The figure is made up of an isosceles triangle and a semi-circle. Find the perimeter of the figure.

2 (7 - 2 π) c m .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(7 + 2 π) c m .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 (7 + 2 π) c m .

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(7 - 2 π) c m .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $2 (7 + 2 π) c m .$
Perimeter of semicircular arc = $π \times \frac{d}{2}$
$= π \times \frac{8}{2}$
$= 4 π c m$

Total Perimeter = Perimeter of semi-circular arc $+$ Sum of outer sides of triangle $A B C$
Total Perimeter $= l (B O C) + C A + A B$
$= 7 + 7 + 4 π$
$= 14 + 4 π$
$= 2 (7 + 2 π)$ cm

Suggest Corrections

Similar questions

Q. Two waves passing through a region are represented by

y_{1} = 5 m m sin [(2 π c m^{- 1}) x - (50 π s^{- 1}) t]

and

y_{2} = 10 m m sin [(π c m^{- 1}) x - (100 π s^{- 1}) t]

Find the vertical displacement of a particle at x = 1 cm at time t = 5.0 ms

Q. The figure below is made up of an isoceles triangle and a semi-circle. Find the periemeter of the figure

Two concentric coils each of radius equal to 2 $π$ cm are placed right angles to each other. If 3A and 4A are the currents flowing through the two coils respectively. The magnetic induction (in Wb m-2) at the center of the coils will be

Q. Two concentric coils, each of radius equal to

2 π cm

, are placed at right angles to each other. Currents of

3 A

and

4 A

, respectively, are flowing through the two coils. The magnetic induction, in

Wb m^{- 2}

, at the center of the coils will be

[μ_{0} = 4 π \times 10^{- 7} Wb

({A m}^{- 1})]

Q. In figure, if triangle

A B C

is an isosceles triangle of perimeter

20

, calculate the approximate area of the circle with center O.

Join BYJU'S Learning Program

The figure is made up of an isosceles triangle and a semi-circle. Find the perimeter of the figure.

The correct option is C 2(7+2π)cm.Perimeter of semicircular arc = π×d2=π×82=4π cmTotal Perimeter = Perimeter of semi-circular arc + Sum of outer sides of triangle ABCTotal Perimeter =l(BOC)+CA+AB=7+7+4π=14+4π=2(7+2π) cm