AP State Syllabus AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions InText Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 13th Lesson Geometrical Constructions InText Questions

**Try These**

Question

Observe the sides, angles and diagonals of quadrilateral BEFD. Name the figures given below and write properties of figures. [Page No. 283]

Solution:

In fig. (1)

\(\overrightarrow{\mathrm{BF}}\) is the bisector of ∠B and ∠F.

In quad BEFD

BE = BD = DF = EF

It is a rhombus

In fig. (2)

BD = BE

FD = FE

∴ BEFD is a kite.

BF is bisector of ∠B and ∠F.

**Try This**

Question

Draw a circle, identify a point on it. Cut arcs on the circle with the length of the radius in succession. How many parts can the circle be divided into ? Give reasons. [Page No. 284]

Solution:

Let P be the centre of the circle.

A is any point on its circumference.

It can be divided into 2π parts

∴ \(\frac{\text { Circumference }}{\text { Radius }}=\frac{2 \pi r}{r}=2 \pi\)

**Think, Discuss and Write**

Question

Can you construct a triangle ABC with BC = 6 cm, ∠B = 60° and AB + AC = 5 cm? If , not, give reasons. (Page No. 286)

Solution:

We can’t construct a triangle with measures ∠B = 60°; BC = 6 cm and AB + AC = 5 cm.

∵ AB + AC < BC

Sum of any two sides of a triangle must be greater than the third side.

**Think, Discuss and Write**

Question

Can you construct the triangle ABC with the same measures by changing the base angle ∠C instead of ∠B ? Draw a rough sketch and construct it.

BC = 4.2cm. ∠C = 30°, AB – AC = 1.6 cm (Page No. 287)

Solution:

Steps of construction:

- Construct ΔBCD where BC = 4.2 cm and ∠C = 30° and AC – AB = 1.6 cm.
- Draw perpendicular bisector of BD which meets \(\overline{\mathrm{CD}}\) produced at A.
- Join B, D.
- ΔABC is the required triangle.

**Try These**

Can you draw the triangle with the same measurements as shown in the figure in alternate way ? (Page No. 289)

[Measurements : ∠B = 6Q°, ∠C = 45° and AB + BC + CA =11 cm]

[Hint: Take ∠YXL = 60°/2 = 30° and ∠XYM = 45°/2 = 22 \(\frac { 1 }{ 2 }\) ]

Solution:

→ Draw XY = 11 cm [AB + BC + CA = 11 cm]

Construct ∠YXP = 30° at X \(\left[\frac{B}{2}=\frac{60^{\circ}}{2}=30^{\circ}\right]\)

Construct ∠XYQ = 22\(\frac { 1 }{ 2 }\) at Y \(\left[\frac{\mathrm{C}}{2}=\frac{45^{\circ}}{2}=22 \frac{1}{2}^{\circ}\right]\)

→ \(\overrightarrow{\mathrm{XP}}\) and \(\overrightarrow{\mathrm{YQ}}\) meet at A.

→ At A, draw \(\overrightarrow{\mathrm{AB}}\) such that ∠XAB = 30° where B is a point on XY.

→ Also draw \(\overrightarrow{\mathrm{AC}}\) such that ∠YAC = 22\(\frac{1}{2}^{\circ}\) where C is a point on XY.

→ Δ ABC is the required triangle.

**Try These**

Question

What happens if the angle in the. circle segment is right angle ? What kind of segment do you obtain ? Draw the figure and give reason. [Page No. 290]

Solution:

If the angle in the circle segment is right angle i.e., 90°, then the angle subtended by it at the centre is 2 x 90° = 180°

Thus the line segment becomes the diameter and the circle segment becomes the semi-circle.