Question

(iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its
perimeter to area.​

Answer 1

Answer:

34:35

Step-by-step explanation:

Perimeter = 2(5+3.5) = 17 cm

Area = 5x3.5 = 17.5 sqcm

Hence req ratio = 17:17.5 = 34:35

Answer 2

$\bf \red{Given}\begin{cases}&\sf{Length=\bf{5\:cm.}} \\ \\ &\sf{Breadth=\bf{3.5\:cm.}}\end{cases}$

To FinD:-

The ratio of its perimeter to area.

SolutioN:-

Diagram:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 5 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 3.5cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

The Perimeter :

We know that,

$\normalsize{\red{\underline{\boxed{\bf{Perimeter_{(rectangle)}=2(Length+Breadth)}}}}}$

where,

• Length = 5 cm
• Breadth = 3.5 cm

Putting the values,

$\normalsize\implies{\sf{Perimeter=2(5+3.5)}}$

$\normalsize\implies{\sf{Perimeter=2\times8.5}}$

$\normalsize\therefore\boxed{\mathfrak{\red{Perimter=17\:cm.}}}$

The Area :

We know that,

$\normalsize{\red{\underline{\boxed{\bf{Area_{(rectangle)}=Length\times\:Breadth}}}}}$

where,

• Length = 5 cm
• Breadth = 3.5 cm

Putting the values,

$\normalsize\implies{\sf{Area=5\times3.5}}$

$\normalsize\therefore\boxed{\mathfrak{\red{Area=17.5\:cm^2.}}}$

The ratio of Perimeter to Area:

$\normalsize\implies{\sf{\dfrac{Perimeter_{(rectangle)}}{Area_{(rectangle)}}=\dfrac{17}{17.5}}}$

By removing the point in 17.5,

$\normalsize\implies{\sf{\dfrac{Perimeter_{(rectangle)}}{Area_{(rectangle)}}=\dfrac{170}{175}}}$

Now we have to bring both the numerator and denominator in its simplest form,

$\normalsize\implies{\sf{\dfrac{Perimeter_{(rectangle)}}{Area_{(rectangle)}}=\dfrac{170\div5}{175\div5}}}$

$\normalsize\implies{\sf{\dfrac{Perimeter_{(rectangle)}}{Area_{(rectangle)}}=\dfrac{34}{35}}}$

$\normalsize\implies{\sf{Perimeter_{(rectangle)}:Area_{(rectangle)}=34:35}}$

$\normalsize\therefore\boxed{\mathfrak{\red{Perimter:Area=34:35.}}}$

The ratio of the Perimeter to the Area of the rectangle is 34 : 35 .