CAT 2025 Lesson : Quadrilaterals - Square

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

$\triangle$APB is a $45^{\mathrm{o}} - 45^{\mathrm{o}} - 90^{\mathrm{o}}$ triangle. ∴ BP $= \dfrac{5}{\sqrt{2}}$ $\triangle$BQC is a $45^{\mathrm{o}} - 45^{\mathrm{o}} - 90^{\mathrm{o}}$ triangle. ∴ BQ $= \dfrac{3}{\sqrt{2}}$ PQ $=$ BP – BQ $= \dfrac{5}{\sqrt{2}} - \dfrac{3}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \sqrt{2}$ The same applies for other sides of PQRS, which is a square.

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

Answer: $2$

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 $=$ ?

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

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 $= R = \dfrac{1}{2}$ OA $= \dfrac{ \text {Diagonal}}{2}= \dfrac{\sqrt{2}}{2}$ $r =$ OA – $R = \dfrac{\sqrt{2}}{2} - \dfrac{1}{2} = \dfrac{\sqrt{2 - 1}}{2}$

$R : r = \dfrac{1}{2} :\dfrac{\sqrt{2 - 1}}{2} =\dfrac{1}{\sqrt{2 - 1}} = \dfrac{1}{\sqrt{2 - 1}} \times \dfrac{\sqrt{2 + 1}}{\sqrt{2 + 1}} : 1$

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

From a square, the $4$ 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 $= 1$ Then, BQ $=$ BR $= \dfrac{1}{\sqrt{2}}$ Area of 4 triangles $= 4 \times \dfrac{1}{2}\times \dfrac{1}{\sqrt2}\times \dfrac{1}{\sqrt2} = 1$ Area of Square $= (\dfrac{1}{\sqrt2} + 1 + \dfrac{1}{\sqrt2})^2 = ( 1 + \sqrt{2})^2 = 3 + 2\sqrt{2}$ Ratio of cut to remaining $= (3 + 2\sqrt{2} - 1) : 1$ =$2(1+\sqrt{2}):1$

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