Tag: maths

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Questions Related to maths

Multiple choice maths how many squares area of rectangular paths comparing areas spaces and boundaries - 2

A room is x meters long, y meters broad, and z meters high; find how many square meters of carpet will be required for the floor and how many square meters of paper for the walls?

  1. xy and 2xy + 2yz

  2. xy and 2xz + 2yz

  3. xy and 2xy

  4. xy and 2xy + 2xz
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Length = x, breadth = y$
Area of the floor $= xy m^2$
Area of four walls $= 2h(l + b) = 2z (x +y) = 2zx + 2yz$

Multiple choice maths how many squares area of rectangular paths comparing areas spaces and boundaries - 2

Find the area of a square inscribed in a circle of radius $\displaystyle 5\sqrt{2}$ cm (in $\displaystyle cm^{2}$)

  1. 75

  2. 100

  3. 125

  4. 150
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The diagonal of the square will be equal to the diameter of the circle.
So,diagonal of the square $ = 2 \times 5 \sqrt {2} = 10 \sqrt {2} $

Diagonal of a square $ = \sqrt {2} \times side $
So, $ 10 \sqrt {2} = \sqrt {2} \times side $
$ => Side  =  10  cm $

Area of the square $ = { side }^{ 2 } = { 10 }^{ 2 } = 100 $ sq cm

Multiple choice maths how many squares area of rectangular paths comparing areas spaces and boundaries - 2

What is the maximum area of a four-sided plane with a perimeter of $12$ inches?

  1. $8$ square inches
  2. $10$ square inches
  3. $6$ square inches
  4. $9$ square inches
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The area is maximum when the four-sided plane is a square.

The perimeter of a square $ = 4 \times$ side $= 12 $ inches
$ \Rightarrow$ side $= 3  $ inches

Now, area of the square $ =$ side $\times$ side $= 3 \times 3 = 9 $ square inches.

Multiple choice maths spaces and boundaries - 1 area of rectangular paths comparing areas spaces and boundaries - 2

Three coins of the same size (radius $1cm$) are placed on table such that each of them touches the other two. The area enclosed by the coins is:

  1. $\left( \cfrac { \pi }{ 2 } -\sqrt { 3 } \right) { cm }^{ 2 }$
  2. $\left( \sqrt { 3 } -\cfrac { \pi }{ 2 } \right) { cm }^{ 2 }$
  3. $\left(2 \sqrt { 3 } -\cfrac { \pi }{ 2 } \right) { cm }^{ 2 }$
  4. $\left(3 \sqrt { 3 } -\cfrac { \pi }{ 2 } \right) { cm }^{ 2 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Radius of each coin $=1\ cm$

With all the three centres an equilateral triangle of side 2 cm is formed.
Area enclosed by coind $=$ Area of equilateral triangle $-3\times$ Area of sector of angle $60^{o}$
                                         $=\dfrac{\sqrt3}{2}(2)^2-3\times\dfrac{60}{360}\times\pi(1)^2$
                                         $=\dfrac{\sqrt3}{4}\times4-3\times\dfrac{1}{6}\times\pi$
                                          $=\left(\sqrt3-\dfrac{\pi}{2}\right)\ cm^2$