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 = 75o
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 = 30o
Z = 60o
XY + YZ + XZ = 10 cm
Step 1: – Draw line segment AB 10 cm = XY + YZ + XZ
Step 2: – Make <Y = 30o from point A , <LAB = 30o
Step 3: – Make <Z = 60o from point B, <MBA = 60o
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) 90o angle
Step 1: – Draw line segment AB
Step 2: – Put midpoint as O therefore ∠AOB = 180o 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 =90o
[∵ Angle subtended by the chord at centre is twice the angle subtended by it on the remaining segment of the circle.]
(ii) 45o angle
Step 1: – Take line segment AB = 5 cm
Step 2: – Draw 45o angle on sash end of segment AB therefore ∠AOB = 90o
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 = 45o = ½ ∠AOB = ½ x 90o
[∴angle scented by a chord on centre of circle is twice the angles subtend by the chord or the remaining circle]
(iii) 120o angle
Step 1: – Draw a segment AB = 5cm
Step 2: – Draw 30o angles at each ends of A & B of AB
∴ ∠AOB = 120o
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 120o
∴ ∠ACB = 120o
∠ADB = ½ ∠AOB = 60o [∵ 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 = 180o
Or, ∠ACB = 180o – 60o
= 120o
Hence, angle subtended by chord AO on the minor segment is 120o