area of the triangle = (frac{1}{2}) × b × l

= (frac{1}{2}) × 3 cm × 4 cm

= 6 cm2

(d) Area of the triangle = (frac{1}{2}) × b × h

= (frac{1}{2}) × 3 cm × 2 cm

= 3 cm2

Ex 11.2 Class 7 Maths Question 3.

Find the missing values:

S.No. |
Base |
Height |
Area of the parallelogram |

(a) | 20 cm | 246 cm2 | |

(6) | 15 cm | 154.5 cm2 | |

(c) | 8.4 cm | 48.72 cm2 | |

(d) | 15.6 | 16.38 cm2 |

Solution:

(a) Area of the parallelogram =b × h

246 = 20 × h

(b) Area of the parallelogram = b × h

154.5 = b × 15

(c) Area of the parallelogram = b × h

48.72 = b × 8.4

(d) Area of the parallelogram = b × h

16.38 = 15.6 × h

Ex 11.2 Class 7 Maths Question 4.

Find the missing values:

Base |
Height |
Area of the triangle |

15 cm | — | 87 cm2 |

— | 31.4 mm | 1256 mm2 |

22 cm | — | 170.5 cm2 |

Solution:

(i) Area of the triangle = (frac{1}{2}) × b × h

So, the height =11.6 cm

(ii) Area of the triangle = (frac{1}{2}) × b × h

So, the required base = 80 mm.

(iii) Area of the triangle = (frac{1}{2}) × b × h

So, the required height = 15.5 cm

Ex 11.2 Class 7 Maths Question 5.

PQRS is a parallelogram. QM is the height of 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 = Base × Height)

= 12 cm × 7.6 cm

= 91.2 cm2

(b) Area of the parallelogram PQRS

Ex 11.2 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 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Solution:

Area of the parallelogram ABCD

= AB × DL

⇒ 1470 cm2 = 35 cm × DL

⇒ (frac{1470}{35}) DL

∴ DL = 42 cm

Area of the parallelogram ABCD = AD × BM

1470 cm2 = 49 cm × BM

⇒ (frac{1470}{49}) = 30 cm

∴ BM = 30 cm

Hence, BM = 30 cm and DL = 42 cm

Ex 11.2 Class 7 Maths Question 7.

∆ABC is right angled at A. 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:

Area of right triangle ABC

Ex 11.2 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 = (frac{1}{2}) × base × height

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