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SEBA Class 9 Mathematics Chapter 11 Constructions
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Constructions
Chapter – 11
| Exercise 11.1 |
Q.1. Construct an angle of 90⁰ at the initial point of a given ray and justify the construction.
Ans: Given: A ray OA.
Required: To construct an angle of 90⁰ at O and justify the construc- tion.
Steps of Construction:
(i) Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
(ii) Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
(iii) Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
(iv) Draw the ray OE passing through C.
Then Draw the ray OF passing through D. Then angle EOA = 60⁰ ∠FOE = 60⁰
(v) Next, taking C and D as centres and with the radius more than 1/2 CD draw arcs to intersect each other, say at G.
(vi) Draw the ray OG. This ray OG is the bisector of the angle ∠FOE, i.e.
Thus,∠GOA=∠GOE+∠EOA = 30⁰ + 60⁰= 90⁰
Justification:
(i) Join BC.
Then, OC = OB = BC (By construction)
∴ ∆COB is an equilateral triangle.
∴ ∠COB = 60⁰∠EOA = 60⁰
(ii) Join CD
Then, OD = OC = CD [By construction]
∴ ∆DOC is an equilateral triangle.
∴ ∠DOC = 60⁰ ∴ ∠FOE = 60⁰
(iii) Join CG and DG
In ∆ ODG and ∠OCG
OD = OC | Radii of the same arc
DG = CG | Arcs of equal radii
OG = OG | Common
∴ ∆ODG ≅ ∆ OCG | SSS Rule
∴ ∠DOG ≅ ∠COG | CPCT
Thus,∠GOA = ∠GOE + ∠EOA = 30⁰ + 60⁰ = 90⁰
Q.2. Construct an angle of 45 deg at the initial point of a given ray and justify the construction.
Ans: Given: A ray OA.
Required: To construct an angle of 45⁰at O and justify the construction.
Steps of Construction:
(i) Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
(ii) Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
(iii) Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
(iv) Draw the ray OE passing through C. Then ∠EOA = 60⁰
(v) Draw the ray OF passing through D.
Then ∠FOE = 60⁰
(vi) Next, taking and D as centres and with radius more than 1/2 CD draw arcs to intersect each other, say at G.
(vii) Draw the ray OG. This ray OG is the bisector of the angle FOE, i.e.,
(viii) Now, taking O as centre and any radius, draw an arc to intersect the rays ÒA and OG, say at H and I respectively.
(ix) Next, taking H and I as centres and with the radius more than 1/2 HI, draw arcs to intersect each other, say at J.
(x) Draw the ray OJ. This ray OJ is the required bisector of the angle GOA.
Justification:
(i) Join BC.
Then, OC = OB = BC (By construction)
∴ ∆COB is an equilateral triangle.
∴ ∠COB = 60⁰ ∴ ∠EOA = 60⁰
(ii) Join CD
Then, OD = OC = CD (By construction)
∴ ∆ DOC is an equilateral triangle.
∴ ∠DCO = 60⁰ ∴ ∠FOE = 60⁰
(iii) Join CG and DG
In ∆ DOG and ∆ OCG,
OD = OC | Radii of the same arc
DG = CG | Arcs of equal radii
OG = OG | Common
∴ ∆ DOG = ∠COG | CPCT
Thus,∠GOA=∠GOE+∠EOA = 30⁰+ 60⁰= 90⁰
(iv) Join HJ and Ij
In ∆OIJ and ∆OHJ,
OI = OH | Radii of the same arc
IJ = HJ | Arcs of equal radii
OJ = OJ | Common
∴ Δ ΟΙJ ≅ Δ ΟΗJ | SSS Rule
∴ ∠OIJ = ∠HOJ | CPCT
Q.3. Construct the angles of the following measurements :
(i) 30⁰
(iii) 15⁰
Ans: (i) 30⁰
Given: A ray OA
Required: To construct an angle of 30º at O.
Steps of Construction:
(i) Taking O as’ centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
(ii) Taking B as centre and with the same radius as before, draw aı arc intersecting the previously drawn arc, say at a point C.
(iii) Draw the ray OE passing through C. Then ∠EOA = 60⁰
(iv) Taking B and C as centres and with the radius more than 1/2 BC, draw arcs to intersect each other, say at D.
(v) Draw the ray OD. This ray OD is the bisector of the angle EOA, i.e.,
(i) Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
(ii) Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
(iii) Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
(iv) Draw the ray OE passing through C. Then ∠EOA = 60⁰.
(v) Draw the ray OF passing through D. Then ∠FOE = 60⁰
(vi) Next, taking C and D as centres and with radius more than ½ CD
draw arcs to intersect each other, say at G.
(vii) Draw the ray OG. This ray OG is the bisector of the angle FOE, i.e..
Thus, ∠GOA = ∠GOE +∠EOA = 30⁰ + 60⁰= 90⁰
(viii) Now, taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I respectively.
(ix) Next taking H and I as centres and with the radius more than ½ HI
draw arcs to intersect each other, say at J.
(x) Draw the ray OJ. This ray OJ is the bisector of the angle GOA.
(xi) Now, taking O as centre and any radius, draw an arc to intersect the rays OA and OJ, say at K and L respectively.
(xii) Next, taking K and L as centres and with the radius more than 1/2 KL, draw arcs to intersect each other, say at M.
(xiii) Draw the ray OM. This ray OM is the bisector of the angle AOJ, i.e.,
i.e., ∠GOJ = ∠JOM = ∠AOM
(iii) 15⁰
Ans: Given: A ray OA
Required: To construct an angle of 15⁰ at O.
Steps of Construction:
(i) Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
(ii) Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
(iii) Draw the ray OE passing through C.
Then ∠EOA = 60⁰
(iv) Now, taking B and C as centres and with the radius more then ½ BC
draw arcs to intersect each other, say at D.
(v) Draw the ray OF intersecting the arc drawn in step 1 at F. This ray OD is the bisector of the angle
(vi) Now , taking B and F as centres and with the radius more than ½ BF
draw arcs to intersect each other, say at G.
(vii)Draw they ray OG is the bisector of the angle AOD, i.e.,
Q.4. Construct the following angles and verify by measuring them by a protractor:
(i) 75⁰
Ans: 75⁰
Given: A ray OA
Required: To construct an angle of 9\75º at O.
Steps of Construction:
(i) Taking Oas centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
(ii) Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
(iii) Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
(iv) Join the ray OE passing through C. Then ∠EOA = 60⁰.
(v) Draw the ray OF passing through D. Then ∠FOE = 60⁰
(vi) Next, taking C and D as centres and with the radius more than ½ CD
draw arcs to intersect each other, say at G.
(vii) Draw the ray OF intersecting the arc of step 1 at H. This ray OG is the bisector of the angle FOE,
(viii) Next , taking C and H as centres and with radius more than 1/2
CH, draw arcs to intersect each other, say at I.
(ix) Draw the rav OL. This rav OI is the bisector of the angle GOE, i.e.,
Thus,∠IOA=∠IOE+∠EOA = 15⁰ + 60⁰ = 75⁰
On measuring the ∠IOA by protractor, we find that ∠IOA = 15⁰.
(ii) 105⁰
Ans: 105⁰
Given: A ray OA
Required: To construct an angle of 105⁰ at O.
Steps of Construction:
(i) Taking O as centre and some radius, draw an arc of a circle, which in- tersects OA, say at a point B.
(ii) Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
(iii) Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
(iv) Draw the ray OE passing through C. Then ∠EOA = 60⁰
(v)Draw the ray OF passing through D. Then ∠FOE = 60⁰
(vi) Next , taking C and H as centres and with radius more than 1/2
CH, draw arcs to intersect each other, say at I.
(vii) Draw the ray OF intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the angle FOE,
Thus, ∠GOA = ∠GOE + ∠EOA = 30⁰ + 60⁰ = 90⁰
(viii) Next , taking C and H as centres and with radius more than 1/2
CH, draw arcs to intersect each other, say at I.
(ix) Draw the ray OI. This ray OI is the bisector of the angle FOG, i.e.,
Thus,∠IOA= ∠IOG + ∠GOA = 15⁰ + 90⁰ = 105⁰
Thus the construction is verified.
(iii) 135⁰
Given: A ray OA.
Required: To construct an angle of 135⁰ at O.
Steps of Construction:
(i) Produce AO to A’ to form ray O A’
(ii) Taking O as centre and some radius, draw an arc of a circle, which in- tersects OA at a point B and O * A’ at a point B’.
(iii) Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.
(iv) Taking C as centre and with the same radius as before, draw an arc inter- secting the arc drawn in step 1, say at D.
(v) Draw the ray OE passing through C. Then ∠EOA = 60⁰
(vi) Draw the ray OF passing through D Then ∠FOE = 60⁰
(vi) Next, taking C and D as centres and with the radius more than ½ CD
draw arcs to intersect each other, say at G.
8. Draw the ray OF intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the angle FOE,
Thus, ∠GOA GOE + ∠EOA = 30⁰ + 60⁰ =90⁰
∴ ΒΟΗ = 90⁰
(ix) Next , taking B and H as centres and with radius more than ½ BH.
draw arcs to intersect each other, say at I.
(x) Draw the ray OI. This ray OI is the bisector of the angel B’OG,
Thus, ∠IOA ∠IOG + ∠GOA
= 45⁰ + 90⁰ = 135⁰
On measuring the ∠IOA by protractor, we find that ∠IOA = 135⁰ Thus the construction is verified.
Q.5. Construct an equilateral triangle, given its side and justify the construction.
Ans: Given: Side (say 6 cm) of an equilateral triangle.
Required: To construct an equilateral triangle and justify the con- struction.
Steps of Construction:
(i) Take a ray AX with initial point A. From AX, cut off AB = 6 cm.
(ii) Taking A as centre and radius (= 6 cm), draw an arc of a circle, which intersects AX, say at a point B.
(iii) Taking B as centre and with the same radius as before, draw an arc in- tersecting the previously drawn arc, say at a point C.
(iv) Draw the ray AE passing through C.
(v) Next, taking B as centre and radius (= cm), draw an arc of a circle, which intersects AX, say at a point A.
(vi) Taking A as centre and with the same radius as in step 5, draw an arc intersecting the previously drawn arc, say at a point C.
(vii) ‘Draw the ray BF passing through C. Then AABC is the required triangle with gives side 6 cm.
Justification:
AB = BC | By construction
AB=AC | By construction
∴ AB = BC CA
∴ ∆ABC is an equilateral triangle
∴ The construction is justified.
| Exercise 11.2 |
Q.1. Construct a triangle ABC in which BC = 7cm ∠ B = 75⁰ and AB + AC = 13cm.
Ans: Given : In ∆ABC, BC = cm, ∠B = 75⁰ and AB + AC = 13cm
Required: To construct the triangle ABC.
Steps of Construction:
(i) Draw the base BC = 7cm.
(ii) At the point B make an angle XBC = 75⁰
(iii) Cut a line segment BD equal to AB+AC (=13 cm) from the ray BX.
(iv) Join DC.
(iv) Make an ∠DCY = ∠BDC
(v) Let CY intersect BX at A.
(vi) Then ABC is the required triangle.
Q.2. Construct a triangle ABC in which BC = 8cm ∠B = 45⁰ and AB – AC = 3.5cm.
Ans: Given In ∆ABC, BC = 8cm ∠B = 45⁰ and AB – AC = 3.5cm
Required: To construct the triangle ABC.
Steps of Construction:
(i) Draw the base BC = 8cm.
(ii) At the point B make an angle XBC =450.
(iii) Cut the line segment BD equal to AB-AC (1 cm) from the ray BX.
(iv) Join DC.
(v) Draw the perpendicular bisector, say PQ of DC.
(vi) Let it intersect BX at a point A.
(vii) Join AC.
(viii)Then ABC is the required triangle.
Q.3. Construct a triangle PQR in which QR = 6 cm, ∠Q = 60⁰ and PR PQ = 2cm.
Ans: Given: In ∆ PQR, QR = 6cm , ∠Q = 90⁰ and PR – PQ = 2cm.
Required: To construct the ∆PQR
Steps of Construction:
(i) Draw the base QR = 6cm.
(ii) At the point Q make an ∠XQR = 60⁰
(iii) Cut segment QS=PR-PQ(=2 cm) from the line QX extended on opposite side of the segment QR.
(iv) Join SR.
(v) Draw the perpendicular bisector LM of SR.
(vi) Let LM intersect QX at P.
(vii) Join PR.
(viii) Then, PQR is the required triangle.
Q.4. Construct a triangle XYZ in which ∠Y= 30⁰, ∠Z = 90⁰ and XY+YZ+ZX = 11cm.
Ans: Given: In triangle XYZ, ∠Y = 30⁰, ∠Z = 90⁰ and XY+YZ+ZX = 1 сm.
Required: To construct the ∆XΥΖ.
Steps of Construction:
(i) Draw a line segment BC = XY+YZ+ZX (= 11 cm)
(ii) Make ∠LBC = ∠Y (= 30⁰) and ∠ MCB = ∠Z (= 90⁰)
(iii) Bisect ∠LBC and ∠MCB. Let these bisectors meet at a point X.
(iv) Draw perpendicular bisectors DE of XB and FG of XC.
(v) Let DE intersect BC at Y and FC intersect BC at Z.
(vi) Join XY and XZ.
(vii) Then XYZ is the required triangle.
Q.5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Ans: Given: In right ∆ABC , base BC = 12cm , ∠B = 90⁰ and AB + AC = 18 cm.
Required: To construct the right triangle ABC.
Steps of Construction:
(i) Draw the base BC = 12cm.
(ii) At point B, make an ∠XBC = 90⁰.
(iii) Cut a line segment BD = AB + AC (= 18cm ) from the ray BX.
(iv) Join DC.
(v) Draw the perpendicular bisector PQ of CD to intersect BD at a point A.
(vi) Join AC.
Then, ABC is the required right triangle.
