Question

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

Brainly stars, best user, genius, MOD

Please answer this.

Amansharma sir

Tenny Sir

Prothala

Lasya

Aryan 0123

Mathdude500

etc... ​

Answer 1

Answer:

600 degree to be exact then I am going to be very helpful for

Answer 2

Given :

In a circle with centre O and radius 1 cm, an arc AB makes an angle 60° at O.

So,

OA = OB = 1 cm

$\angle \: AOB = {60}^{ \circ} \: = \theta \:$

Area of region AOB is = R

Again,

C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R

We have to find :

• The length of OC

Solution :

In triangle OCD =>

${\begin{array}{cc} OC = OD \\ \\ \angle \: COD = \angle \: DOC = {60}^{ \circ} \\ \\ \therefore \: \angle \: OCD = \angle \: ODC \end{array}}$

So it is an equilateral triangle.

$\orange{ \boxed{\boxed{\begin{array}{cc} \sf \: area \: of \: an \: equilateral \: triangle \: \: is \: : \\ \\ \Delta_{COD} = \frac{ \sqrt{3} }{4} \times {OC}^{2} \end{array}}}}$

Again,

$\pink{ \boxed{\boxed{\begin{array}{cc} \sf \: area \: of \: region \: AOB \: \:i s : \\ \\ R = \frac{ \theta}{ {360}^{ \circ} } \times \pi {(OA)}^{2} \\ \\ = \frac{60}{360} \times \pi \times {1}^{2} \\ \\ = \frac{\pi}{6} \: {cm}^{2} \end{array}}}}$

But, according to the question,

${ \boxed{\boxed{\begin{array}{cc}\Delta_{COD} = \frac{R}{2} \\ \\ \implies \: \frac{ \sqrt{3} }{4} {(OC)}^{2} = \frac{ \frac{\pi}{6} }{2} = \frac{\pi}{6\times 2} \\ \\ \implies \: OC {}^{2} = \frac{\pi\times 4}{6 \times 2 \times \sqrt{3}} \\ \\ \implies \: {OC}^{2} = 0.6045 \: \\ \\ \implies \: OC = \sqrt{0.6045} \\ \\ \therefore \: OC = 0.777 \: \: cm\end{array}}}}$

So length of OC = 0.777 cm.