**Telangana**** SCERT Solution Class IX (9) Math Chapter 13 Geometrical Constructions Exercise 13.2**

**(1.) Construct ΔABC in which BC = 7 cm, ****∠****B = 75° and AB + AC = 12 cm**

**Solution:**

**(1)** Given, BC = 7 cm

B = 75^{o}

AB + Ac = 12 cm

We have AB + AC = 12 cm

Now, extend BA upto D so that

BD = 12 cm

∴ BD = AB + AD = 12 cm

∴ AD = AC

Since A is equidistant from point C & D draw ⊥ bisects of /CD locate A on BD

∴ △ABC is the required triangle.

**(3.) Construct Δ XYZ in which ****∠****Y = 30° , ****∠****Z = 60° and XY + YZ + ZX = 10 cm.**

**Solution:**

Given, XYZ is Δ

Y = 30^{o}

Z = 60^{o}

XY + YZ + XZ = 10 cm

Step 1: – Draw line segment AB 10 cm = XY + YZ + XZ

Step 2: – Make <Y = 30^{o} from point A , <LAB = 30^{o}

Step 3: – Make <Z = 60^{o} from point B, <MBA = 60^{o}

Step 4: – Bisect <CAB & <NBA, they must each other, thus at point X.

Step 5: – Make perpendicular bisect of XA which intersect AB at point Y

Step 6: – Make perpendicular bisects of XB which intersect AB at point Z

Step 7: – Join XY & XZ

∴ △XYZ is the required triangle.

**(4.) Construct a right triangle whose base is 7.5 cm. and sum of its hypotenuse and other side is 15 cm.**

**Solution:**

Given,

ABC is a right angled triangle at point A.

Sum of hypotenuse and height is 15 cm

Draw base AB = 7.5 cm line segment.

Base is 7.5 cm

∴ BC + AC = 15cm [hypotenuse + height]

Draw AD = 15cm on perpendicular AC produced.

∴ AD = AC + CD = 15 cm

∴ BC = CD

∴ C is equal font from point O & point B

∴ Perpendicular bisector of BD cu AD at point C to locate point C

∴ △ABC is the required triangle

**(5.) Construct a segment of a circle on a chord of length 5cm. containing the following angles.**

**(i) 90° (ii) 45° (iii) 120°**

**Solution:**

**(i)** 90^{o} angle

Step 1: – Draw line segment AB

Step 2: – Put midpoint as O therefore ∠AOB = 180^{o} sine AB is a straight line

Step 3: – Draw a circle of radius OA around AB

Step 4: – Join AC and AB on any point circle ∠ACB =90^{o}

[∵ Angle subtended by the chord at centre is twice the angle subtended by it on the remaining segment of the circle.]

**(ii)** 45^{o} angle

Step 1: – Take line segment AB = 5 cm

Step 2: – Draw 45^{o} angle on sash end of segment AB therefore ∠AOB = 90^{o}

Step 3: – AO and BO meets at O centre of circle

Step 4: – Draw a circle taking O as centre and OA is radius

Step 5: – Join AC and BC at any point on the major segment.

∴ ACB = 45^{o} = ½ ∠AOB = ½ x 90^{o}

[∴angle scented by a chord on centre of circle is twice the angles subtend by the chord or the remaining circle]

**(iii)** 120^{o} angle

Step 1: – Draw a segment AB = 5cm

Step 2: – Draw 30^{o} angles at each ends of A & B of AB

∴ ∠AOB = 120^{o}

Step 3: – Join AO & BO to get the centre of circle

Step 4: – Draw a circle of radius OA around segment AB

Step 5: – Join AC and BC at any point on the minor segment to get 120^{o}

∴ ∠ACB = 120^{o}

∠ADB = ½ ∠AOB = 60^{o} [∵ angle subtended by the chord AB at centre O is twice the angle substance by it on the major segment]

Now, ABCD is a cyclic quadrilateral.

∴ Opposite angles are supplementary

∴ ∠ACB + ∠ADB = 180^{o}

Or, ∠ACB = 180^{o} – 60^{o}

= 120^{o}

Hence, angle subtended by chord AO on the minor segment is 120^{o}