Question

Draw an equilateral triangle of side 5 cm. Draw its inscribed circle. Measure the radius of the inscribed circle.

Answer 1

Hint: Take the triangle as ABC. Draw base BC as 5 cm and construct the equilateral triangle with side 5 cm. Take angle bisectors of B and C. Find the center of the circle, and with the center O, draw a circle inscribed in triangle ABC.

Complete step-by-step answer:
Let us name the equilateral triangle as ABC.
First draw the line segment of 5 cm. Let us take BC = 5 cm. Then with center as B and C, draw two arcs of 5 cm length. These arcs intersect at point A.
Thus we got an equilateral triangle ABC by joining the points AB and AC.
As it’s an equilateral triangle, we know all the sides are equal which is 5 cm. Similarly all the angles of triangle ABC is similar, i.e. the angle is $${60}^{\circ }.$$
Let us draw angle bisectors from B and C of the triangle and they meet at point O, i.e. angle bisectors of $$\mathrm{\angle }B$$ and $$\mathrm{\angle }C$$ intersect each other at O. From point O, we can say that $$OL\mathrm{\perp }BC$$, i.e. OL is perpendicular to the line segment BC.
Now with center O and radius as OL, draw a circle which touches the sides of $$\mathrm{\Delta }ABC.$$

Thus we got the required figure, where an equilateral triangle of side 5 cm is drawn and a circle is inscribed in it.
Thus if we measure the radius of the inscribed circle, OL = 1.4 cm.
Hence radius, OL = 1.4 cm.

Note: We can also take angle bisectors and A and C or even B and C. We will get the same center as given in the figure. Draw OL perpendicular to BC in order to get the radius, then the construction of circle becomes easier and the circle should touch the sides of the $$\mathrm{\Delta }ABC.$$