CAT 2025 Lesson : Circles - Circumference & Area

Where O is the centre of the circle, if the length of the minor arc AB $= 22$ cm and $\angle$AOB $= 90^\mathrm{o}$, then what is the area of the shaded region (in cm$^{2}$)?

Solution

Let $r$ be the radius of the circle.

Circumference of the minor arc $= \dfrac{\theta^\mathrm{o}}{360^\mathrm{o}} 2 \pi r = 22$

⇒ $\dfrac{90}{360} \times 2 \times \dfrac{22}{7} \times r =22$
⇒ $r = 14$

Area of sector OAB $= \dfrac{\theta^\mathrm{o}}{360^\mathrm{o}} \pi r^{2} =$ $\dfrac{90}{360} \times \dfrac{22}{7} \times 14^{2} = 154$ cm$^{2}$ -----(1)

Area of $\triangle$AOB $= \dfrac{1}{2} bh = \dfrac{1}{2} \times 14 \times 14 = 98$ cm$^{2}$ -----(2)

Area of shaded region $= (1) - (2) = 154 - 98 = 56$ cm$^{2}$

Answer: $56$

In the following figure, two circles, each with radius of $7$ cm and centres at A and B are drawn such that each circle passes through the centre of the other circle. What is the area of the shaded region?
(1) $\dfrac{7}{3}\times (22-7\sqrt{3})$      (2) $\dfrac{7}{3}\times (44-21\sqrt{3})$     (3)$\dfrac{7}{2}\times (33-14\sqrt{3})$     (4) $\dfrac{7}{6}\times (88-21\sqrt{3})$

Solution

The shaded regions in the two figures below includes the regions for which we need to find the area. However, each of these figures additionally include the rhombus ACBD. Therefore, we can add up the area of the two sectors and then subtract twice the area of the rhombus ACBD.


In the figure below, radius $=$ AB $=$ AC $=$ AD $=$ BC $=$ BD $= 7$cm. Therefore, $\triangle$ACB and $\triangle$ADB are two congruent equilateral triangles. Also, the $2$ sectors – Sector ACD in circle with centre A and Sector BCD in circle with centre B are equal in area. The angle formed by these sectors at the centre is $120^\mathrm{o}$.


Sum of area of sectors ACD and BCD $= 2 \times \dfrac{\theta^\mathrm{o}}{360^\mathrm{o}} \pi r^{2} =$ $2 \times \dfrac{120}{360} \times \dfrac{22}{7} \times 7^{2} = \dfrac{308}{3}$ cm$^{2}$

Area of ACBD $= 2 \times$ Area ( $\triangle$ ACB) $= 2 \times \dfrac{\sqrt{3}}{4} \times 7^{2} = \dfrac{49\sqrt{3}}{2}$

Area of required shaded region = Sum of Areas of $2$ sectors $-$ $2 \times$ Area (ACBD) $= \dfrac{308}{3} - 2 \times \dfrac{49\sqrt{3}}{2} =$ $\dfrac{7}{3}(44 - 21\sqrt{3})$

Answer: (2) $\dfrac{7}{3}(44 - 21\sqrt{3})$

Three horses are grazing within a semi-circular field. In the diagram given below, AB is the diameter of the semi-circular field with centre at O. Horses are tied up at P, R and S such that PO and RO are the radii of semi-circles with centres at P and R respectively, and S is the centre of the circle touching the two semi-circles with diameters AO and OB. The horses tied at P and R can graze within the respective semi-circles and the horse tied at S can graze within the circle centred at S. The percentage of the area of the semi-circle with diameter AB that cannot be grazed by the horses is nearest to
[CAT 2003L]

(1) $20$                     (2) $28$                     (3) $36$                     (4) $40$                    

Solution

Let the radius of each smaller semicircles centred at P and R be $1$ each. Therefore, in the adjacent figure, when OS is extended to intersect the larger semi-circle at C, we get the radius OC $= 2$. Let the radius of the smaller semi-circle be $r$. In right angled $\triangle$POS, OS $=$ OC $-$ SC $=$ $2 - r$ PS $=$ $1 + r$ and OP $=$ $1$ PO$^{2}$ + OS$^{2} =$ PS$^{2}$ $1 + (2 - r)^{2} = (r + 1)^{2}$⇒ $1 + 4 - 4r + r^{2} =$ $r^{2} + 2r + 1$ ⇒ $6r = 4$ ⇒ $r = \dfrac{2}{3}$ Total area of semi-circle with diameter AB $= \dfrac{1}{2} \times \pi 2^{2} = 2 \pi$

Area of $2$ smaller semi-circles $= 2 \times \dfrac{1}{2} \times \pi 1^{2} = \pi$

Area of smaller circle $= \pi \times \left( \dfrac{2}{3} \right)^{2} =$ $\dfrac{4}{9}\pi$
Percentage of ungrazed area $= \dfrac{2 \pi - \pi - \dfrac{4}{9}\pi}{2 \pi} =$ $\dfrac{\dfrac{5}{9}\pi}{2 \pi} =$ $\dfrac{5}{18}$ ~ $0.272727$ ~ $28\%$

Answer: (2) $28$