# Geometrical Constructions Solution of TS & AP Board Class 9 Mathematics

###### Exercise 13.1

Question 1.

Construct the following angles at the initial point of a given ray and justify the construction.

90o

Construction of angle of 90°

Steps of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as center and any radius, draw an arc, cutting OA at C.

Step 3: With center c and same radius (as in step 2) draw an arc cutting arc at D.

Step 4: With D as center and the same radius, draw an arc cutting the arc cutting at E.

Step 5: With D and E as centers and any convenient radius (more than  DE). Draw to two arcs intersecting at P.

Step 6: Join OP. Then ∠AOP = 90°

Justification: -

By construction, OC = CD = OD

Therefore, Î”OCD is an equilateral triangle. So, ∠COD = 60°

Again OD = DE = OE

Therefore, Î”ODE is also an equilateral triangle. So, ∠DOE = 60°

Since, OP bisects ∠DOE, so ∠POD = 30°.

Now,

∠AOP = ∠COD + ∠DOP

= 60° + 30°

= 90°

Question 2.

Construct the following angles at the initial point of a given ray and justify the construction.

45o

Construction of angle of 45°

Steps of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as center and any radius, draw an arc, cutting OA at B.

Step 3: With center B and same radius (as in step 2), cut the previous drawn arc at C.

Step 4: With C as center and the same radius, draw an arc cutting the arc drawn in step 2 cutting at D.

Step 5: With D and E as centers and any convenient radius (more than  DE). Draw to two arcs intersecting at E.

Step 6: Join OE. Then ∠AOE = 90°

Step 7: Draw the bisector ‘OF’ of ∠AOE. Then, ∠AOF = 45°

Justification: -

By construction, ∠AOE = 90° and OF is the bisector of ∠AOE.

Therefore,

∠AOF = ∠AOE

= 45°

Question 3.

Construct the following angles using ruler and compass and verify by measuring them by a protractor.

30o

Steps of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as centre and any radius, draw an arc, cutting OA at C.

Step 3: With centre C and same radius (as in step 2). Draw an arc, cutting the arc of step 2 in D.

Step 4: With C and D as centres, and any convenient radius (more than CD), draw two arcs intersecting at B. join OB. Then ∠AOB = 30°

Verification:

On measuring ∠AOB, with the protractor,

we find ∠ AOB = 30° .

Question 4.

Construct the following angles using ruler and compass and verify by measuring them by a protractor.

Steps of construction:

Step 1: Draw an angle AOB = 90°

Step 2: Draw the bisector OC of ∠AOB, then ∠AOC = 45°

Step 3:Bisect ∠AOC, such that ∠AOD = ∠COD = 22.5°

Thus ∠AOD = 22.5°

Verification:

On measuring ∠AOD, with the protractor,

we find ∠ AOD = 22.5° .

Question 5.

Construct the following angles using ruler and compass and verify by measuring them by a protractor.

15o

Steps of construction:

Step 5: Draw a ray OA.

Step 6: With its initial point O as centre and any radius, draw an arc, cutting OA at C.

Step 7: With centre C and same radius (as in step 2). Draw an arc, cutting the arc of step 2 in D.

Step 8: With C and D as centres, and any convenient radius (more than CD), draw two arcs intersecting at B. join OB. Then ∠AOB = 30°

Step 9: Bisect ∠AOB intersecting at D.

Thus ∠ AOD is required angle.

Verification:

On measuring ∠AOB, with the protractor,

we find ∠ AOB = 15°.

Thus ∠AOD = 15°

Question 6.

Construct the following angles using ruler and compass and verify by measuring them by a protractor.

75o

Step of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as center and any radius, draw an arc, cutting OA at C.

Step 3: With center c and same radius (as in step 2) draw an arc cutting arc at D.

Step 4: With D as center and the same radius, draw an arc cutting the arc cutting at E.

Step 5: With D and E as centers and any convenient radius (more than  DE). Draw to two arcs intersecting at P.

Step 6: Join OP. Then ∠AOP = 90°

Step 7:Bisect ∠BOP so that ∠BOQ =  ∠BOP

(∠AOP - ∠AOB)

(90° -60°) =  × 30° = 15°

So, we obtain

∠AOQ = ∠AOB + ∠BOQ

= 60° + 15° = 75°

Verification:

On measuring ∠AOQ, with the protractor,

we find ∠ AOQ = 75° .

Question 7.

Construct the following angles using ruler and compass and verify by measuring them by a protractor.

105o

Steps of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as center and any radius, draw an arc, cutting OA at C.

Step 3: With center c and same radius (as in step 2) draw an arc cutting arc at D.

Step 4: With D as center and the same radius, draw an arc cutting the arc cutting at E.

Step 5: With D and E as centers and any convenient radius (more than  DE). Draw to two arcs intersecting at P.

Step 6: Draw ∠AOB = 120° and ∠POB = 90°

Step 7: Bisect angle POB,

Then ∠AOQ is the required angle of 105°.

Question 8.

Construct the following angles using ruler and compass and verify by measuring them by a protractor.

135o

Step of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as center and any radius, draw an arc, cutting OA at C.

Step 3: With center c and same radius (as in step 2) draw an arc cutting arc at D.

Step 4: With D as center and the same radius, draw an arc cutting the arc cutting at E.

Step 5: With D and E as centers and any convenient radius (more than  DE). Draw to two arcs intersecting at P.

Step 6: Join OP. Then ∠AOP = 90°

Step 7: Bisect ∠AOP towards 180°

Question 9.

Construct an equilateral triangle, given its side of length of 4.5 cm and justify the construction.

Let us draw an equilateral triangle of side 4.5 cm.

Step of construction:

Step 1: Draw BC = 4.5 cm

Step 2: With B and C as centres and radii equal to BC = 4.5 cm, draw two arcs on the same side of BC, intersecting each other at A.

Step 3: Join AB and AC.

Î”ABC is the required equilateral triangle.

Justification:

Since by construction:

AB = BC = CA = 4.5 cm

Therefore ∆ ABC is an equilateral triangle.

Question 10.

Construct an isosceles triangle, given its base and base angle and justify the construction.

[Hint: You can take any measure of side and angle]

Let us assume the base to be 5.5cm and base angle to be 50°

∴, AB = 5.5 cm and ∠B = 50°

We know that,

In an isosceles triangle, opposite sides are equal and opposite angles are equal.

So, ∠B = ∠A = 50°

and AC = BC

Steps of construction:

Step 1: Draw base AB = 5.5cm.

Step 2: At vertex B, Draw a ray constructing angle of 50°.

Step 3: Now draw another ray at A constructing an angle of 50°

Step 4: Mark the point of intersection as C.

ABC is the required triangle.

###### Exercise 13.2

Question 1.

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

Given: base BC = 7cm, AB + AC = 12 cm and

∠B = 75° of Î” ABC.

Required: To construct a Î”ABC

Steps of construction:

Step 1: Draw a segment BC of length of 7 c

Step 2: At vertex B, construct ∠B = 75° and produce a ray BP.

Step 3: Mark an arc on ray BP cutting at D such that BD = 12cm.

Step 4: Draw segment CD.

Step 5: Construct the perpendicular bisector of segment CD.

Step 6: Name the point of intersection of ray BP and the perpendicular bisector of CD as A.

Step 7: Draw segment AB.

Î”ABC is the required triangle.

Question 2.

Construct PQR in which QR = 8 cm, ∠Q = 60° and PQ - PR = 3.5 cm.

Given: Base QR = 8 cm, PQ-PR = 3.5cm and

∠Q = 60° of Î” PQR.

Required: To construct a triangle PQR.

Steps of construction:

Step 1: Draw a segment QR of length 8cm.

Step 2: Draw ray QL such that ∠Q = 60°

Step 3: Mark an arc on opposite ray QL i.e. QS cutting at D such that QD = 3.5cm.

Step 4: Draw segment RD.

Step 5: Construct the perpendicular bisector of segment RD.

Step 6: Name the point of intersection of ray QL and the perpendicular bisector of RD as P.

Step 7: Draw segment PR.

Î”PQR is the required triangle.

Question 3.

Construct ∆XYZ in which ∠Y = 30°, ∠Z = 60° and XY + YZ + ZX = 10 cm.

Given: ∠Y = 30°, ∠Z = 60° and perimeter of Î”XYZ = 10 cm

Steps of construction:

Step 1: Step 1: raw a line segment QR of 10.5cm.

Step 2: From point Q draw a ray QD at 30° and from R draw a ray RE at 60°.

Step 3: Draw an angle bisector of Q and R, two angle bisectors intersect each other at point X.

Step 4: Draw a line bisector of QX and XR respectively these two-line bisectors intersect at point Y and Z

Step 5: Join XY AND XZ.

Step 6: Î” XYZ is required triangle.

Question 4.

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

Given base(BC) = 7.5cm and AB + AC = 15cm and ∠B = 90°

Steps of construction:

Step 1: Draw the base BC = 7.5cm

Step 2: Make an ∠XBC = 90° at the point B of base BC.

Step 3: Cut the line segment BD equals to AB + AC i.e. 15cm from the ray

XB.

Step 4: Join DC and make an angle bisector of ∠DCB.

Step 5: Let Y intersect BX at A.

Thus, Î”ABC is the required triangle.

Question 5.

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

90°

Given an angle of 90° and chord 5cm

Steps of construction:

Rough image:

Explanation:

x + x + 90° = 180°

[using sum of all angles in a triangle = 180°]

⇒ 2x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 900°

⇒ x =  = 45°

Step 1: Draw a line segment AB = 5cm

Step 2: Draw an angle of 45° on point A and B to intersect at O.

Step 3: With centre ‘O’ and radius OA and OB, draw the circle.

Step 4: Mark a point ‘C’ on the arc of the circle. Join AC and BC.

we get ∠CAB = 90°.

Thus, ACB is the required circle segment.

Question 6.

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

45°

Given an angle of 45° and chord 5cm

Steps of construction:

Step 1: Draw a line segment AB = 5cm

Step 2: Draw an angle of 45° on point A and B to intersect at O.

Step 3: With centre ‘O’ and radius OA and OB, draw the circle.

Step 4: Mark a point ‘C’ on the arc of the circle. Join AC and BC.

we get ∠ACB = 45°.

Thus, ACB is the required circle segment.

Question 7.

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

120°

Given an angle of 120° and chord 5cm

Rough Image :

Explanation:

x + x + 120° = 180°

[using sum of all angles in a triangle = 180°]

⇒ 2x + 120° = 180°

⇒ 2x = 180° - 120°

⇒ 2x = 60°

⇒ x =  = 30°

Steps of construction:

Step 1: Draw a line segment AB = 5cm

Step 2: Draw an angle of 30° on point A and B to intersect at O.

Step 3: With centre ‘O’ and radius OA and OB, draw the circle.

Step 4: Mark a point ‘C’ under the chord AB and on the arc of the

circle . Join AC and BC.

we get M∠A0B = 240°.

Thus, ACB is the required circle segment.

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