1. 

Let a = 13, b = 14, and c = 15.
Then, s = 1⁄2(a + b + c)
∴ s = 21
∴ (s  a) = 8, (s  b) = 7 and (s  c) = 6
∴ Area = √
Area = √
2. 

base = 5 cm and hypotenuse = 13 cm.
Applying pythagoras theorem,
we find the third side of the triangle = √
∴ Area of triangle = ½ x b x h
⇒ A = ½ x 5 x 12 = 30 cm²
3. 

Area of an equilateral triangle = √
= √
4. 

Distance covered = number of revolutions x circumference of the wheel.
⇒ 550 = 1000 x 22/7 x d (d = diameter of wheel)
∴ d = (550 x 7)/(1000 x 22) = 0.175 m
∴ d = 17.5 cm.
5. 

Area of square = 272.25 cm²
∴ side = √
Perimeter = 4 x 16.5 = 66 cm
This will be the circumference of the wheel
⇒ 2 x 22/7 x r = 66
∴ r = (66 x 7)/(2 x 22) = 10.5 cm.
6. 

2(l + b)/b = 5/1
2l + 2b = 5b
3b = 2l
b = (2/3) l
Then, Area = 216 cm
l x b = 216
l x (2/3) l = 216
l = 324
l = 18 cm.
7. 

Let the radius of smaller circle = r
⇒ Radius of bigger circle = 3r/2
Area of smaller circle = πr²
Area of bigger circle = 9πr²/4
∴ Ratio of area of bigger circle to that of smaller circle.
⇒ (9πr²/4)/(πr²) = 9/4
8. 

Area of square = 14 x 14 = 196 m²
The area that the cow can graze is a sector in the square and the radius of the sector is 10.5 m,
which is the length of the rope and angle of the sector is 90°.
Area of sector = (90/360) (22/7) (10.5 x 10.5) = 86.625 m²
Area that the cow cannot graze = 196 – 86.625 = 109.375 m².
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