In a circle with center $O$ and radius $1 \mathrm{~cm}$ , an arc $A B$ makes an angle $60$ degrees at $O$ . Let $R$ be the region bounded by the radii $\mathrm{OA}, \mathrm{OB}$ and the arc $AB$ . If $C$ and $D$ are two points on $OA$ and $OB$ , respectively, such that $\mathrm{OC}=\mathrm{OD}$ and the area of triangle $OCD$ is half that of $R$ , then the length of $OC$ , in cm , is

$\left(\frac{\pi}{4}\right)^{\frac{1}{2}}$

$\left(\frac{\pi}{3 \sqrt{3}}\right)^{\frac{1}{2}}$

$\left(\frac{\pi}{6}\right)^{\frac{1}{2}}$

$\left(\frac{\pi}{4 \sqrt{3}}\right)^{\frac{1}{2}}$

✅ Correct Option: 2

We have a circle with center O and radius 1 cm. Arc AB creates a 60° angle at the center O. This means we're dealing with a sector - think of it like a "slice of pie" where the crust is the arc AB and the two edges are the radii OA and OB.

Points C and D lie on radii OA and OB respectively, with OC = OD.

Region R is a sector with central angle 60°.

Area of sector = $\frac{\text{central angle}}{360°} \times \pi r^2$

Area of R = $\frac{60°}{360°} \times \pi (1)^2$

$= \frac{1}{6} \times \pi$

$= \frac{\pi}{6}$

The problem states that the area of triangle OCD is half that of region R.

Area of triangle OCD = $\frac{1}{2} \times \frac{\pi}{6}$

$= \frac{\pi}{12}$

Since C is on OA and D is on OB, the angle COD is the same as the original central angle = 60°.

For triangle OCD:

OC = OD (given)
Angle COD = 60°
Area = $\frac{1}{2} \times OC \times OD \times \sin(60°)$

Since OC = OD:

Area = $\frac{1}{2} \times OC^2 \times \sin(60°)$

Since $\sin(60°) = \frac{\sqrt{3}}{2}$ :

Area = $\frac{1}{2} \times OC^2 \times \frac{\sqrt{3}}{2}$

$= \frac{OC^2 \sqrt{3}}{4}$

We know the area of triangle OCD is $\frac{\pi}{12}$ :

$\frac{OC^2 \sqrt{3}}{4} = \frac{\pi}{12}$

$OC^2 \sqrt{3} = \frac{\pi}{12} \times 4$

$= \frac{\pi}{3}$

$OC^2 = \frac{\pi}{3\sqrt{3}}$

$OC = \sqrt{\frac{\pi}{3\sqrt{3}}}$

Therefore: $OC = \sqrt{\frac{\pi}{3\sqrt{3}}}$ cm

CAT 2023 Slot 3

Let $\triangle ABC$ be an isosceles triangle such that $AB$ and $AC$ are of equal length. $AD$ is the altitude from $A$ on $BC$ and $BE$ is the altitude from $B$ on $AC$ . If $AD$ and $BE$ intersect at $O$ such that $\angle AOB = 105^\circ$ , then $AD/BE$ equals

CAT 2023 Slot 1

In a right-angled triangle $\triangle A B C$ , the altitude $A B$ is $5 \mathrm{~cm}$ , and the base $B C$ is $12 \mathrm{~cm}$ . $P$ and $Q$ are two points on $B C$ such that the areas of $\triangle \mathrm{ABP}, \triangle \mathrm{ABQ}$ and $\triangle \mathrm{ABC}$ are in arithmetic progression. If the area of $\triangle \mathrm{ABC}$ is $1.5$ times the area of $\triangle \mathrm{ABP}$ , the length of PQ in cm , is

CAT 2024 Slot 3

The midpoints of sides $AB$ , $BC$ , and $AC$ in $\triangle ABC$ are $M$ , $N$ , and $P$ , respectively. The medians drawn from $A$ , $B$ , and $C$ intersect the line segments $MP$ , $MN$ and $NP$ at $X$ , $Y$ , and $Z$ , respectively. If the area of $\triangle ABC$ is $1440$ sq cm, then the area, in sq cm, of $\triangle XYZ$ is

CAT 2024 Slot 1

ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of in circle of $\triangle$ ADE is

CAT 2018 Slot 1

Given an equilateral triangle $\mathrm{T} 1$ with side $24 \mathrm{~cm}$ , a second triangle $\mathrm{T} 2$ is formed by joining the midpoints of the sides of $\mathrm{T} 1$ . Then a third triangle $\mathrm{T} 3$ is formed by joining the midpoints of the sides of $\mathrm{T} 2$ . If this process of forming triangles is continued, the sum of the areas, in sq cm , of infinitely many such triangles $\mathrm{T} 1, \mathrm{~T} 2, \mathrm{~T} 3, \ldots$ will be

CAT 2017 Slot 1

From a triangle $ABC$ with sides of lengths $40$ ft, $25$ ft and $35$ ft, a triangular portion $GBC$ is cut off where $G$ is the centroid of $ABC$ . The area, in sq ft, of the remaining portion of triangle $ABC$ is

Solution

Related questions:

Keyboard Shortcuts

CAT Past Year Questions (Topic-Wise)