# ML Aggarwal Mensuration Exe-14.1 Class 6 ICSE Maths Solutions

ML Aggarwal Mensuration Exe-14.1 Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-14.1 Questions for Mensuration as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.

## ML Aggarwal Mensuration Exe-14.1 Class 6 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 6th Chapter-14 Mensuration Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-14.1 Questions Edition 2023-2024

### Mensuration Exe-14.1

ML Aggarwal Class 6 ICSE Maths Solutions

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#### Question 1. Find the perimeter of each of the following figures:

Perimeter = Sum of all the sides.
(i) Perimeter = 5 cm + 3 cm + 2 cm + 7 cm = 17 cm
(ii) Perimeters = 31 cm + 38 cm + 48 cm + 38 cm = 155 cm
(iii) Perimeter = 19 cm + 19 cm + 19 cm + 19 cm = 76 cm
(iv) Perimeter = 7 cm + 7 cm + 7 cm + 7 cm + 7 cm = 35 cm

### Mensuration Exe-14.1

ML Aggarwal Class 6 ICSE Maths Solutions

Page-289

#### Question 2. Find the perimeter of each of the following shapes:

(i) A triangle of sides 3 cm, 4 cm and 6 cm.
(ii) A triangle of sides 7 cm, 5.4 cm and 10.2 cm.
(iii) An equilateral triangle of side 11 cm.
(iv) An isosceles triangle with equal sides 10 cm each and third side 7 cm.

(i) Perimeter of the triangle with sides 3 cm, 4 cm and 6 cm
= 3 cm + 4 cm + 6 cm
= 13 cm
(ii) Perimeter of the triangle with sides 7 cm, 5.4 cm, 10.2 cm
= 7 cm + 5.4 cm + 10.2 cm
= 22.6 cm
(iii) Perimeter of an equilateral triangle
= 3 × length of a side
= 3 × 11 cm
= 33 cm
(iv) Perimeter of isosceles triangle
= 10 cm + 10 cm + 7 cm
= 27 cm

#### Question 3. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?

Length of the tape required
= Perimeter of the rectangular box
= 2 × (Length + Breadth)
= 2 × (40 cm + 10 cm)
= 2 × (50 cm)
= 100 cm
= 1 m

#### Question 4. A Table-Top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Perimeter of the table-top
= 2 × (Length + Breadth)
= 2 × (2 m 25 cm + 1 m 50 cm)
= 2 × (2.25 m + 1.50 m)
= 2 × (3.75 m)
= 7.5 m

#### Question 5. The perimeter of rectangle is 58 m. If its length is 17 m, find its breadth.

Given : Perimeter of Rectangle = 58 m

Length of Rectangle = 17 m

Length = 17 m

Perimeter = 58 m

58 = 2 (I + b)

58 = 2(17 + b)

58 = 34 + 2b

58 – 34 = 2b

24 = 2b

24/2 = b

Therefore , The Breadth of the Rectangle is 12 m.

#### Question 6. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?

Perimeter of the rectangle
= 2 × (Length + Breadth)
= 2 × (0.7 km + 0.5 km)
= 2 × (1.2 km)
= 2.4 km
Length of the wire needed
= 4 × Perimeter of the rectangle
= 4 × (2.4 km)
= 9.6 km

#### Question 7. Find the perimeter of a regular hexagon with each side measuring 7.5 m.

The perimeter of a regular hexagon
= 6 × Length of a side
= 6 × 7.5 m
= 45 m

#### Question 8. The lengths of two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the length of its third side?

Perimeter of a triangle = 12 cm + 14 cm + l
⇒ 36 cm = 12 cm + 14 cm + l
⇒ 36 cm = 26 cm + l
⇒ l = 36 cm – 26 cm
⇒ l = 10 cm

#### Question 9. The perimeter of a regular pentagon is 100 cm. How long is its each side?

Perimeter of the regular pentagon = 5 × Length of a side
⇒ Length of one (each) side

= Perimeter of the regular pentagon/5

= 100/5 cm

= 20 cm

#### Question 10. A piece of string is 30 cm long. What will be the length of each side if the string is used to form:

(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?

(a) Perimeter of the square = 4 × Length of aside
⇒ Length of a side
= Perimeter of the square/4 = 30/4 cm
= 7.5 cm

(b) Perimeter of the equilateral triangle = 3 × Length of a side
⇒ Length of a side
= Perimeter of the equilateral triangle/3
= 30/4 cm = 10 cm

(c) Perimeter of the regular hexagon = 6 × Length of a side
⇒ Length of a side

= Perimeter of the regular hexagon/6

= 30/6 cm

= 5 cm

#### Question 11. Find the cost of fencing a rectangular park of length 225 m and breadth 115 m at the rate of ₹13 per metre.

Perimeter of the rectangular park
= 2 × (Length + Breadth)
= 2 × (225 m + 115 m)
= 2 × (340 m)
= 680 m
∴ Cost of fencing the rectangular park
at the rate of ₹13 per metre = ₹13 × 680 m = ₹8840

#### Question 12. Meera went to a rectangular park 140 m long and 90 m wide. She took 5 complete rounds on its boundary. What is the distance covered by her?

Length (l) = 140 m
Width (b) = 90 m

∴ Perimeter of park = 2(l + b)
= 2 (140 + 90)
= 2(230)
= 460 m
She takes 5 complete round,
therefore distance covered by her = 5 × 460m = 2300 m

#### Question 13. Pinky runs 8 times around a rectangular park with length 80 m and breadth 55 m while Pankaj runs 7 times around a square park of side 75 cm. Who covers more distance and by how much?

Perimeter of rectangular park = 2 × (Length + Breadth)
= 2 × (80 + 55)
= 270 m
Pinky runs 8 times = 8 × 270 = 2160 m
And, perimeter of square park = 4 × Length of a side
= 4 × 75
= 300
Pankaj runs 7 times = 7 × 300 = 2100 m
∴ Pinky covers more distance i.e., 2160 – 2100 = 60 m

#### Question 14. A rectangular park is 104 m long and 56 m wide. Anjali walks around it at the rate of 4 km/h. What time will she take in making 5 rounds of the park ?

Given:

A rectangular park is 104 m long and 56 m wide.

Distance covered in one round is given by the perimeter of the park

Perimeter of the park is given by:

2 (L + B)

2(104+56)

2×160

320 m

Distance in 5 rounds:

5×320

1600m

The Speed of Anjali is 4 km/h

4000 m ———– 60min

1 m —————— 60/4000

1600 m ————- 60×1600/4000

1600 m ————- 24 min

Time taken by Anjali to complete the 5 rounds is 24 min.

#### Question 15. A wire is in the shape of a regular pentagon of side 12 cm. It is rebent into the shape of a rectangle whose length is 3/2 times its breadth. Find the length and the breadth of the rectangle.

a regular pentagon has all sides equal

so its perimeter becomes 5 × side = 5 × 12 = 60

now on changing the shape

its perimeter remains the same

we are given that its length is 3/2 times its breadth

so perimeter of rectangle = 2(l+b)

here,

2(l+b)=60

l+b=60/2=30

put l = (3/2)b……..(given)

3/2b+b=30

1.5b+b=30

2.5b=30

b=30/2.5=b=12

put b = 12 in l = (3/2)b

l = 3/2×12 = 3×6 = 18

so, length = 18, and breadth = 12

—  : End of ML Aggarwal Mensuration Exe-14.1 Class 6 ICSE Maths Solutions :–

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