Geometrical Constructions Class 9th Mathematics AP Board Solution

Class 9th Mathematics AP Board Solution

Exercise 13.1
Question 1.

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

90o


Answer:

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, ΔOCD is an equilateral triangle. So, ∠COD = 60°


Again OD = DE = OE


Therefore, ΔODE 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


Answer:

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


Answer:

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.



Answer:

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


Answer:

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


Answer:

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


Answer:

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


Answer:

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.


Answer:

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.



ΔABC 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]


Answer:

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.


Answer:

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

∠B = 75° of Δ ABC.


Required: To construct a ΔABC


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.



ΔABC is the required triangle.



Question 2.

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


Answer:

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

∠Q = 60° of Δ 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.



ΔPQR is the required triangle.



Question 3.

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


Answer:

Given: ∠Y = 30°, ∠Z = 60° and perimeter of ΔXYZ = 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: Δ 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.


Answer:

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, ΔABC is the required triangle.



Question 5.

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

90°


Answer:

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°


Answer:

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°


Answer:

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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