Rd Sharma 2021 Solutions for Class 9 Maths Chapter 16 Construction are provided here with simple step-by-step explanations. These solutions for Construction are extremely popular among Class 9 students for Maths Construction Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 9 Maths Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 16.10:

#### Question 7:

Using ruler and compasses only, draw a right angle.

#### Answer:

We have to draw a right angle using ruler and compasses only

We use the following algorithm for the construction

Steps of construction

STEP1: Draw a ray BC.

STEP2: With B as a centre, and taking convenient radius, draw an arc, intersecting the ray BC at point N.

STEP3: With N as a centre, and taking the same radius, draw an arc cutting the previous arc at M.

STEP4: With M as a centre, and the same radius, draw an arc cutting the arc drawn in STEP2 at L.

STEP5: With M and L as centre and the same radius draw arcs intersecting at a point say A.

STEP6: Draw the ray BA.

ABC is a right angle.

#### Page No 16.10:

#### Question 8:

Using ruler and compasses only, draw an angle of measure 135°.

#### Answer:

We have to draw an angle of 135° using ruler and compasses only

We follow the following algorithm for the construction

Steps of construction

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.

#### Page No 16.10:

#### Question 9:

Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.

#### Answer:

Using protractor we have to draw an angle of 72° and considering this angle given draw an angle of 36° and 54°

For this construction we will follow the following algorithm

Steps of Construction

STEP1: Draw ∠ABC of measure 72°, using protractor.

STEP2: Bisect ∠ABC. Let BD be the bisecting ray.

∠ABD = ∠DBC = 36°

STEP3: Bisect ∠ABD. Let BE be the bisecting ray.** **

So,

Thus,

Thus, ∠DBC = 36° and ∠EBC = 54° are the required angles.

#### Page No 16.10:

#### Question 10:

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

(i) 45° (ii) 90°

#### Answer:

(i) Construction of 45$\xb0$.

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

(ii) Construction of 90$\xb0$.

(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°

#### Page No 16.10:

#### Question 11:

Construct the angles of the following measurements:

(i) 30°

(ii) 75°

(iii) 105°

(iv) 135°

(v) 15°

(vi) $22{\frac{1}{2}}^{\xb0}$

#### Answer:

(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) 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 75º.

(iii) 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 105º.

(iv) 135°

We follow the following algorithm for the construction

Steps of construction

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.

(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.

(v) 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.

(vi)

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.

(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.

#### Page No 16.18:

#### Question 1:

Construct a Δ *ABC* in which *BC* = 3.6 cm, *AB* + *AC* = 4.8 cm and ∠*B* = 60°.

#### Answer:

We are asked to construct a triangle ABC in which, and

We will follow the following algorithm for the construction

Steps of construction

STEP1: Draw BC of 3.6 cm

STEP2: Draw.

STEP3: From ray BE, cut BD of 4.8 cm.

STEP4: Draw segment DC.

STEP5: Draw the perpendicular bisector of DC, intersecting the ray BE at A

STEP6: Draw segment AC.

ΔABC is the required angle.

#### Page No 16.18:

#### Question 2:

Construct a Δ *ABC* in which *AB *+ *AC* = 5.6 cm, *BC* = 4.5 cm and ∠*B* = 45°.

#### Answer:

We are asked to construct a triangle ABC in which, and

We will follow the following algorithm for the construction

Steps of construction

STEP1: Draw BC of 4.5 cm.

STEP2: Draw

STEP3: From ray BE, cut BD of 5.6 cm.

STEP4: Draw segment DC.

STEP5: Draw the perpendicular bisector of DC, intersecting the ray BE at A

STEP6: Draw segment AC.

ΔABC is the required angle.

#### Page No 16.18:

#### Question 3:

Construct a Δ ABC in which BC = 3.4 cm, AB − AC = 1.5 cm and ∠B = 45°.

#### Answer:

We are asked to construct a triangle ABC in which , and

We will follow the following algorithm for the construction

Steps of construction

STEP1: Draw BC of 3.4 cm.

STEP2: Draw CBE =

STEP3: From the ray BE, cut BD of 1.5 cm.

STEP4: Draw segment DC.

STEP5: Draw the perpendicular bisector of DC, intersecting the ray BE at A

STEP6: Draw AC.

ΔABC is the required angle.

#### Page No 16.18:

#### Question 4:

Using ruler and compasses only, construct a Δ*ABC*, given base *BC* = 7 cm, ∠*ABC* = 60° and *AB* + *AC* = 12 cm.

#### Answer:

We are asked to construct a triangle ABC in which BC = 7 cm, ∠B = 60° and AB + AC = 12 cm

We will follow the following algorithm for the construction

Steps of construction

STEP1: Draw BC of length 7 cm.

STEP2: With B as a centre, and convenient radius draw an arc cutting BC at N.

STEP3: With N as a centre, and same radius draw an arc cutting the previous arc at M.

STEP4: Draw BM and produce it to P.

STEP5: From BP, cut BR = 12 cm.

STEP6: Draw segment RC.

STEP7: Draw perpendicular bisector of RC. It cuts the ray BR at A.

STEP8: Draw AC.

ΔABC is the required triangle.

#### Page No 16.18:

#### Question 5:

Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.

#### Answer:

We are asked to draw triangle ABC whose base angles are 60$\xb0$ and 90$\xb0$ and its perimeter is 10 cm.

We will follow the following algorithm for the construction

Steps of construction

STEP1: Draw a line segment DE of 10 cm.

STEP2: Draw ∠EDM = ∠B = 90° and ∠DEN = ∠C = 60°.

STEP3: Draw the angle bisectors of ∠EDM and ∠DEN and mark their point of intersection as A.

STEP4: Draw the perpendicular bisectors of AD and AE. They meet DE at B and C respectively.

STEP5: Draw AB and AC to get the desired ΔABC.

#### Page No 16.18:

#### Question 6:

Construct a triangle *ABC *such that *BC* = 6 cm, *AB* = 6 cm and median *AD* = 4 cm.

#### Answer:

We are given and median

We have to construct the triangle ABC

We will follow the following algorithm for the construction

Steps of construction

STEP1: Draw a line segment BC of length 6 cm.

STEP2: Draw the perpendicular bisector of line BC cutting the line at point D, such that BD = 3 cm.

STEP3: With centre B and radius = 6 cm, draw an arc.

STEP4: With centre D and radius = 4 cm, draw an arc intersecting the previous arc at A.

STEP5: Draw AC to get the desired ΔABC.

#### Page No 16.18:

#### Question 7:

Construct a right triangle *ABC* whose base *BC* is 6 cm and the sum of hypotenuse *AC* and other side *AB* is 10 cm.

#### Answer:

We have to construct a right triangle ABC with base and the sum of hypotenuse and the other side is 10 cm

We will follow the following algorithm for the construction

Steps of construction

STEP1: Draw a line segment BC of length 6 cm.

STEP2: Draw CBE =

STEP3: From BE, cut BD of length 10 cm.

STEP4: Draw the line segment DC.

STEP5: Draw the perpendicular bisector of DC, intersecting the ray BE at point A

STEP6: Draw AC to get the desired ΔABC.

#### Page No 16.18:

#### Question 8:

Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.

#### Answer:

We are asked to draw triangle ABC whose base angles are 60° and 45° and its perimeter is 6.4 cm

We will follow the following algorithm for the construction

Steps of construction

STEP1: Draw DE of 6.4 cm.

STEP2: Draw ∠EDP =∠B = 60° and ∠DER = ∠C = 45°.

STEP3: Draw the angle bisectors of ∠EDP and ∠DER and mark their point of intersection as A.

STEP4: Draw the perpendicular bisectors of AD and AE. They meet DE at point B and Point C respectively.

STEP5: Draw AB and AC to get the required ΔABC.

#### Page No 16.18:

#### Question 9:

Using ruler and compasses only, construct a Δ *ABC* from the following data:

*AB* + *BC* + *CA* = 12 cm, ∠*B* = 45° and ∠*C* = 60°.

#### Answer:

We are asked to draw triangle ABC whose base angles are 60° and 45° and its perimeter is 12 cm

We will follow the following algorithm for the construction

Steps of construction

STEP1: Draw DE of length 12 cm.

STEP2: Construct ∠EDP =∠C = 60° and ∠DER =∠B = 45°, using regular methods.

STEP3: Draw the angle bisectors of ∠EDP and ∠DER and mark their point of intersection as A.

STEP4: Draw the perpendicular bisectors of AE and AD. They meet DE at point B and Point C respectively.

STEP5: Draw AB and AC to get the required ΔABC.

#### Page No 16.18:

#### Question 10:

Construct a triangle *XYZ* in which ∠*Y* = 30°, ∠*Z* = 90° and* XY* + *YZ* + *ZX* = 11.

#### Answer:

We are asked to construct a triangle XYZ such that

We will follow the following algorithm for the construction

Steps of construction-

STEP1: Draw a line segment AB of length 11 cm.

STEP2: Draw BAC = Y = and ABD =Z =

STEP3: Bisect BAC and ABD and mark the point of intersection of these angle bisectors by X.

STEP4: Draw perpendicular bisectors of AX and XB. They meet AB at point Y and Z, respectively.

STEP5: Draw XY and XZ to get the desired ΔXYZ.

#### Page No 16.19:

#### Question 1:

With the help of a ruler and compasses, it is possible to construct an angle1 of

(a) 35°

(b) 40°

(c) $37\frac{1\xb0}{2}$

(d) $47\frac{1\xb0}{2}$

#### Answer:

Using a ruler and compass, construct an angle of 150°.

Bisect the angle to get an angle of 75°.

Again bisect the angle to get an angle of 37.5°.

Hence, the correct option is (c).

#### Page No 16.19:

#### Question 2:

With the help of a ruler and compasses it is not possible to construct an angle of

(a) $37\frac{1\xb0}{2}$

(b) 40°

(c) $22\frac{1\xb0}{2}$

(d) $67\frac{1\xb0}{2}$

#### Answer:

(a) Using a ruler and compass, construct an angle of 150°.

Bisect the angle to get an angle of 75°.

Again bisect the angle to get an angle of 37.5°.

(b) It is not possible to construct an angle of 40° with the help of a ruler and compass.

(c) Using a ruler and compass, construct an angle of 90°.

Bisect the angle to get an angle of 45°.

Again bisect the angle to get an angle of 22.5°.

(d) Using a ruler and compass, construct an angle of 270°.

Bisect the angle to get an angle of 135°.

Again bisect the angle to get an angle of 67.5°.

Hence, the correct option is (b).

#### Page No 16.19:

#### Question 3:

The construction of a triangle *ABC *in which *AB *= 4 cm, ∠*A *= 60° is not possible when difference of *BC *and *AC *is equal to

(a) 3.5 cm

(b) 4.5 cm

(c) 3 cm

(d) 2.5 cm

#### Answer:

We know,

In a triangle, sum of any two sides must be greater than the third side.

$AB+AC>BC\phantom{\rule{0ex}{0ex}}\Rightarrow AB>BC-AC\phantom{\rule{0ex}{0ex}}\Rightarrow 4>BC-AC\left(\because AB=4\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 4BC-AC$

Thus, difference of *BC *and *AC *cannot be greater than 4 cm.

Hence, the correct option is (b).

#### Page No 16.19:

#### Question 4:

The construction of a triangle *ABC*, given *BC *= 6 cm, ∠*B *= 45° is not possible when difference of *AB *and *AC *is equal to

(a) 6.9 cm

(b) 5.2 cm

(c) 5 cm

(d) 4 cm

#### Answer:

We know,

In a triangle, sum of any two sides must be greater than the third side.

$AB+BC>AC\phantom{\rule{0ex}{0ex}}\Rightarrow BC>AC-AB\phantom{\rule{0ex}{0ex}}\Rightarrow 6>AC-AB\left(\because BC=6\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 6AC-AB$

Thus, difference of *AB *and *AC *cannot be greater than 6 cm.

Hence, the correct option is (a).

#### Page No 16.19:

#### Question 5:

The construction of a triangle *ABC*, given that *BC *= 3 cm, ∠*C *= 60° is possible when differences of *AB *and *AC *is equal to

(a) 3.2 cm

(b) 3.1 cm

(c) 3 cm

(d) 2.8 cm

#### Answer:

We know,

In a triangle, sum of any two sides must be greater than the third side.

$AB+BC>AC\phantom{\rule{0ex}{0ex}}\Rightarrow BC>AC-AB\phantom{\rule{0ex}{0ex}}\Rightarrow 3>AC-AB\left(\because BC=3\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 3AC-AB$

Thus, differences of *AB *and *AC *can be 2.8 cm.

Hence, the correct option is (d).

#### Page No 16.19:

#### Question 6:

The construction of a triangle *ABC*, given that *AB *= 5 cm, ∠*A *= 45° is possible when *BC *+ *CA *is equal to

(a) 5 cm

(b) 4.5 cm

(c) 8 cm

(d) 4 cm

#### Answer:

We know,

In a triangle, sum of any two sides must be greater than the third side.

Since, *AB *= 5 cm

Thus, *BC *+ *CA >* 5 cm

Hence, the correct option is (c).

#### Page No 16.3:

#### Question 1:

Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.

#### Answer:

We are given a line segment of length 8.6 cm

We have to bisect it and measure each part.

We follow the following steps of construction

STEP1: Draw a line segment AB = 8.6 cm by using a graduated ruler.

STEP 2: With A as centre and radius more than half of AB, draw arcs, one on each side of AB.

STEP3: With B as centre and the same radius as in step II, draw arcs cutting the arcs drawn in step II at E and F respectively.

STEP 4: Draw the line segment with E and F as end-points. Suppose it meets AB at M. Then M bisects the line segment AB.

By measuring AM and MB, we find that

AM = MB = 4.3cm

#### Page No 16.3:

#### Question 2:

Draw a line segment *AB* of length 5.8 cm. Draw the perpendicular bisector of this line segment.

#### Answer:

We are given the line segment of length 5.8 cm

We are asked to draw the perpendicular bisector of the line segment

We will follow the following algorithm for the construction

We follow the following steps for constructing the perpendicular bisector of PQ.

STEP1: Draw a line segment PQ=5.8 cm by using a graduated ruler.

STEP2: With P as centre and radius more than half of PQ, draw two arcs, one on each side of PQ.

STEP3: With Q as centre and the same radius as in step II, draw arcs cutting the arcs drawn in the previous step at Land N respectively.

STEP4: Draw the line segment with L and N end-points.

The line segment LN is the required perpendicular bisector of Q.

#### Page No 16.4:

#### Question 3:

Draw a circle with centre at point *O* and radius 5 cm. Draw its chord *AB*, draw the perpendicular bisector of line segment *AB*. Does it pass through the centre of the circle?

#### Answer:

We are asked to draw the circle centered at *O* of radius 5 cm with its chord *AB*

We will follow the following algorithm for the construction

We follow the following steps:

STEP1: Draw a circle with centre at point O and radius 5 cm.

STEP2: Draw its cord AB.

STEP3: With A as centre and radius more than half of AB, draw two arcs, one on each side of AB.

STEP4: With B as centre and the same radius as in step3, draw arcs cutting the arcs drawn in the previous step at C and D respectively.

STEP5: Draw the line segment with C and D as end-points.

The line segment CD is the required perpendicular bisector of AB. Since CD is perpendicular bisector of AB which is chord of circle, hence it passes through the centre of the circle.

#### Page No 16.4:

#### Question 4:

Draw a circle with centre at point *O.* Draw its two chords *AB* and *CD* such that *AB* is not parallel to *CD*. Draw the perpendicular bisectors of *AB* and *CD*. At what point do they intersect?

#### Answer:

We are asked to draw a circle centered at *O* and two chords *AB* and *CD*

We will follow the following algorithm for the construction

Steps of construction

STEP1: With centre O, draw a circle of any radius.

STEP2: Draw any two chords AB and CD, such that the two chords are not parallel.

STEP3: With centre B and taking any radius (more than half of AB), draw two arcs, one on each side of the chord AB.

STEP4: With centre A, and taking the same radius, draw two arcs, one on each side of the chord AB, cutting the previous arcs in E and F respectively.

STEP5: Draw a line segment with E and F as end-points. It passes through centre O.

STEP6: With centre C and taking any radius (more than half of CD), draw two arcs, one on each side of the chord CD.

STEP7: With centre D, and taking the same radius as in STEP 6, draw two arcs, one on each side of the chord CD, cutting the previous arcs in G and H respectively.

STEP8: Draw a line segment with G and H as end-points. This also passes through centre O. It is clear that perpendicular bisectors EF and GH intersect at point O, which is the centre of the circle.

#### Page No 16.4:

#### Question 5:

Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.

#### Answer:

We are asked to bisect the line segment of length 10 cm and again bisect the half of its original line segment

We follow the following steps for the construction

Steps of construction

STEP1: Draw line segment AB having length 10 cm.

STEP2: With centre A and radius greater than half of AB, draw two arcs one on each side of AB.

STEP3: With centre B and taking same radius, draw two arcs one on each side of AB, intersecting the previous two arcs at C and D respectively.

STEP4: Draw a line segment having end-points C and D. Segment CD is the perpendicular bisector of AB. Let CD intersects AB at M.

STEP 5: with centre A and radius greater than half of AM, draw two arcs one on each side of AM.

STEP6: With centre M and taking same radius, draw two arcs one on each side of AM intersecting the previous two arcs at E and F respectively.

STEP7: Draw a line segment having end-points E and F. Segment EF is the perpendicular bisector of AM. Let EF intersects AM at N.

STEP8: Measure the length of AN.

Length of line segment AN must be 2.5 cm

#### Page No 16.4:

#### Question 6:

Draw a line segment *AB *and bisect it. Bisect one of the equal parts to obtain a line segment of length $\frac{1}{2}$ (*AB*)

#### Answer:

We are asked to draw a line segment of any length, bisect it and again bisect it such that it is equals to

We will follow certain algorithm to construct the figure

Steps of construction

STEP1: Draw line segment AB of any length.

STEP2: With centre A and radius greater than half of AB, draw two arcs one on each side of AB.

STEP3: With centre B and taking same radius, draw two arcs one on each side of AB, intersecting the previous two arcs at C and D respectively.

STEP4: Draw a line segment having end-points C and D. Segment CD is the perpendicular bisector of AB. Let CD intersects AB at M.

STEP5: With centre A and radius greater than half of AM, draw two arcs one on each side of AM.

STEP6: With centre M and taking same radius, draw two arcs one on each side of AM, intersecting the previous two arcs at E and F respectively.

STEP7: Draw a line segment having end-points E and F. Segment EF is the perpendicular bisector of AM. Let EF intersects AM at N.

.

Disclaimer: In the question, instead of $\frac{1}{2}\left(AB\right)$, there should be $\frac{1}{4}\left(AB\right)$.

#### Page No 16.4:

#### Question 7:

Draw a line segment AB and by ruler and compasses, obtain a line segment of length $\frac{3}{4}$ (*AB*).

#### Answer:

We are asked to draw a line segment AB of any length and get a line segment of

Length

We will follow the following algorithm for the construction

Steps of construction

STEP1: Draw line segment AB of any length.

STEP2: With centre A and radius greater than half of AB, draw two arcs one on each side of AB.

STEP3: With centre B and taking same radius, draw two arcs one on each side of AB, intersecting the previous two arcs at C and D respectively.

STEP4: Draw a line segment having end-points C and D. Segment CD is the perpendicular bisector of AB. Let CD intersects AB at M.

STEP5: With centre A and radius greater than half of AM, draw two arcs one on each side of AM.

STEP6: With centre M and taking same radius, draw two arcs one on each side of AM, intersecting the previous two arcs at E and F respectively.

STEP7: Draw a line segment having end-points E and F. Segment EF is the perpendicular bisector of AM. Let EF intersects AM at N

Here,

#### Page No 16.9:

#### Question 1:

Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠*BAC*.

#### Answer:

We are asked to draw an angle BAC and construct another angle equal to angle BAC

We follow a certain algorithm to construct the angle

Steps of Construction

STEP1: Draw an angle of any measure, and label it as ∠BAC. We have to construct another angle equal to this angle.

STEP2: Draw a ray LN.

STEP3: With centre A and radius equal to AB, draw an arc cutting the ray AC at point D.

STEP4: With centre L and taking the same radius as AB, draw an arc cutting the ray LN at point P.

STEP5: With centre P, and radius equal to BD, draw an arc intersecting the arc drawn in step 4, at point M.

STEP6: Draw ray LM.

#### Page No 16.9:

#### Question 2:

Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.

#### Answer:

We are asked to construct an angle and bisect it and measure each angle

We will follow a certain algorithm to construct

Steps of construction

STEP1: Draw an obtuse angle of any measure. For example take 100 degrees. And label it as ∠BAC.

STEP2: With centre as A, and radius of any measure, draw an arc intersecting the ray AB and AC, at L and M respectively.

STEP3: With centre M and radius greater than half of LM, draw an arc inside the cone.

STEP4: With centre L and taking the same radius as in STEP 3, draw another arc intersecting the previous one at N.

STEP5: Draw the ray AN. This is the angle bisector of BAC.

After measurement, we can find out that, $\angle \mathrm{BAN}=\angle \mathrm{NAC}=50\xb0$.

#### Page No 16.9:

#### Question 3:

Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.

#### Answer:

We are asked to draw an angle of 108°, using protractor and obtain an angle of 54°

We will follow certain algorithm for this construction

Steps of construction

STEP1: Using the protractor, draw ∠ABC of measure 108°.

STEP2: With centre B, and taking any radius, draw an arc, intersecting the ray BA and the ray BC at point D and E respectively.

STEP 3: With D and E as centre and radius greater than half of DE, draw arcs to intersect each other at say F.

STEP 4: Draw the ray BF. This is the angle bisector of ∠ABC. So ∠*FBC* = 54°.

STEP 5: With centre B and taking any radius, draw an arc intersecting the ray BF and the ray BC at G and H respectively.

STEP 6: Draw a ray MN.

STEP 7: With centre M and taking the same radius as in STEP 5, draw an arc intersecting the ray MN at point P.

STEP 8: With centre P and taking radius equal to HG, draw an arc intersecting the arc drawn in STEP 7, at point L.

STEP 9: Draw the ray ML.

∠LMN is the desired angle of measure 54°.

#### Page No 16.9:

#### Question 4:

Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.

#### Answer:

We are asked to construct an angle of 90° using protractor and bisect it

We will follow the following algorithm to construct this figure

Steps of Construction

STEP1: Using the protractor, draw a right angle and label it as ∠ABC.

STEP2: With centre as B, draw an arc of any radius, intersecting the ray BA at D and the ray BC at E.

STEP3: With centre as D, and taking radius greater than half of DE, draw an arc inside the ∠ABC.

STEP4: With centre as E, and taking the same radius, draw another arc, intersecting the previous arc at F.

STEP5: Draw the ray BF.

#### Page No 16.9:

#### Question 5:

Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.

#### Answer:

We are asked to draw a linear pair of angles and bisect each of them and verify that the two bisecting rays are perpendicular to each other

We follow the following algorithm for construct

Steps of construction

STEP1: Draw a linear pair of angles and label them as ABD and DBC.

STEP2: Draw an arc of sufficient radius, intersecting the ray BA, ray BD and the ray BC at points G, H and I respectively.

STEP3: With centre I and H and radius greater than half of HI draw arcs intersecting each other at point say F.

STEP4: With centre G and H as centres and radius as more than half of GH, draw arcs intersecting each other at point say E.

STEP5: Draw the ray BE and the ray BF.

After the measurement, it can be verified that EBF is a right angle.

Hence the ray BE and the ray BF are perpendicular to each other.

#### Page No 16.9:

#### Question 6:

Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line.

#### Answer:

We are asked to draw a pair of vertically opposite angles bisect each of them and verify that the two of them are in the sane line

We will follow the following algorithm for construction

Steps of construction

STEP1: Draw a pair of vertically opposite angles, ABC and DBE.

STEP2: With B as a centre, and taking any radius, draw an arc intersecting the ray BD and the ray BE at points F and G, respectively.

STEP3: With G as a centre, and radius greater than half of FG, draw an arc inside of DBE.

STEP4: With F as a centre, and taking the same radius as in STEP 3, draw an arc intersecting the arc drawn in STEP 3, at N.

STE 5: Draw the ray BN.

STEP6: With B as a centre, and taking any radius, draw an arc intersecting the ray BA and the ray BC at points H and K, respectively.

STEP7: With H as a centre, and radius greater than half of HK, draw an arc inside of ABC.

STEP8: With K as a centre, and taking the same radius as in STEP 7, draw an arc intersecting the arc drawn in STEP 7, at M

STEP9: Draw the ray BM

After measuring the MBN, it can be verified that the bisecting rays BM and BN are in the same line.

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