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

- Class 7 Maths Perimeter and Area Exercise 11.1
- Class 7 Maths Perimeter and Area Exercise 11.2
- Class 7 Maths Perimeter and Area Exercise 11.3
- Class 7 Maths Perimeter and Area Exercise 11.4

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

Question 1.

Find the area of each of the following parallelograms:

Solution:

(a) Area of the parallelogram

= base × altitude = 7 cm × 4 cm

= 28 cm^{2}

(b) Area of the parallelogram

= base × altitude = 5 cm × 3 cm

= 15 cm^{2}

(c) Area of the parallelogram

= base × altitude = 2.5 cm × 3.5 cm

= 8.75 cm^{2}

(d) Area of the parallelogram

= base × altitude = 5 cm × 4.8 cm

= 24.0 cm^{2}

(e) Area of the parallelogram

= base × altitude = 2 cm × 4.4 cm

= 8.8 cm^{2}

Question 2.

Find the area of each of the following triangles:

Solution:

Area of the triangle = × b × h

= × 4 cm × 3 cm

= 6m^{2}

(b) Area of the triangle = × b × h

= × 5 cm × 3.2 cm

= 8.0 cm^{2}

(c) Area of the triangle = × b × l

= × 3 cm × 4 cm

= 6 cm^{2}

(d) Area of the triangle = × b × h

= × 3 cm × 2 cm

= 3 cm^{2}

Question 3.

Find the missing values:

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

(a) | 20 cm | 246 cm^{2} | |

(6) | 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 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

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 the triangle = × b × h

So, the height =11.6 cm

(ii) Area of the triangle = × b × h

So, the required base = 80 mm.

(iii) Area of the triangle = × b × h

So, the required height = 15.5 cm

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 cm^{2}

(b) Area of the parallelogram PQRS

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

⇒ DL

∴ DL = 42 cm

Area of the parallelogram ABCD = AD × BM

1470 cm^{2} = 49 cm × BM

⇒ = 30 cm

∴ BM = 30 cm

Hence, BM = 30 cm and DL = 42 cm

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

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 = × base × height

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