Question

Draw the figure below according to the given specifications:

Answer 1

The given figure is:

We know that sum of interior angles on the same side of the transversal are supplementary.

∴∠A + ∠B = 180°

⇒ 50° + ∠B = 180°

⇒ ∠B = 180° − 50° = 130°

The given rhombus can be drawn as follows:

1) Draw a line segment AB of length 5 cm.

2) At point A, draw ∠TAB = 50°.

3) Taking A as the centre and 5 cm as the radius, draw an arc on ray AT. Name the point of intersection as D.

4) At point B, draw ∠UBA = 130°.

5) Taking B as the centre and 5 cm as the radius, draw an arc on ray BU. Name the point of intersection as C.

6) Join CD.

Quadrilateral ABCD is the required rhombus.

The incircle of this rhombus can be made as follows:

1) Draw angle bisectors of ∠A and ∠B. Let them intersect at point O.

2) With O as the centre, draw OK perpendicular to AB. Name the point where it intersects AB as P.

3) With O as centre and OP as the radius, draw a circle. The circle so drawn touches all the sides of the triangle.

The given figure shows the required incircle of the given rhombus.