Perimeter of rhombus
= 2 ? d1 2 + d2 2
= 2 ?(4.8)2 + (1.4)2
= 2 ?23.04 + 1.96
= 2 ?25
= 2 x 5 = 10 cm
Area of rhombus =(d1 x d2)/2
=(1 x 1.5)/2 m2
= 0.75 m2
Area of the field =1215/135 = 9 hec
= 90000 m2 [1 hec =10000 m2]
? Side of the field = ?90000 = 300 m
Perimeter of the field = 4 x 300 = 1200 m
Now, cost of putting a fence around field = (1200 x 75)/100 = ? 900
Ratio of area of 2 squares = (ratio of perimeter of 2 squares)2
= (24/12)2
= 4
Required increment = 2a + [a2 / 100] %
= 2 x 25 + [(252)/100)] %
= 50 + (625/100)%
= 56.25%
Let the diagonals of the squares be 3x and 2x.
? Ratio of their areas = [(1/2) (3x2)] / [(1/2) (2x2)] = 9/4
Let First angle = 3k
Second angle = 4k
Third angle = 5k and
Fourth angle = 8k
We know 3k + 4k + 5k + 8k = 360°
? 20k = 360°
? k = 18 °
Measure of smallest angle = 3k
=3 x 18 °
= 54 °
Area of square room = (10)2 = 100 sq m
= 100 x (100)2 sq cm
= 100 x 100 x 100 sq cm
Now, area of tiles = (50)2 = 50 x 50 sq cm
? Number of tiles needed = (Area of square room) / (Area of tile)
= (100 x 100 x 100) / (50 x 50)
= 400
Hence, 400 tiles will be needed.
Let the third side be x.
According to the question,
x + 20 = 26
? x = 26 - 20 = 6 cm
? Each equal side = 20/2 = 10 cm
Let length of rectangle = 5k
and breadth of rectangle = 3k
According to the quecation,
5k - 3k = 8
? 2k =8
? k = 4
? Lenght = 5k = 5 x 4 = 20 m
Breadth = 3k = 3 x 4 = 12 m
? Required area = Lenght x Breadth
= 20 x 12 = 240 sq m
Let length = 2x and breadth = x
According to the question,
2x - x = 5
? x = 5
? Required perimeter = 2(2x + x)
= 6x
= 30 cm
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