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- 11. If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

A.40 cm

B.45 cm

C.50 cm

D.52 cm

Answer & Explanation

Answer: Option C

Explanation:**Let x and y be the length and breadth of the rectangle respectively.**

Then, x - 4 = y + 3 or x - y = 7 ----(1)

Area of the rectangle =xy; Area of the square = (x - 4)

(x - 4) (y + 3) =xy <=> 3x -

Solving 1 & 2

Perimeter of the rectangle = 2 (x +

Then, x - 4 = y + 3 or x - y = 7 ----(1)

Area of the rectangle =xy; Area of the square = (x - 4)

*(y*+ 3)(x - 4) (y + 3) =xy <=> 3x -

*4y*= 12 ----(2)Solving 1 & 2

*,*we get x = 16 and*y*= 9.Perimeter of the rectangle = 2 (x +

*y)*= [2 (16 + 9)] cm = 50 cm.- 12. A room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq. m is Rs. 270 and the cost of papering the four walls at Rs. 10 per m
^{2}is Rs. 1420. If a door and 2 windows occupy 8 sq. m, find the height of the room.

A.4 m

B.4.5 m

C.5 m

D.5.5 m

Answer & Explanation

Answer: Option C

Explanation:**Let breadth = x m, length = 3x m, height = H metres.**

Area of the floor = (Total cost of carpeting)/(Rate/m

x

So, breadth = 6 m and length =(3/2)×6 = 9 m.

Now, papered area = (1420/10)m

Area of 1 door and 2 windows = 8 m

Total area of 4 walls = (142 + 8) m

2×(9+ 6)× H = 150 <=> H = 150/30 = 5 m.

Area of the floor = (Total cost of carpeting)/(Rate/m

^{2})=(270/5)m^{2}=54m^{2}.x

*×*(3x/2) = 54 =>x^{2}= (54×2/3) = 36 <=>*x*= 6.So, breadth = 6 m and length =(3/2)×6 = 9 m.

Now, papered area = (1420/10)m

^{2}= 142 m^{2}.Area of 1 door and 2 windows = 8 m

^{2}.Total area of 4 walls = (142 + 8) m

^{2}= 150 m^{2}2×(9+ 6)× H = 150 <=> H = 150/30 = 5 m.

- 13. a ratio between the area of a square of side a and an equilateral triangle of side a is

A.2:√3

B.1:√3

C.√3:1

D.4:√3

Answer & Explanation

Answer: Option D

Explanation:**Area of square/area of triangle = a**

= 4/√3 i.e 4:√3

^{2}/(√3/4)a^{2}= 4/√3 i.e 4:√3

- 14. Length and Breadth of a rectangle is 7 m and 3.5 m respectively. Find the area of circle of maximum radius

A.9.625

B.9.74

C.9.8

D.9.725

Answer & Explanation

Answer: Option A

Explanation:**area of circle = πb**

= 22/7 × 3.5

= 9.625sq.m

^{2}/4= 22/7 × 3.5

^{2}/4= 9.625sq.m

- 15. Find the area of a rhombus one side of which measures 20 cm and one diagonal is 24 cm.

A.370 cm

^{2}B.365 cm

^{2}C.380 cm

^{2}D.384 cm

^{2}
Answer & Explanation

Answer: Option D

Explanation: Let other diagonal = 2x cm.Since diagonals of a rhombus bisect each other at right angles, we have:

(20)

^{2}= (12)

^{2}+ (x)

^{2}=>

*x*=√(20)

^{2}– (12)

^{2}= √256= 16 cm. _I

So, other diagonal = 32 cm.

Area of rhombus = (1/2) x (Product of diagonals) =(1/2× 24 x 32) cm

^{2}= 384 cm

^{2}

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