Answer: (A).

Explanation:

Radius of the circle (r) = 28 cm.

Length of the wire (Circumference) = $MF#% \text{2} \pi r = \text{2} \pi \text{28} = \text{176 cm} $MF#%

Let side of the square be 'a' cm.

Perimeter of square (4a) = Circumference of the circle = 176 cm

or, 4a = 176 cm

or, a = 44 cm

Thus, side of the square is 44 cm.

Answer: (C).

Explanation:

Let length of the rectangular field be x m.

∴ breadth = 75% of x = $MF#% \dfrac{3x}{4} $MF#%

Perimeter of rectangular field = 1050

or, 2(length + breadth) = 1050

or, $MF#% \text{2} \left( \text{x} + \dfrac{3x}{4} \right) = 1050 $MF#%

or, 7x = 1050 * 2

or, x = 300 m

∴ Length = 300 m

$MF#% Breadth = \dfrac{3x}{4} = \dfrac{3 * 300}{4} = 225 m $MF#%

∴ Area = length * Breadth = 300 * 225 = 67500 $MF#% m^2 $MF#%

Answer: (C).

Explanation:

When diagonal is given, Area of square = (diagonal)² divided by 2

∴ $MF#% \text{Area} = \dfrac{30 * 30}{2} = \text{450 } m^2 $MF#%

Answer: (B).

Explanation:

Length of the playground (L1) = 60 m

Breadth of the playground (B1) = 40 m

∴ area of the playground = length * breadth = 60 * 40 = 2400 sq. m

Also,

Length of the playground including length of the path (L2) = 60 + 2 = 62 m

Breadth of the playground including length of the path (B2) = 40 + 2 = 42 m

∴ area of the playground including area of the path = 62 * 42 = 2604 sq. m

Thus, area of the path = 2604 - 2400 = 204 sq. m

Answer: (B).

Explanation:

Length (l) = 20 m, Breadth (b) = 15 m and Height (h) = 5 m

Total area of the hall to be covered with mat = 2(lb + bh + hl)

= 2(20 * 15 + 15 * 5 + 5 * 20)

=2(300 + 75 + 100)

= 2 * 475

= 950 sq. m

Total expenditure = 60 * 950

= Rs. 57000

Answer: (B).

Explanation:

Let radius of the outer circle be R and radius of the inner circle be r.

Circumference of the outer circle = $MF#% 2 \pi R $MF#%

Circumference of the inner circle = $MF#% 2 \pi r $MF#%

But, $MF#% 2 \pi R - 2 \pi r = 44 $MF#%

or, R - r = $MF#% \dfrac{44}{2 \pi} = 44 * \dfrac{7}{44} = \text{7 m} $MF#%

Thus, width of the road = 7 m.

Answer: (C).

Explanation:

Length of the rectangular box (l) = 8 cm

Breadth of the rectangular box (b) = 6 cm

Height of the rectangualar box (h) = 2 cm

The maximum length of a pencil that can be kept in a rectangular box = $MF#% \sqrt{l^2 + b^2 + h^2} $MF#%

= $MF#% \sqrt{8^2 + 6^2 + 2^2} =\sqrt{64 + 36 + 4} = \sqrt{104} = 2 \sqrt{26} $MF#%

Answer: (B).

Explanation:

Area of the parallelogram = Length of the base * Perpendicular height

= 54 * 24 = 1296 m.

Total cost of levelling = $MF#% \text(Rs. ) \dfrac{1296}{10} * \text{50} = \text{Rs. 6480} $MF#%

Answer: (B).

Explanation:

Perimeter of square = 4 * Side

Perimeter of first square = 24 m.

Side of first square = $MF#% \dfrac{24}{4} = 6 $MF#%

∴ Area of first square = $MF#% \left( \text{Side} \right)^2 = \left( \text{6} \right)^2 = \text{36} $MF#%

Perimeter of second square = 32 m.

Side of second square = $MF#% \dfrac{32}{4} = 8 $MF#%

∴ Area of second square = $MF#% \left( \text{Side} \right)^2 = \left( \text{8} \right)^2 = \text{64} $MF#%

Sum of the areas of two squares = 36 + 64 = 100 m²

∴ Side of square = $MF#% \sqrt{100} = \text{10 m} $MF#%

∴ Perimeter of square = 4 * 10 = 40 m.

Answer: (A).

Explanation:

The three sides of the triangle are

a = 3 cm, b = 4 cm and c = 5 cm

∴ $MF#% \text{S} = \dfrac{1}{2} \left(\text{a + b + c} \right) = \dfrac{1}{2} \left(\text{3 + 4 + 5} \right) = \text{6 cm} $MF#%

∴ $MF#% \text{Area of a triangle =}\sqrt{s(s-a)(s-b)(s-c)} $MF#%

= $MF#% \sqrt{6(6-3)(6-4)(6-5)} $MF#%

= $MF#% \sqrt{6*3*2*1} = \text{6 sq.cm} $MF#%