AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 4 Textbook Questions and Answers.
AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 4
Question 1.
Find the area of the following rhombuses.
Solution:
Area = \(\frac { 1 }{ 2 }\)d_{1}d_{2}
d_{1} = 5 + 5 = 10cm
d_{2} = 2 + 2 = 4cm
A = \(\frac { 1 }{ 2 }\) × 10 × 4 = 20 cm^{2}
d_{1} = 3 + 3 = 6 cm
d_{2} = 4 + 4 = 8cm
Area = \(\frac { 1 }{ 2 }\)d_{1}d_{2}
= \(\frac { 1 }{ 2 }\) × 6 × 8
= 24 cm^{2}
Question 2.
Find the missing values.
Diagonal – 1 (d_{1}) | Diagonal – 2 (d_{2}) | Area of the rhombus |
12cm | 16cm | |
27mm | 2025mm^{2} | |
14m | 57.6m |
Solution:
Diagonal – 1 (d_{1}) | Diagonal – 2 (d_{2}) | Area of the rhombus |
12cm | 16cm | \(\frac{1}{2}\) × 12 × 16 = 96cm^{2} |
27mm | \(\frac{2025 \times 2}{27}\) = 150mm | 2025mm^{2} |
14m | 57.6m | \(\frac{1}{2}\) × 24 × 57.6 = 691.2 m^{2} |
Question 3.
If length of diagonal of a rhombus whose area 216 sq. cm. is 24 cm. when find the length of second diagonal.
Solution:
Given: Length of one diagonal d_{1} = 24cm, d_{2} =?
Area = \(\frac { 1 }{ 2 }\)d_{1}d_{2} = 216
\(\frac { 1 }{ 2 }\) × d_{1} × d_{2} = 216
\(\frac { 1 }{ 2 }\) × 24 × d_{2} = 216
d_{2} = \(\frac{216}{12}\) = 18
Question 4.
The floor of a building consists of3 000 tiles which are rhombus shaped. The diagonals of each of the tiles are 45 cm and 30 cm. Find the total cost of polishing the floor, if cost per m^{2} is Rs.2.50.
Solution:
Diagonals of each (shape / rhombus) tiles
d_{1} = 45cm, d_{2} = 30cm
Total tiles = 3000
Total area = 3000 × Area of each tile = 3000 × \(\frac { 1 }{ 2 }\) × d_{1} × d_{2}
= 3000 x \(\frac { 1 }{ 2 }\) × 45 × 30 = 2025000 cm^{2}
= \(\frac{2025000}{100 \times 100}\) m^{2} = 202.5m^{2}
∴ Cost of polishing the floor at the rate of Rs. 2.50 per a square metre 202.5 × 2.50 = Rs. 506.25