Question

find yhe area of rectangle whose length is x and breadth is 1/3 of it's length​

Answer 1

Answer:

$\sf{The \ area \ of \ rectangle \ is \ \dfrac{x^{2}}{3} \ unit^{2}}$

Given:

• Length (l)=x cm

• Breadth (b)=1/3 of length.

To find :

• Are of the rectangle.

Solution:

$\sf{According \ to \ the \ given \ condition. }$

$\sf{Breadth=\dfrac{x}{3}}$

$\boxed{\sf{Area \ of \ rectangle=Length\times \ Breadth}}$

$\sf{\therefore{Area \ of \ rectangle=(x)\times\dfrac{x}{3}}}$

$\sf{\therefore{Area \ of \ rectangle=\dfrac{x^{2}}{3}}}$

$\sf\purple{\tt{\therefore{The \ area \ of \ rectangle \ is \ \dfrac{x^{2}}{3} \ unit^{2}}}}$

$\sf\blue{\underline{\underline{Extra \ Information}}}$

$\sf{\leadsto{Perimeter \ of \ rectangle=Length\times \ Breadth}}$

$\sf{\leadsto{Length=\dfrac{Perimeter}{2}- \ Breadth}}$

$\sf{\leadsto{Breadth=\dfrac{Perimeter}{2}- \ Length}}$

$\sf{\leadsto{Area \ of \ rectangle=Length\times \ Breadth}}$

$\sf{\leadsto{Length=\dfrac{Area}{Breadth}}}$

$\sf{\leadsto{Breadth=\dfrac{Area}{Length}}}$

$\sf{\leadsto{Diagonal \ of \ rectangle=\sqrt{l^{2}+b^{2}}}}$