AP State Syllabus AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 13th Lesson Geometrical Constructions Exercise 13.2

Question 1.

Construct AABC in which BC = 7 cm,∠B = 75° and AB + AC =12 cm.

Solution:

A.

Steps:

- Draw a line segment BC = 7 cm.
- Erect ∠B = 75°
- MarkapointDon \(\overrightarrow{\mathrm{BX}}\) suchthat BD = AB + AC.
- Join D, C and draw the perpendicular bisector of \(\overline{\mathrm{CD}}\) meeting BD at A.
- Join A to C to form the ΔABC.

Question 2.

Construct ΔPQR in which QR = 8 cm, ∠B = 60° and AB – AC = 3.5 cm.

Read ∠Q = 60°and PQ – PR = 3.5 cm

Solution:

A.

Steps: I

- Draw QR = 8 cm.
- Construct ∠RQX = 300 at Q.
- Mark a point S on \(\overrightarrow{\mathrm{QX}}\) such that QS = PQ – PR = 3.5 cm.
- Join S, R and draw the perpendicular bisector to \(\overline{\mathrm{QR}}\) meeting \(\overrightarrow{\mathrm{QX}}\) at P.
- Join P, R to form the ΔPQR.

Question 3.

Construct ΔXYZ in which ∠Y = 30 °; ∠Z = 60 ° and XY + YZ + ZX = 10 cm.

Solution:

A.

Steps:

- Draw a line segment AB = XY + YZ + ZX = 10 cm.
- Construct ∠BAP = \(\frac { 1 }{ 2 }\) ∠Y at A and ∠ABQ = \(\frac { 1 }{ 2 }\) ∠Z at B meeting at X.
- Draw the perpendicular bisectors to XA and XB meeting \(\overline{\mathrm{AB}}\) at Y and Z respectively.
- Join X to Y and Z to form the ΔXYZ.

Question 4.

Construct a right triangle whose base is 7.5 cm and sum of its hypotenuse and otherside is 15 cm.

Solution:

A.

Steps:

- Draw BC = 7.5 cm.
- Construct ∠CBX = 90°
- Mark a point D on \(\overrightarrow{\mathrm{BX}}\) such that BD = 15 cm.
- Join C, D. ‘
- Draw the perpendicular bisectors of \(\overline{\mathrm{CD}}\) meeting BD at A.
- Join A, C to form the ΔABC.

Question 5.

Construct a segment of a circle on a chord of length 5 cm containing the following angles i) 90° ii) 45° iii) 120°

Solution:

i) 90°

A.

Steps:

- Draw a rough sketch of ∠BAC = 90° and ∠BOC = 180°.
- Draw a line segment BC = 5 cm.
- Draw the perpendicular bisector of BC meeting \(\overline{\mathrm{BC}}\) at O
- Draw an arc of radius OB or OC with centre O.
- Mark any point A on the arc and join it with B and C.
- ∠BAC = 90°

ii) 45°

Steps:

- Draw a line segment BC = 5 cm.
- Construct ABOC such that BC = 5 cm, ∠B = 45° = ∠C.
- Draw a circle segment of radius OB or OC with centre ’O’.
- Mark any point A on the segment and join it with B and C.
- ∠BAC = 45°

iii) 120°

Steps:

- Draw a line segment AB = 5 cm. ,
- Construct ΔAOB in which ∠A = 30°; ∠B = 30°; AB = 5 cm.
- With ‘O’ as centre draw a circle segment.
- On the opposite side make any point C and join it with B and C.
- ∠ACB = 120°