The answer would be -5 1/4
(a) 1.65 + 1.36 then add 79 + 118.17
(b)mr. sanchez class and mr. kellys calss
(c) 3.01
answer: 3.01
<h3>G I V E N :</h3>
find the cost of the painting of a rectangular box of length 6m breadth 4m and height 3m at the rate of rs.25 per meter square
<h3>S O L U T I O N :</h3>
According to the question, first we need to get the T.S.A. or Total Surface Area of the cuboid then dividing it by ₹ 0.25 we can get the cost. After solving the concept will be more clear.
Let's start :
Here,
l represents length
b represents breadth
h represents height
Now, putting the values of all these we get
- T.S.A. = 2(6 × 4 + 4 × 3 + 3 × 6)
- T.S.A. = 2(24 + 12 + 18) m²
Now, we are given that cost of every square metre costs ₹ 0.25
So, for 108 m² it will cost = ₹ 108 × 0.25 = ₹ 27
<u>Hence, cost of painting the </u><u>rectang</u><u>ular box is ₹ 27</u>
Answer:
in the figure four quadrants each of radius 2 m are removed from a rectangle.
I have attached the required figure, assuming the sides of a rectangle based on given measurement of quadrants.
Let the sides of a rectangle be 2 m × 4 m.
The radius of quadrant is 2 m ( given )
From given, we have
Area of figure = Area of 4 quadrants - Area of rectangle .
= 4 × π/4 × r² - l × w
= 4 × π/4 × 2² - 2 × 4
= 3.14 × 4 - 8
= 12.4 - 8
= 4.4 m²
Therefore, the area of the figure is 4.4 m²
Perimeter of figure = Perimeter of 4 quadrants - Perimeter of rectangle
= 4 × [ 0.5πr + 2r ] - 2 ( l + w )
= 4 × [ 0.5 × 3.14 × 2 + 2 × 2 ] - 2 ( 4 + 2 )
= 4 × [ 3.14 + 4 ] - 2 (6)
= 28.56 - 12
= 16.56 m
the perimeter of the figure is 16.56 m