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A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.

A.

140 m

B.

148 m

C.

150 m

D.

200 m

Answer with explanation

Answer: Option DExplanation

Let the triangle and parallelogram have common base b,

let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then

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A copper wire when bent in the form of a square encloses an area of 121 cm^{2}. If the same wire is bent in the form of a circle, it encloses an area equal to?

A.

121 cm^{2}

B.

144 cm^{2}

C.

154 cm^{2}

D.

168 cm^{2}

Answer with explanation

Answer: Option CExplanation

As is given, area of square = a^{2} = 121 cm^{2}

So, side of square, a = 11 cm

length of the wire = Perimeter of square of side a = 4a = 4*11 = 44cm

Let the radius of circle be r cm. Then,

Perimeter of circle = Length of wire

=> 2πr = 44 cm

=> r = 44/2π = 7 (π = 22/7)

Now, Area of circle = πr^{2} = (22/7) *7*7 = 154 cm^{2}

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If each side pair of opposite sides of a square is increased by 20 m, the ratio of the length and breadth of the rectangular so formed becomes 5:3. The area of the old square is?

A.

990m²

B.

900m²

C.

930m²

D.

945m²

Answer with explanation

Answer: Option BExplanation

(x+20) / x = 5 / 3

3x + 60 = 5x

x = 30m; Area = 900m²

Workspace

Two small circular parks of diameters 6 m and 8 m are to be replaced by a bigger circular park. What would be the radius of this new park, in meter, if the new park occupies the same space as the two small parks (in meter)?

A.

5

B.

10

C.

15

D.

20

Answer with explanation

Answer: Option AExplanation

Area of the new circular park = sum of the areas of the 2 smaller parks

⇒ π (6/2)^{2} + π (8/2)^{2} = π(9+ 16) = 25π ⇒ 25 π = π R^{2}. ∴ R^{2} = 25 ⇒ R = 5 m

Workspace

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.

20

B.

30

C.

50

D.

40

Answer with explanation

Answer: Option CExplanation

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

Then, x – 4 = y + 3 or x – y = 7 —-(i)

Area of the rectangle =xy; Area of the square = (x – 4) (y + 3)

(x – 4) (y + 3) =xy <=> 3x – 4y = 12 —-(ii)

Solving (i) and (ii), we get x = 16 and y = 9.

Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

Workspace

The area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field ?

A.

18 meter

B.

20 meter

C.

22 meter

D.

25 meter

Answer with explanation

Answer: Option AExplanation

Let breadth =x metres.

Then length =(115x/100)metres.

Workspace

If the diameters of a circle is increased by 100% . Its area is increased by ?

A.

100%

B.

200%

C.

300%

D.

400%

Answer with explanation

Answer: Option CExplanation

Area of circle = πd^{2}/4

diameters increased by 100%, then new area = π(2d)^{2} /4 => πd^{2}

Then, New area = (πd^{2} – πd^{2}/4) => 3πd^{2}/4

increase percent =[(3πd^{2}/4) / ( πd^{2}/4) ]*100 %

= [(3πd^{2}/4) x (4/πd^{2})]*100 %

= 300%

Workspace

The area of a rectangle is equal to the area of a square whose diagonal is 12√2 metre. The difference between the length and the breadth of the rectangle is 7 metre. What is the perimeter of rectangle ? (in metre).

A.

68 metre

B.

50 metre

C.

62 metre

D.

64 metre

Answer with explanation

Answer: Option BExplanation

d = a√2

12√2 = a√2

a = 12

l * b = a² = (12²) = 144

l – b = 7 ; l = b + 7

(b + 7)*(b) = 144

b² + 7b – 144 = 0

b = 9; l = 16

2(l + b) = 2(16 + 9) = 50m

Workspace

If the length is increased by 25%, by what percent the width of a rectangle should be decreased so as to keep the area same.

A.

25%

B.

10%

C.

20%

D.

30%

Answer with explanation

Answer: Option CExplanation

Let the original length be l and the width be b

Therefore, the area = l*b

Now, as the length is increased by 25%, the new length is (1.25*l) and let the new width be x.

As the area is same, 1.25*l*x = l*b

x = b/1.25 = 0.8b

Therefore, the width is to be decreased by 20%.

Workspace

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