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CAT 2025 Lesson : Quadrilaterals - Square

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3.5 Square

Square is a rectangle and a rhombus, i.e., all sides are equal and all angles equal 90o\bm{90^{\mathrm{o}}}.

1) Diagonals are equal and bisect each other at 90o90^{\mathrm{o}}.

2) The diagonals bisect the angles of the square.

3) Where the length of a square's side is ss,

Diagonal =d=2×s= d = \sqrt2 \times s

Perimeter =4s= 4s

Area =s2=12×d2= s^2 = \dfrac{1}{2} \times d^2

4) For a given perimeter of a 4-sided figure, a square maximises the area.

And, for a given area of a 4-sided figure, a square minimises the perimeter.

5) A square can have an inscribed circle as well as a circumscribed circle. O is the incentre and the circumcentre.

6) Inradius =s2= \dfrac{s}{2} and Circumradius =12=2s2= \dfrac{1}{2}= \dfrac{\sqrt{2}s}{2}

7) If the mid-points of adjacent sides of a square are joined, we get another square.

Example 16

In a rectangle ABCD where AB = 5 and BC = 3, the internal angle bisectors for the 4 internal angles are drawn. The angle bisectors from A & B, B & C, C & D and D& A meet at P, Q, R and S respectively. What is the area of PQRS?

Solution

△\triangleAPB is a 45o−45o−90o 45^{\mathrm{o}} - 45^{\mathrm{o}} - 90^{\mathrm{o}} triangle. ∴ BP =52= \dfrac{5}{\sqrt{2}}

△\triangleBQC is a 45o−45o−90o 45^{\mathrm{o}} - 45^{\mathrm{o}} - 90^{\mathrm{o}} triangle. ∴ BQ =32= \dfrac{3}{\sqrt{2}}

PQ == BP – BQ =52−32=22=2= \dfrac{5}{\sqrt{2}} - \dfrac{3}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \sqrt{2}

The same applies for other sides of PQRS, which is a square.

Area of Square =PQ2=(2)2=2= \mathrm {PQ^2} = (\sqrt{2})^2 = 2

Answer: 22


Example 17

From the 4 vertices of a square, 4 arcs of radii R are cut such that they touch each other along with a side of the square. A fifth smaller circle of radius r is drawn in the area enclosed by the arcs such that the smaller circle touches each of the arcs at exactly one point. Then, R : r == ?

(1)(2−1):1(2)(2+1):1(1) (\sqrt{2} - 1 ) : 1 (2)(\sqrt{2} + 1 ) : 1
(3)2:(2+1)(4)2:(2−1) (3) \sqrt{2} : (\sqrt{2} + 1 ) (4) \sqrt{2} : (\sqrt{2} - 1 )

Solution

As we are required to find the ratio, we can assume the length of a side of the square to be 1.

Two arcs would meet at the mid-point of a side of the square.

∴ Radius of arc =R=12= R = \dfrac{1}{2}

OA =Diagonal2=22= \dfrac{ \text {Diagonal}}{2}= \dfrac{\sqrt{2}}{2}

r=r = OA – R=22−12=2−12R = \dfrac{\sqrt{2}}{2} - \dfrac{1}{2} = \dfrac{\sqrt{2 - 1}}{2}

R:r=12:2−12=12−1=12−1×2+12+1:1 R : r = \dfrac{1}{2} :\dfrac{\sqrt{2 - 1}}{2} =\dfrac{1}{\sqrt{2 - 1}} = \dfrac{1}{\sqrt{2 - 1}} \times \dfrac{\sqrt{2 + 1}}{\sqrt{2 + 1}} : 1

Answer: (22) (2+1):1(\sqrt{2} + 1 ) : 1


Example 18

From a square, the 44 corners are cut to form a regular octagon. What is the ratio of the area cut to the area remaining?

Solution

From the 4 corners, 4 isosceles right-angled triangles are cut such that their diagonals are sides of the octagon.

Let PQ == QR =1= 1

Then, BQ == BR =12= \dfrac{1}{\sqrt{2}}

Area of 4 triangles =4×12×12×12=1= 4 \times \dfrac{1}{2}\times \dfrac{1}{\sqrt2}\times \dfrac{1}{\sqrt2} = 1

Area of Square =(12+1+12)2=(1+2)2=3+22= (\dfrac{1}{\sqrt2} + 1 + \dfrac{1}{\sqrt2})^2 = ( 1 + \sqrt{2})^2 = 3 + 2\sqrt{2}

Ratio of cut to remaining =(3+22−1):1= (3 + 2\sqrt{2} - 1) : 1 =2(1+2):1 2(1+\sqrt{2}):1


Answer: 2(1+2):12(1+\sqrt{2}):1


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