JIPMAT 2024 (QA) - A circle, square, equilateral triangle and regular pentagon have the same areas. Arrange the following in ascending order based on their perimeters.(A) Circle(B) Square(C) Equilateral triangle(D) Regular pentagon | PYQs + Solutions

JIPMAT 2024

Geometry

Quadrilaterals

Hard

A circle, square, equilateral triangle and regular pentagon have the same areas. Arrange the following in ascending order based on their perimeters.
(A) Circle
(B) Square
(C) Equilateral triangle
(D) Regular pentagon

(C), (D), (B), (A)

(A), (B), (C), (D)

(A), (D), (B), (C)

(D), (B), (C), (A)

Correct Option: 3

Let me solve this simply.

1) Let's say Area =

A

for all shapes.

2) Finding perimeters:

\newline

* Circle (A):

\newline

Area =

πr^2

\newline

radius =

r = \sqrt{\frac{A}{π}}

\newline

perimeter =

2πr = 2\sqrt{πA}

≈

3.5\sqrt{A}

* Square (B):

\newline

Area =

s^2

\newline

side =

s = \sqrt{A}

\newline

perimeter =

4s = 4\sqrt{A}

* Equilateral Triangle (C):

\newline

Area =

\frac{\sqrt{3}}{4}s^2

\newline

side =

s = 2\sqrt{\frac{A}{\sqrt{3}}}

\newline

perimeter =

3s = 6\sqrt{\frac{A}{\sqrt{3}}}

≈

4.6\sqrt{A}

* Regular Pentagon (D):

\newline

perimeter ≈

3.8\sqrt{A}

\newline

3) Arranging perimeters from smallest to largest:

\newline

Circle (

3.5\sqrt{A}

) < Pentagon (

3.8\sqrt{A}

) < Square (

4\sqrt{A}

) < Triangle (

4.6\sqrt{A}

)

Therefore, the order is (A), (D), (B), (C).