Mensuration (Class VIII)

Let area of the 1st square = 9x sq.m and area of the 2nd square = 25x sq.m.
So, 9x + 25x = 306
x = 9
Then area of the 1st square = 81 sq.metres
Area of the 2nd square = 225 sq.metres
So, side of the first square = 9 m and second square = 15 m.
Now, differences between sides = 15 - 9 = 6.

It is given that,

Side of square = 5 cm and let the length of rectangle = l cm and breadth of rectangle = b cm. 
So, l - b = area of square,
= l - b = (5)²,

= l - b = 25 ---------------------(1) and,

Perimeter of rectangle = 30 cm, = 2(l + b) = 30,

= l + b = 15 --------------------(2),

After solving equation (1) and (2), we get

l = 20 cm and b = 5 cm,

So, area of rectangle = 20*5 = 100 cm²

It is given that:
Diameter of the wheel = 28 cm, so radius = 28/2 = 14 cm.
Now, circumference of wheel = 2pier = (22214)/7 = 176 cm.
Number of revolutions in 352 m = 352*100/176 = 200

Let breadth =x cm and length = 3x cm Then, original area = x3x = 3x² cm², After changing into dimensions, Area = (3x + 4)(x – 3) = 3x² – 5x - 12 So, 3x² – (3x² – 5x – 12) = 77, 5x + 12 = 77 x = 13, Length = 3*13 = 39 cm.

Let the length of rectangle = 8x cm and breadth = 7x cm.
Perimeter of rectangle = circumference of circle = 2(8x + 7x) = 60
x = 2
Then, length = 16 cm and breadth = 14 cm.
Now, area of the rectangle = 224 cm².

Let length = x metres, then breadth = 4x/5 metres.

Perimeter = 270 metres

=> 2[x + 4x/5] = 270

=> x = 75 m

Length = 75 m and breadth = 4/5*250 = 60 m

Area = 75*60 = 4500 m²,

50% of area = 4500*50/100 = 2250 m²

The sector has taken out from a circle. Therefore, radii of both will be same.
So, area of sector = (pie*r²angle)/360 degree = 154/6 = (22r²60)/7360 = 154/6
r = 7 cm
Therefore, radius of circle = 7 cm.
 

Perimeter of ground = 2(130 + 90) = 440 m
Because the distance among each soldier is 11 meters, so the number of soldiers = 440/11 = 40.

Total are of the sheet = 4030 = 1200 cm² 
Length of sheet without margin = 40 - 5 - 5 = 30 cm
Breadth of sheet without margin = 30 - 5 - 5 = 20 cm
Area of sheet without margin = 3020 = 600 cm²
Now, area of margin = 1200 - 600 = 600 cm².

The horse is tightened on a corner of a square field. Therefore, angle = 90 degrees. Because, (pie x r x r x 90 degree) = 693/2, = (22 x r² x 90)/7 x 360 = 693/2, r = 21. So, length of the rope is 21 m.

Let height = x cm.
Then, base = 4x cm
Now, area = 169 sq. cm
b x h = 169,
4x * x = 169,
x = 6.5
So, height = 6.5 cm, then base = 4*6.5 = 26 cm.

It is given that length = 70 m.
Then, breadth = area/length
Breadth = 3150/70 = 45 m
Now, perimeter of the rectangle = 2(70 + 45) = 230 m
So, the quantity of fencing that will be required = 230 - 10 = 220 m.

Let the length = l m and breadth = b m.

So, the length of diagonal = 54*20/60 = 18 m,

Then, l² + b² = 18²,

l² + b² = 324 ----------------------------------(1)

Total distance covered with the rate 63 m/min along with its sides = 

l + b = 63*20/60 = 21,

l + b = 21

On squaring both sides,

l² + b² + 2lb = 441 ---------------------------(2),

On solving equation 1 and 2,

2lb = 117

lb = 58.5

So, area of the field = 58.5 m².

Let the length of rectangle = 10 and breadth = 10, then area = 10 x 10 = 100.
Now, if 4% in excess in length and 3% in deficit in breadth, then area = 10.4 x 9.7 = 100.88.
Increase in area = 100.88 - 100 = 0.88
Now, percentage error = 0.88%

 Area of hall = 2520 = 500 sq. metres
Length of hall with veranda = 25 + 5 = 30 m
Breadth of hall with veranda = 20 + 5 = 25 m
Area of hall with veranda = 3025 = 750 sq. metres
Then, area of veranda = 750 - 500 = 250 sq. metres
Cost of flooring = 250*9.50 = Rs. 2375