NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Q1 ) Construct an angle of ${90}^{?}$ at the initial point of a given ray and justify the construction.

Answer :

Steps of construction:



1. Taking O as centre and some radius, draw an arc of a circle which intersects OA say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then, $\mathrm{?}EOA$ = ${60}^{?}$

5. Draw the ray OF passing through D. Then, $\mathrm{?}FOE$ = ${60}^{?}$.

6. Next, taking C and D as centres and with the radius more then ($\frac{1}{2}$)CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the $\mathrm{?}FOE$

$?$ $\mathrm{?}FOG$ = $\mathrm{?}EOG$ = (1/2)$\mathrm{?}FOE$ = ($\frac{1}{2}$ )${60}^{?}$ = ${30}^{?}$

Thus, $\mathrm{?}GOA$ = $\mathrm{?}GOE$ + $\mathrm{?}EOA$
$?$ $\mathrm{?}GOA$ = ${60}^{?}$ + ${30}^{?}$ = ${90}^{?}$.


Justification:
i)Join BC.
Then, OC = OB = BC ...(By construction)
$?$ $\mathrm{?}COB$ is an equilateral triangle.

Thus, $\mathrm{?}COB$ = ${60}^{?}$
Also, $\mathrm{?}EOA$ = ${60}^{?}$


ii)Join CD.
Then, OD = OC = CD ...(By construction)
$?$ $\mathrm{?}DOC$ is an equilateral triangle.
Thus, $\mathrm{?}DOC$ = ${60}^{?}$
Also, $\mathrm{?}FOE$ = ${60}^{?}$


iii)Now, join CG and DG.

In $\mathrm{?}ODG$ and $\mathrm{?}OCG$, we have,

OD = OC ...((Radii of the same arc)

DC = CG .
(Arcs of equal radii)

and OG = OG ...(Given)

By SSS criterion, we get,

$\mathrm{?}ODG$ ${\displaystyle ?}$ $\mathrm{?}OCG$

Hence, $\mathrm{?}DOG$ = $\mathrm{?}COG$ ...(By CPCT)

So, $\mathrm{?}FOG$ = $\mathrm{?}EOG$ = ($\frac{1}{2}$ )$\mathrm{?}FOE$ = $\frac{1}{2}$${60}^{?}$ = ${30}^{?}$

Thus, $\mathrm{?}GOA$ = $\mathrm{?}GOE$ + $\mathrm{?}EOA$
$?$ $\mathrm{?}GOA$ = ${60}^{?}$ + ${30}^{?}$ = ${90}^{?}$.

Q2 ) Construct an angle of ${45}^{?}$ at the initial point of a given ray and justify the construction.

Answer :

Steps of construction:



1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then, $\mathrm{?}EOA$ = ${60}^{?}$

5. Draw the ray OF passing through D. Then, $\mathrm{?}FOE$ = ${60}^{?}$.

6. Next, taking C and D as centres and with the radius more then ($\frac{1}{2}$ )CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the $\mathrm{?}FOE$

$?$ $\mathrm{?}FOG$ = $\mathrm{?}EOG$ = (1/2)$\mathrm{?}FOE$ = ($\frac{1}{2}$ )${60}^{?}$ = ${30}^{?}$
$?$ $\mathrm{?}GOA$ = $\mathrm{?}GOE$ + $\mathrm{?}EOA$
$?$ $\mathrm{?}GOA$ = ${60}^{?}$ + ${30}^{?}$ = ${90}^{?}$.

8. Now taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I, respectively.

9. Next, taking H and I as centres and with the radius more than ($\frac{1}{2}$ )HI, draw arcs to intersect each other, say at J.

10. Draw the raw OJ. This ray OJ is the required bisector of the $\mathrm{?}GOA$.

Thus, $\mathrm{?}GOJ$ = $\mathrm{?}AOJ$ = (1/2)$\mathrm{?}GOA$ = ($\frac{1}{2}$ )${90}^{?}$ = ${45}^{?}$


Justification:
i)Join BC.
Then, OC = OB = BC ...(By construction)
$?$ , $\mathrm{?}COB$ is an equilateral triangle.
Thus, $\mathrm{?}COB$ = ${60}^{?}$
Also, $\mathrm{?}EOA$ = ${60}^{?}$

ii)Join CD.
Then, OD = OC = CD ...(By construction)
Therefore, $\mathrm{?}DOC$ is an equilateral triangle.
Thus, $\mathrm{?}DOC$ = ${60}^{?}$
Also, $\mathrm{?}FOE$ = ${60}^{?}$

iii)Now, join CG and DG.
In $\mathrm{?}ODG$ and $\mathrm{?}OCG$, we have,
OD = OC ...((Radii of the same arc)
DC = CG ...(Arcs of equal radii)
and OG = OG ...(Given)
By SSS criterion, we get,
$\mathrm{?}ODG$ ${\displaystyle ?}$ $\mathrm{?}OCG$
Hence, $\mathrm{?}DOG$ = $\mathrm{?}COG$ ...(By CPCT)
So, $\mathrm{?}FOG$ = $\mathrm{?}EOG$ = ($\frac{1}{2}$ )$\mathrm{?}FOE$ = (1/2)${60}^{?}$ = ${30}^{?}$
Thus, $\mathrm{?}GOA$ = $\mathrm{?}GOE$ + $\mathrm{?}EOA$
i.e., $\mathrm{?}GOA$ = ${60}^{?}$ + ${30}^{?}$ = ${90}^{?}$.

iv) Now, join HJ and IJ.
In $\mathrm{?}OIJ$ and $\mathrm{?}OHJ$, we have,
OI = OH ...((Radii of the same arc)
IJ = HJ ...(Arcs of equal radii)
and OJ = OJ ...(Given)
By SSS criterion, we get,
$\mathrm{?}OIJ$ ${\displaystyle ?}$ $\mathrm{?}OHJ$
Hence, $\mathrm{?}IOJ$ = $\mathrm{?}HOJ$ ...(By CPCT)
So, $\mathrm{?}AOJ$ = $\mathrm{?}GOJ$ = (1/2)$\mathrm{?}GOA$ = ($\frac{1}{2}$ )${90}^{?}$ = ${45}^{?}$.

Q3 ) Construct the angles of the following measurements:
I)${30}^{?}$
II) $22\frac{1}{2}{)}^{?}$
III)${15}^{?}$

Answer :

I) Steps of construction:



1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Draw the ray OE passing through C. Then, $\mathrm{?}EOA$ = ${60}^{?}$.

4. Taking B and C as centres and with the radius more than ($\frac{1}{2}$ )BC, draw arcs to intersect each other, say at D.

5. Draw the ray OD, this ray OD is the bisector of the $\mathrm{?}EOA$ = ${60}^{?}$.
$?$ $\mathrm{?}EOD$ = $\mathrm{?}AOD$ = ($\frac{1}{2}$ )$\mathrm{?}EOA$ = ($\frac{1}{2}$ )${60}^{?}$ = ${30}^{?}$


II) Steps of construction:

1.Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.


3. Taking C as centre and with the same radius as before, drawn an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then, $\mathrm{?}EOA$ = ${60}^{?}$

5. Draw the ray OF passing through D. Then, $\mathrm{?}FOE$ = ${60}^{?}$.

6. Next, taking C and D as centres and with the radius more then ($\frac{1}{2}$ )CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the $\mathrm{?}FOE$
$?$ $\mathrm{?}FOG$ = $\mathrm{?}EOG$ = ($\frac{1}{2}$ )$\mathrm{?}FOE$ = $\frac{1}{2}$${60}^{?}$ = ${30}^{?}$
Thus, $\mathrm{?}GOA$ = $\mathrm{?}GOE$ + $\mathrm{?}EOA$
$?$ $\mathrm{?}GOA$ = ${60}^{?}$ + ${30}^{?}$ = ${90}^{?}$.

8. Now taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I, respectively.

9. Next, taking H and I as centres and with the radius more than $\frac{1}{2}$HI, draw arcs to intersect each other, say at J.

10. Draw the raw OJ. This ray OJ is the required bisector of the $\mathrm{?}GOA$.
Thus, $\mathrm{?}GOJ$ = $\mathrm{?}AOJ$ = $\frac{1}{2}$$\mathrm{?}GOA$ = $\frac{1}{2}$${90}^{?}$ = ${45}^{?}$.

11. Now, taking O as centre and any radius, drawn an arc to intersect the rays OA and OJ, say at K and L, respectively.

12. Next, taking K and L as centres and with the radius more than $\frac{1}{2}$KL, draw arcs to intersect each other, say at H.

13. Draw the ray OM. This ray OM is the bisector of the $\mathrm{?}AOJ$
$?$ $\mathrm{?}JOM$ = $\mathrm{?}AOM$ = (1/2) $\mathrm{?}AOJ$ = $\frac{1}{2}$× ${45}^{?}$ = $22{\frac{1}{2}}^{?}$


III) Steps of construction:

1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an intersecting the previously drawn arc, say at a point c.


3. Draw the ray OE passing through C. Then, $\mathrm{?}EOA$ = ${60}^{?}$

4. Next, taking B and C as centres and with the radius more then $\frac{1}{2}$BC, draw arcs to intersect each other, say at D.

5. Draw the ray OD intersecting the arc drawn in step 1 at F. This ray OD is the bisector of the the $\mathrm{?}EOA$.
$?$ $\mathrm{?}EOD$ = $\mathrm{?}AOD$ = $\frac{1}{2}$$\mathrm{?}EOA$ = $\frac{1}{2}$${60}^{?}$ = ${30}^{?}$

6. Now, taking B and F as centers and with the radius more then $\frac{1}{2}$ BF, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the $\mathrm{?}AOD$.
$?$ $\mathrm{?}DOG$ = $\mathrm{?}AOG$ = $\frac{1}{2}$$\mathrm{?}AOD$ = $\frac{1}{2}$${30}^{?}$ = ${15}^{?}$

Q4 ) Construct the following angles and verify by measuring then by a protractor:
I)${75}^{?}$
II) ${105}^{?}$
III)${135}^{?}$

Answer :

I) Steps of construction:



1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then, $\mathrm{?}EOA$ = ${60}^{?}$
5. Draw the ray OF passing through D. Then, $\mathrm{?}FOE$ = ${60}^{?}$.

6. Next, taking C and D as centres and with the radius more then $\frac{1}{2}$CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG intersecting the arc of step 1 at H.. This ray OG is the bisector of the $\mathrm{?}FOE$
$?$ $\mathrm{?}FOG$ = $\mathrm{?}EOG$ = $\frac{1}{2}$$\mathrm{?}FOE$ = $\frac{1}{2}$×${60}^{?}$ = ${30}^{?}$


8. Next, taking C and H as centres and with the radius more than $\frac{1}{2}$CH, draw arcs to intersect each other, say at I.

9. Draw the ray OI. This ray OI is the bisector of the $\mathrm{?}GOE$
$?$ $\mathrm{?}GOI$ = $\mathrm{?}EOI$ = $\frac{1}{2}$$\mathrm{?}GOE$ = $\frac{1}{2}$${30}^{?}$ = ${15}^{?}$
Thus, $\mathrm{?}IOA$ = $\mathrm{?}IOE$ + $\mathrm{?}EOA$
$?$ $\mathrm{?}IOA$ = ${15}^{?}$ + ${60}^{?}$ = ${75}^{?}$

On measuring the IOA by protractor, we find that $\mathrm{?}IOA$ = ${75}^{?}$.
Thus, the construction is verified.


II) Steps of construction:


1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then, $\mathrm{?}EOA$ = ${60}^{?}$

5. Draw the ray OF passing through D. Then, $\mathrm{?}FOE$ = ${60}^{?}$.

6. Next, taking C and D as centres and with the radius more then $\frac{1}{2}$CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG intersecting the arc of step 1 at H.. This ray OG is the bisector of the $\mathrm{?}FOE$
i.e., $\mathrm{?}FOG$ = $\mathrm{?}EOG$ = $\frac{1}{2}$$\mathrm{?}FOE$ = $\frac{1}{2}$×${60}^{?}$ = ${30}^{?}$
Thus, $\mathrm{?}GOA$ = $\mathrm{?}GOE$ + $\mathrm{?}EOA$
$?$ $\mathrm{?}GOA$ = ${60}^{?}$ + ${30}^{?}$ = ${90}^{?}$.

8. Next, taking H and D as centres and with the radius more then ($\frac{1}{2}$HD, draw arcs to intersect each other, say at I.

9. Draw the ray OI. This ray OI is the bisector of the $\mathrm{?}FOG$
$?$ $\mathrm{?}FOI$ = $\mathrm{?}GOI$ = $\frac{1}{2}$$\mathrm{?}FOG$ = $\frac{1}{2}$×${30}^{?}$ = ${15}^{?}$
Thus, $\mathrm{?}IOA$ = $\mathrm{?}IOG$ + $\mathrm{?}GOA$
$?$ $\mathrm{?}IOA$ = ${15}^{?}$ + ${90}^{?}$ = ${105}^{?}$
On measuring the IOA by protractor, we find that $\mathrm{?}FOA$ = ${105}^{?}$.
Thus, the construction is verified.


III) Steps of construction:



1. Produce AO to A' to form ray OA'.

2. Taking O as centre and some radius, draw an arc of a circle which intersects OA at a point B and OA' at a point B'.

3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.

4. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at O.

5. Draw the ray OE passing through C. Then, $\mathrm{?}EOA$ = ${60}^{?}$

6. Draw the ray OF passing through D. Then, $\mathrm{?}FOE$ = ${60}^{?}$.

7. Next, taking C and D as centres and with the radius more then $\frac{1}{2}$CD, draw arcs to intersect each other, say at G.

8. Draw the ray OG intersecting the arc of step 1 at H.. This ray OG is the bisector of the $\mathrm{?}FOE$
i.e., $\mathrm{?}FOG$ = $\mathrm{?}EOG$ = $\frac{1}{2}$$\mathrm{?}FOE$ = $\frac{1}{2}$×${60}^{?}$ = ${30}^{?}$
Thus, $\mathrm{?}GOA$ = $\mathrm{?}GOE$ + $\mathrm{?}EOA$
$?$ $\mathrm{?}GOA$ = ${60}^{?}$ + ${30}^{?}$ = ${90}^{?}$.
$?$ $\mathrm{?}{B}^{?}OH$ = ${90}^{?}$

9. Next, taking B' and H as centres and with the radius more then $\frac{1}{2}$B'H, draw arcs to intersect each other, say at I.

10. Draw the ray OI. This ray OI is the bisector of the $\mathrm{?}{B}^{?}OG$
$?$ $\mathrm{?}{B}^{?}OI$ = $\mathrm{?}GOI$ = $\frac{1}{2}$$\mathrm{?}{B}^{?}OG$ = $\frac{1}{2}$×${90}^{?}$ = ${45}^{?}$
Thus, $\mathrm{?}IOA$ = $\mathrm{?}IOG$ + $\mathrm{?}GOA$
$?$ $\mathrm{?}IOA$ = ${45}^{?}$ + ${90}^{?}$ = ${135}^{?}$
On measuring the IOA by protractor, we find that $\mathrm{?}IOA$ = ${135}^{?}$.
Thus, the construction is verified.

Q5 ) Construct an equilateral triangle, given its side and justify the construction.

Answer :

Steps of construction:


1. Take a ray AX with initial point A. From AX, cut off AB = 4cm.

2. Taking A as centre and radius (= 4 cm), draw an arc of a circle, which intersects AX, say at a point B.

3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

4. Draw the ray AE passing through C.

5. Next, taking B as centre and radius (= 4 cm), draw an arc of a circle, which intersects AX, say at a point A.

6. 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.

7. Draw the ray BF passing through C. Then, ABC is the required triangle with gives side 4 cm.


Justification:
AB = BC ...(By construction)
AB = AC ...(By construction)
Thus, AB = BC = CA
Therefore, $\mathrm{?}ABC$ is an equilateral triangle.
Hence, The construction is justified.

Q1 ) Construct a $\mathrm{?}ABC$ in which BC = 7 cm, $\mathrm{?}B$ = ${75}^{?}$ and AB + AC = 13cm.

Answer :

Given: In $\mathrm{?}ABC$ in which BC = 7 cm, $\mathrm{?}B$ = ${75}^{?}$ and AB + AC = 13cm.
Steps of construction:


1. Draw the base BC = 7 cm.

2. At the point B make an $\mathrm{?}XBC$ = ${75}^{?}$.

3. Cut a line segment BD equal to AB + AC = 13 cm from the ray BX.

4. Join DC.

5. Make $\mathrm{?}DCY$ = $\mathrm{?}BDC$.

6. Let CY intersect BX at A.
Hence, ABC is the required triangle.

Q2 ) Construct a $\mathrm{?}ABC$ in which BC = 8 cm, $\mathrm{?}B$ = ${45}^{?}$ and AB - AC = 35cm.

Answer :

Given: In $\mathrm{?}ABC$ in which BC = 8 cm, $\mathrm{?}B$ = ${45}^{?}$ and AB - AC = 35cm.
Steps of construction:


1. Draw the base BC = 8cm

2. At the point B make an $\mathrm{?}XBC$ = ${45}^{?}$.

3. Cut the line segment BD equal to AB - AC = 3.5 cm from the ray BX.

4. Join DC.

5. Draw the perpendicular bisector, say PQ of DC.

6. Let it intersect BX at a point A.

7. Join AC.
Hence, ABC is the required triangle.

Q3 ) Construct a $\mathrm{?}ABC$ in which QR = 6 cm, $\mathrm{?}Q$ = ${60}^{?}$ and PR - PQ = 2cm.

Answer :

Given: In $\mathrm{?}ABC$ in which QR = 6 cm, $\mathrm{?}Q$ = ${60}^{?}$ and PR - PQ = 2cm.
Steps of construction:


1. Draw the base QR = 6 cm.

2. At the point Q make an $\mathrm{?}XQR$ = ${60}^{?}$.

3. Cut line segment QS = PR - PQ (= 2 cm) from the line QX extended on opposite side of line segment QR.

4. Let LM intersect QX at P.

5. Join PR.
Hence, ABC is the required triangle.

Q4 ) Construct a $\mathrm{?}XYZ$ in which $\mathrm{?}Y$ = ${30}^{?}$, $\mathrm{?}Z$ = ${90}^{?}$ and XY + YZ + ZX = 11cm.

Answer :

Given:
In $\mathrm{?}XYZ$ in which $\mathrm{?}Y$ = ${30}^{?}$, $\mathrm{?}Z$ = ${90}^{?}$ and XY + YZ + ZX = 11cm.
Steps of construction:



1. Draw a line segment BC = XY + YZ + ZX = 11cm.

2. Make $\mathrm{?}LBC$ = $\mathrm{?}Y$ = 30° and $\mathrm{?}MCB$ = $\mathrm{?}Z$ = 90°.

3. Bisect $\mathrm{?}LBC$ and $\mathrm{?}MCB$. Let these bisectors meet at a point X.

4. Draw perpendicular bisectors DE of XB and FG of XC.

5. Let DE intersect BC at Y and FC intersect BC at Z.

6. Join XY and XZ.
Then, $\mathrm{?}XYZ$ is the required triangle.

Q5 ) Construct right triangles whose base is 12 cm and sum its hypotenuse and other side is 18 cm.

Answer :

Given: In $\mathrm{?}ABC$, base BC = 12 cm, $\mathrm{?}B$ = ${90}^{?}$ and AB + BC = 18 cm.

Steps of construction:



1. Draw the base BC =12cm.

2. At the point B, make an $\mathrm{?}XBC$ = ${90}^{?}$.

3. Cut a line segment BD = AB + AC = 18 cm from the ray BX.

4. Join DC.

5. Draw the perpendicular bisector PQ of CD to intersect BD at a point A.

6. Join AC.
Then, $\mathrm{?}ABC$ is the required right triangle.