Let the height of the parallelogram be x. cm. Then, base = (2x) cm.
Area of square = 40 x 40
= 1600 sq.cm
Given that the areas of Square and Rectangle are equal
=> Area of rectangle = 1600 Sq.cm
We know that, Area of rectangle = L x B
Given L = 64 cm
Breadth of rectangle = 1600/64 = 25 cm
Perimeter of the rectangle = 2(L + B) = 2(64+25) = 178 cm.
Area of 4 walls = 2(l+b)h
=2(10+7) x 5 = 170 sq m
Area of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) = 12 sq m
area to be planted = 170 -12 = 158 sq m
Cost of painting = Rs. 158 x 3 = Rs. 474
required area = (area of an equilateral triangle of side 7 cm)- (3 * area of sector with à = 60 degrees and r = 3.5cm)
sq cm
= sq cm
= 1.967 sq cm
=> 2l + 2b = 5b => 3b = 2l
b=(2/3)l
Then, Area = 216 cm2
=> l x b = 216 => l x (2/3)l =216
l = 18 cm.
Let the side of the square be 's' cm
length of rectangle = (s+5) cm
breadth of rectangle = (s-3)cm
(s+5) (s-3) =
- 5s - 3s - 15 =
2s = 15
Perimeter of rectangle = 2(L+B) = 2(s+5 + s?3) = 2(2s + 2)
= 2(15 + 2) = 34 cm
From statement (A),
20b = (1/2) × b × h
h = 40 cm.
Let breadth = x metres, length = 3x metres, height = H metres.
Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m
So, breadth = 6 m and length = = 9 m.
Now, papered area = (1720/10) = 172 sq.m
Area of 1 door and 2 windows = 8 sq.m
Total area of 4 walls = (172 + 8) sq.m = 180 sq.m
speed = 12 km/h =
distance covered =
time taken = distance /speed =
area of cross roads = (55 x 4) + (35 x 4)- (4 x 4) = 344sq m
cost of graveling = 344 x (75/100) = Rs. 258
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