Tag: area of rectangular paths

Questions Related to area of rectangular paths

Calculate area of the shape obtained by joining an equilateral triangle to one of the end of rectangle and a semicircle to the other end of rectangle. The sides of equilateral triangle is $3$ cm and one of the sides of rectangle is $4$ cm.

  1. $28.12\ cm^{2}$

  2. $30.02\ cm^{2}$

  3. $35.02\ cm^{2}$

  4. $40.02\ cm^{2}$


Correct Option: B

Consider a rectangle of perimeter $24\ cm$ with sides $a$ and $b$. Then the total number of square grids formed in that rectangle is given by

  1. $a\times b$

  2. $a+b$

  3. $a-b$

  4. $\dfrac{a}{b}$


Correct Option: A
Explanation:
Area of rectangle$=a\times b$
Area of $1$ square grid$=1 cm^2$
No. of square grids$=\cfrac { a\times b }{ 1 } =a\times b$

Form a rectangle with $8$ square grids where each square grid measure $1\ cm^{2}$. Find the total area of the rectangle.

  1. $8\ cm^{2}$

  2. $12\ cm^{2}$

  3. $16\ cm^{2}$

  4. $20\ cm^{2}$


Correct Option: A
Explanation:

Area of rectangle$=8\times $ area of $1$ square grid$=8\times 1=8 cm^2$

Let there be a rectangle of area $24\ cm^{2}$. Then find the total number of square grids made in this rectangle.(Each square grid measures $1\ cm^{2}$)

  1. $3$

  2. $6$

  3. $12$

  4. $24$


Correct Option: D
Explanation:

Number of square grids$=\cfrac { Area\quad of\quad rectangle }{ area\quad of\quad square\quad grid } =\cfrac { 24 }{ 1 } =24$

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. $\left(3 \sqrt { 3 } -\cfrac { \pi }{ 2 } \right) { cm }^{ 2 }$


Correct Option: B
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 $60^{o}$
                                         $=\dfrac{\sqrt3}{2}(2)^2-3\times\dfrac{60}{360}\times\pi(1)^2$
                                         $=\dfrac{\sqrt3}{4}\times4-3\times\dfrac{1}{6}\times\pi$
                                          $=\left(\sqrt3-\dfrac{\pi}{2}\right)\ cm^2$