Haryana State Board HBSE 6th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 Textbook Exercise Questions and Answers.

## Haryana Board 6th Class Maths Solutions Chapter 10 Mensuration Exercise 10.2

Question 1.

Find the area of the following figures :

Solution:

(a) This figure is made up of line- segments. Moreover, it is covered by full squares. Here, fully filled squares = 9.

∴ Total area = 9 sq. cm.

(b) This figure is also made up of line- segments. Moreover, it is covered by full squares. Here, fully filled squares 5

∴ Total area = 5 sq. cm.

(c) This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only.

(i) Fully filled squares = 2.

(ii) Half-filled squares = 4.

∴ Area covered by full squares

= 2 x 1 = 2 sq. cm

and area covered by half squares

= 4 x \(\frac{1}{2}\) = 2 sq. cm

Hence, total area = 2 + 2 = 4 sq. cm.

(d) This figure is made up of line- segments. Moreover, it is covered by full squares. This makes our job simple.

Here, fully filled squares = 8

∴ Total area = 8 x 1 = 8 sq. cm.

(e) This figure is made up of line-segments. Moreover, it is covered by full squares. Here, fully filled squares = 10

∴ Total area = 10 x 1 = 10 sq. cm.

(f) This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only.

Here, (i) fully filled squares = 2

(ii) half-filled squares = 4

∴ Area covered by full squares

= 2 x 1 = 2 sq. cm

Area covered by half squares = 4 x \(\frac{1}{2}\) = 2 sq. cm

Hence, total area = 2 + 2 = 4 sq. cm.

(g) This figure is also made up of line- segments. Moreover, it is covered by full squares and half squares only.

Here, (i) fully filled squares = 4

(ii) half-filled squares = 4

∴ Area covered by full squares

= 4 x 1 = 4 sq. cm

Area covered by half squares

= 4 x \(\frac{1}{2}\) = 2 sq. cm

Hence, total area = 4 + 2 = 6 sq. cm.

(h) This figure is made up of line- segments. Moreover, it is covered by full squares only. This makes our job simple.

Here, fully filled squares = 5

∴ Total area = 5 x 1 sq. cm = 5 sq cm.

(i) This figure is made up of line-segments. Moreover, it is covered by full squares only. This makes our job simple.

Here, fully filled squares = 9

∴ Total area = 9 x 1 sq. cm = 9 sq. cm.

(j) This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only.

Here, (i) fully filled squares = 2

(ii) half-filled squares = 4

∴ Area covered by full squares = 4 x \(\frac{1}{2}\) = 2 sq. cm

Hence,

total area = (2 + 2) sq. cm :

= 4 sq. cm.

(k) This figure is made up of line segments. Moreover, it is covered by full squares and half squares only.

Here, (i) fully filled squares = 4

(ii) half filled squares = 2

∴ Area covered by full squares

= 4 x 1 = 4 sq. cm

Area covered by half squares

= 2 x \(\frac{1}{2}\) = 1 sq. cm ,

Hence,

total area = (4+1) sq. cm

= 5 sq. cm.

(l)

Cover | Number | Area estimated (sq. cm) |

(i) Fully-filled squares | 4 | 4 |

(ii) Half-filled squares | 2 | 2 x \(\frac{1}{2}\) = 1 |

(Hi) More than half-filled squares | 3 | 3 |

(iv) Less than half-filled squares | 3 | 0 |

Total area = 4 + 1 + 3 + 0 = 8sq. cm.

(m)

Cover | Number | Area estimated (sq. cm) |

(i) Fully filled squares | 8 | 8 |

(ii) Half-filled squares | 2 | 2 x \(\frac{1}{2}\) = 1 |

(Hi) More than half-filled squares | 4 | 4 |

(iv) Less than half-filled squares | 6 | 0 |

Total area = 8 + 1 + 4 + 0 = 13 sq. cm.

(n)

Cover | Number | Area estimated (sq. cm) |

(i) Fully filled squares | 13 | 13 |

(ii) Half-filled squares | 2 | 2 x \(\frac{1}{2}\) = 1 |

(Hi) More than half-filled squares | 3 | 3 |

(iv) Less than half-filled squares | 5 | • 0 |

∴ Total area = 13 + 1 + 3 + 0 = 17 sq. cm.

Question 2.

Use tracing paper and centimeter graph paper to compare the areas of the following pair of figures :

Solution:

(a)

Cover | Number | Area estimated (sq. cm) |

(i) Fully filled squares (√) | 3 | 3 |

(ii) Half-filled squares (.) | 2 | 2 x \(\frac{1}{2}\) = 1 |

(iii) More than half-filled squares (Δ) | 8 | 8 |

(iv) Less than half-filled squares (x) | 7 | 0 |

∴ Total area = 3 + 1 + 8 + 0 = 12 sq. cm.

(b)

Cover | Number | Area estimated (sq. cm) |

(i) Fully filled squares (√) | 6 | 6 |

(ii) Half-filled squares (.) | 3 | 3 x \(\frac{1}{2}\) = 1\(\frac{1}{2}\) |

(iii) More than half-filled squares (Δ) | 7 | 7 |

(iv) Less than half-filled squares (x) | 6 | 0 |

Total area = 6 + 1\(\frac{1}{2}\) + 7 + 0 = 14\(\frac{1}{2}\) sq. cm.

Thus, area in Fig.(b) is greater than the area in Fig.(a).