Home » class 9 Maths » NCERT Solutions for Class 9 Maths: Chapter 11 Constructions

# NCERT Solutions for Class 9 Maths: Chapter 11 Constructions

Exercise 11.1 : Solutions of Questions on Page Number : 191

Q1 : Construct an angle of 90° at the initial point of a given ray and justify the construction.
The below given steps will be followed to construct an angle of 90°.
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) Join PU, which is the required ray making 90° with the given ray PQ. Justification of Construction:
We can justify the construction, if we can prove ∠UPQ = 90°.
For this, join PS and PT. We have, ∠SPQ = ∠TPS = 60°. In (iii) and (iv) steps of this construction, PU was drawn as the bisector of ∠TPS.
∴ ∠UPS = ∠TPS Also, ∠UPQ = ∠SPQ + ∠UPS
= 60° + 30°
= 90°

Q2 : Construct an angle of 45° at the initial point of a given ray and justify the construction.
The below given steps will be followed to construct an angle of 45°.
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) Join PU. Let it intersect the arc at point V.
(vi) From R and V, draw arcs with radius more than RV to intersect each other at W. Join PW.
PW is the required ray making 45° with PQ. Justification of Construction:
We can justify the construction, if we can prove ∠WPQ = 45°.
For this, join PS and PT. We have, ∠SPQ = ∠TPS = 60°. In (iii) and (iv) steps of this construction, PU was drawn as the bisector of ∠TPS.
∴ ∠UPS = ∠TPS Also, ∠UPQ = ∠SPQ + ∠UPS
= 60° + 30°
= 90°
In step (vi) of this construction, PW was constructed as the bisector of ∠UPQ.
∴ ∠WPQ = ∠UPQ Q3 : Construct the angles of the following measurements:
(i) 30° (ii) (iii) 15°
(i)30°
The below given steps will be followed to construct an angle of 30°.
Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc of a circle which intersects PQ at R.
Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.
Step III: Taking R and S as centre and with radius more than RS, draw arcs to intersect each other at T. Join PT which is the required ray making 30° with the given ray PQ. (ii) The below given steps will be followed to construct an angle of .
(1) Take the given ray PQ. Draw an arc of some radius, taking point P as its centre, which intersects PQ at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join PU. Let it intersect the arc at point V.
(6) From R and V, draw arcs with radius more than RV to intersect each other at W. Join PW.
(7) Let it intersect the arc at X. Taking X and R as centre and radius more than RX, draw arcs to intersect each other at Y.
Joint PY which is the required ray making with the given ray PQ. (iii) 15°
The below given steps will be followed to construct an angle of 15°.
Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc of a circle which intersects PQ at R.
Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.
Step III: Taking R and S as centre and with radius more than RS, draw arcs to intersect each other at T. Join PT.
Step IV: Let it intersect the arc at U. Taking U and R as centre and with radius more than RU, draw an arc to intersect each other at V. Join PV which is the required ray making 15° with the given ray PQ. Q4 : Construct the following angles and verify by measuring them by a protractor:
(i) 75° (ii) 105° (iii) 135°
(i) 75°
The below given steps will be followed to construct an angle of 75°.
(1) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join PU. Let it intersect the arc at V. Taking S and V as centre, draw arcs with radius more than SV. Let those intersect each other at W. Join PW which is the required ray making 75° with the given ray PQ. The angle so formed can be measured with the help of a protractor. It comes to be 75 º.
(ii) 105°
The below given steps will be followed to construct an angle of 105°.
(1) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join PU. Let it intersect the arc at V. Taking T and V as centre, draw arcs with radius more than TV. Let these arcs intersect each other at W. Join PW which is the required ray making 105° with the given ray PQ. The angle so formed can be measured with the help of a protractor. It comes to be 105 º.
(iii) 135°
The below given steps will be followed to construct an angle of 135°.
(1) Take the given ray PQ. Extend PQ on the opposite side of Q. Draw a semi-circle of some radius taking point P as its centre, which intersects PQ at R and W.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join PU. Let it intersect the arc at V. Taking V and W as centre and with radius more than VW, draw arcs to intersect each other at X. Join PX, which is the required ray making 135°with the given line PQ. The angle so formed can be measured with the help of a protractor. It comes to be 135 º.

Q5 : Construct an equilateral triangle, given its side and justify the construction
Let us draw an equilateral triangle of side 5 cm. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be 5 cm. We also know that each angle of an equilateral triangle is 60º.
The below given steps will be followed to draw an equilateral triangle of 5 cm side.
Step I: Draw a line segment AB of 5 cm length. Draw an arc of some radius, while taking A as its centre. Let it intersect AB at P.
Step II: Taking P as centre, draw an arc to intersect the previous arc at E. Join AE.
Step III: Taking A as centre, draw an arc of 5 cm radius, which intersects extended line segment AE at C. Join AC and BC. ΔABC is the required equilateral triangle of side 5 cm. Justification of Construction:
We can justify the construction by showing ABC as an equilateral triangle i.e., AB = BC = AC = 5 cm and ∠ A = ∠ B = ∠ C = 60°.
In ΔABC, we have AC = AB = 5 cm and ∠ A = 60°.
Since AC = AB,
∠ B = ∠ C (Angles opposite to equal sides of a triangle)
In ΔABC,
∠ A + ∠ B + ∠ C = 180° (Angle sum property of a triangle)
⇒ 60° + ∠ C + ∠ C = 180°
⇒ 60° + 2 ∠ C = 180°
⇒ 2 ∠ C = 180° – 60° = 120°
⇒ ∠ C = 60°
∴ ∠ B = ∠ C = 60°
We have, ∠ A = ∠ B = ∠ C = 60° … (1)
⇒ ∠ A = ∠ B and ∠ A = ∠ C
⇒ BC = AC and BC = AB (Sides opposite to equal angles of a triangle)
⇒ AB = BC = AC = 5 cm … (2)
From equations (1) and (2), ΔABC is an equilateral triangle.

Exercise 11.2 : Solutions of Questions on Page Number : 195

Q1 : Construct a triangle ABC in which BC = 7 cm, ∠ B = 75° and AB + AC = 13 cm.
The below given steps will be followed to construct the required triangle.
Step I: Draw a line segment BC of 7 cm. At point B, draw an angle of 75°, say ∠ XBC.
Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX.
Step III: Join DC and make an angle DCY equal to ∠ BDC.
Step IV: Let CY intersect BX at A. ΔABC is the required triangle. Q2 : Construct a triangle ABC in which BC = 8 cm, ∠ B = 45° and AB – AC = 3.5 cm.
The below given steps will be followed to draw the required triangle.
Step I: Draw the line segment BC = 8 cm and at point B, make an angle of 45°, say ∠ XBC.
Step II: Cut the line segment BD = 3.5 cm (equal to AB – AC) on ray BX.
Step III: Join DC and draw the perpendicular bisector PQ of DC.
Step IV: Let it intersect BX at point A. Join AC. ΔABC is the required triangle. Q3 : Construct a triangle PQR in which QR = 6 cm, ∠ Q = 60° and PR – PQ = 2 cm
The below given steps will be followed to construct the required triangle.
Step I: Draw line segment QR of 6 cm. At point Q, draw an angle of 60°, say ∠ XQR.
Step II: Cut a line segment QS of 2 cm from the line segment QT extended in the opposite side of line segment XQ. (As PR > PQ and PR – PQ = 2 cm). Join SR.
Step III: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P. Join PQ, PR.
ΔPQR is the required triangle. Q4 : Construct a triangle XYZ in which ∠ Y = 30°, ∠ Z = 90° and XY + YZ + ZX = 11 cm.
The below given steps will be followed to construct the required triangle.
Step I: Draw a line segment AB of 11 cm.
(As XY + YZ + ZX = 11 cm)
Step II: Construct an angle, ∠ PAB, of 30° at point A and an angle, ∠ QBA, of 90° at point B.
Step III: Bisect ∠ PAB and ∠ QBA. Let these bisectors intersect each other at point X.
Step IV: Draw perpendicular bisector ST of AX and UV of BX.
Step V: Let ST intersect AB at Y and UV intersect AB at Z.
Join XY, XZ.
ΔXYZ is the required triangle. Q5 : Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. 