# NCERT solution for class 9 maths constructions ( Chapter 11) #### Solution for Exercise 11.1

1. Construct an angle of $$90^\circ$$ at the initial point of a given ray and justify the construction.

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, $$\angle{EOA}$$ = $$60^\circ$$
5. Draw the ray OF passing through D. Then, $$\angle{FOE}$$ = $$60^\circ$$.
6. Next, taking C and D as centres and with the radius more then (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 $$\angle{FOE}$$
i.e., $$\angle{FOG}$$ = $$\angle{EOG}$$ = (1/2)$$\angle{FOE}$$ = (1/2)$$60^\circ$$ = $$30^\circ$$
Thus, $$\angle{GOA}$$ = $$\angle{GOE}$$ + $$\angle{EOA}$$
i.e., $$\angle{GOA}$$ = $$60^\circ$$ + $$30^\circ$$ = $$90^\circ$$.

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

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

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

2. Construct an angle of $$45^\circ$$ at the initial point of a given ray and justify the construction.

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, $$\angle{EOA}$$ = $$60^\circ$$
5. Draw the ray OF passing through D. Then, $$\angle{FOE}$$ = $$60^\circ$$.
6. Next, taking C and D as centres and with the radius more then (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 $$\angle{FOE}$$
i.e., $$\angle{FOG}$$ = $$\angle{EOG}$$ = (1/2)$$\angle{FOE}$$ = (1/2)$$60^\circ$$ = $$30^\circ$$
Thus, $$\angle{GOA}$$ = $$\angle{GOE}$$ + $$\angle{EOA}$$
i.e., $$\angle{GOA}$$ = $$60^\circ$$ + $$30^\circ$$ = $$90^\circ$$.
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 (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 $$\angle{GOA}$$.
Thus, $$\angle{GOJ}$$ = $$\angle{AOJ}$$ = (1/2)$$\angle{GOA}$$ = (1/2)$$90^\circ$$ = $$45^\circ$$

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

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

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

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

3. Construct the angles of the following measurements:
I)$$30^\circ$$
II) $$22(1/2)^\circ$$
III)$$15^\circ$$

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, $$\angle{EOA}$$ = $$60^\circ$$.
4. Taking B and C as centres and with the radius more than (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 $$\angle{EOA}$$ = $$60^\circ$$.
i.e., $$\angle{EOD}$$ = $$\angle{AOD}$$ = (1/2)$$\angle{EOA}$$ = (1/2)$$60^\circ$$ = $$30^\circ$$

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, $$\angle{EOA}$$ = $$60^\circ$$
5. Draw the ray OF passing through D. Then, $$\angle{FOE}$$ = $$60^\circ$$.
6. Next, taking C and D as centres and with the radius more then (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 $$\angle{FOE}$$
i.e., $$\angle{FOG}$$ = $$\angle{EOG}$$ = (1/2)$$\angle{FOE}$$ = (1/2)$$60^\circ$$ = $$30^\circ$$
Thus, $$\angle{GOA}$$ = $$\angle{GOE}$$ + $$\angle{EOA}$$
i.e., $$\angle{GOA}$$ = $$60^\circ$$ + $$30^\circ$$ = $$90^\circ$$.
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 (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 $$\angle{GOA}$$.
Thus, $$\angle{GOJ}$$ = $$\angle{AOJ}$$ = (1/2)$$\angle{GOA}$$ = (1/2)$$90^\circ$$ = $$45^\circ$$.
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 (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 $$\angle{AOJ}$$
i.e., $$\angle{JOM}$$ = $$\angle{AOM}$$ = (1/2) $$\angle{AOJ}$$ = (1/2) $$45^\circ$$ = $$22(1/2)^\circ$$

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, $$\angle{EOA}$$ = $$60^\circ$$
4. Next, taking B and C as centres and with the radius more then (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 $$\angle{EOA}$$.
i.e., $$\angle{EOD}$$ = $$\angle{AOD}$$ = (1/2)$$\angle{EOA}$$ = (1/2)$$60^\circ$$ = $$30^\circ$$
6. Now, taking B and F as centers and with the radius more then (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 $$\angle{AOD}$$.
i.e., $$\angle{DOG}$$ = $$\angle{AOG}$$ = (1/2)$$\angle{AOD}$$ = (1/2)$$30^\circ$$ = $$15^\circ$$

4. Construct the following angles and verify by measuring then by a protractor:
I)$$75^\circ$$
II) $$105^\circ$$
III)$$135^\circ$$

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, $$\angle{EOA}$$ = $$60^\circ$$
5. Draw the ray OF passing through D. Then, $$\angle{FOE}$$ = $$60^\circ$$.
6. Next, taking C and D as centres and with the radius more then (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 $$\angle{FOE}$$
i.e., $$\angle{FOG}$$ = $$\angle{EOG}$$ = (1/2)$$\angle{FOE}$$ = (1/2)$$60^\circ$$ = $$30^\circ$$
8. Next, taking C and H as centres and with the radius more than (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 $$\angle{GOE}$$
i.e., $$\angle{GOI}$$ = $$\angle{EOI}$$ = (1/2)$$\angle{GOE}$$ = (1/2)$$30^\circ$$ = $$15^\circ$$
Thus, $$\angle{IOA}$$ = $$\angle{IOE}$$ + $$\angle{EOA}$$
i.e., $$\angle{IOA}$$ = $$15^\circ$$ + $$60^\circ$$ = $$75^\circ$$
On measuring the IOA by protractor, we find that $$\angle{IOA}$$ = $$75^\circ$$.
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, $$\angle{EOA}$$ = $$60^\circ$$
5. Draw the ray OF passing through D. Then, $$\angle{FOE}$$ = $$60^\circ$$.
6. Next, taking C and D as centres and with the radius more then (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 $$\angle{FOE}$$
i.e., $$\angle{FOG}$$ = $$\angle{EOG}$$ = (1/2)$$\angle{FOE}$$ = (1/2)$$60^\circ$$ = $$30^\circ$$
Thus, $$\angle{GOA}$$ = $$\angle{GOE}$$ + $$\angle{EOA}$$
i.e., $$\angle{GOA}$$ = $$60^\circ$$ + $$30^\circ$$ = $$90^\circ$$.
8. Next, taking H and D as centres and with the radius more then (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 $$\angle{FOG}$$
i.e., $$\angle{FOI}$$ = $$\angle{GOI}$$ = (1/2)$$\angle{FOG}$$ = (1/2)$$30^\circ$$ = $$15^\circ$$
Thus, $$\angle{IOA}$$ = $$\angle{IOG}$$ + $$\angle{GOA}$$
i.e., $$\angle{IOA}$$ = $$15^\circ$$ + $$90^\circ$$ = $$105^\circ$$
On measuring the IOA by protractor, we find that $$\angle{FOA}$$ = $$105^\circ$$.
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, $$\angle{EOA}$$ = $$60^\circ$$
6. Draw the ray OF passing through D. Then, $$\angle{FOE}$$ = $$60^\circ$$.
7. Next, taking C and D as centres and with the radius more then (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 $$\angle{FOE}$$
i.e., $$\angle{FOG}$$ = $$\angle{EOG}$$ = (1/2)$$\angle{FOE}$$ = (1/2)$$60^\circ$$ = $$30^\circ$$
Thus, $$\angle{GOA}$$ = $$\angle{GOE}$$ + $$\angle{EOA}$$
i.e., $$\angle{GOA}$$ = $$60^\circ$$ + $$30^\circ$$ = $$90^\circ$$.
i.e., $$\angle{B'OH}$$ = $$90^\circ$$
9. Next, taking B' and H as centres and with the radius more then (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 $$\angle{B'OG}$$
i.e., $$\angle{B'OI}$$ = $$\angle{GOI}$$ = (1/2)$$\angle{B'OG}$$ = (1/2)$$90^\circ$$ = $$45^\circ$$
Thus, $$\angle{IOA}$$ = $$\angle{IOG}$$ + $$\angle{GOA}$$
i.e., $$\angle{IOA}$$ = $$45^\circ$$ + $$90^\circ$$ = $$135^\circ$$
On measuring the IOA by protractor, we find that $$\angle{IOA}$$ = $$135^\circ$$.
Thus, the construction is verified.

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

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, $$\triangle{ABC}$$ is an equilateral triangle.
Hence, The construction is justified.

#### Solution for Exercise 11.2

1. Construct a $$\triangle{ABC}$$ in which BC = 7 cm, $$\angle{B}$$ = $$75^\circ$$ and AB + AC = 13cm.

Given: In $$\triangle{ABC}$$ in which BC = 7 cm, $$\angle{B}$$ = $$75^\circ$$ and AB + AC = 13cm.
Steps of construction: 1. Draw the base BC = 7 cm.
2. At the point B make an $$\angle{XBC}$$ = $$75^\circ$$.
3. Cut a line segment BD equal to AB + AC = 13 cm from the ray BX.
4. Join DC.
5. Make $$\angle{DCY}$$ = $$\angle{BDC}$$.
6. Let CY intersect BX at A.
Hence, ABC is the required triangle.

2. Construct a $$\triangle{ABC}$$ in which BC = 8 cm, $$\angle{B}$$ = $$45^\circ$$ and AB - AC = 35cm.

Given: In $$\triangle{ABC}$$ in which BC = 8 cm, $$\angle{B}$$ = $$45^\circ$$ and AB - AC = 35cm.
Steps of construction: 1. Draw the base BC = 8cm
2. At the point B make an $$\angle{XBC}$$ = $$45^\circ$$.
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.

3. Construct a $$\triangle{ABC}$$ in which QR = 6 cm, $$\angle{Q}$$ = $$60^\circ$$ and PR - PQ = 2cm.

Given: In $$\triangle{ABC}$$ in which QR = 6 cm, $$\angle{Q}$$ = $$60^\circ$$ and PR - PQ = 2cm.
Steps of construction: 1. Draw the base QR = 6 cm.
2. At the point Q make an $$\angle{XQR}$$ = $$60^\circ$$.
3. Cut line segment QS = PR - PQ (= 2 cm) from the line QX extended on opposite side of line segment QR.
6. Let LM intersect QX at P.
7. Join PR.
Hence, ABC is the required triangle.

4. Construct a $$\triangle{XYZ}$$ in which $$\angle{Y}$$ = $$30^\circ$$, $$\angle{Z}$$ = $$90^\circ$$ and XY + YZ + ZX = 11cm.

Given: In $$\triangle{XYZ}$$ in which $$\angle{Y}$$ = $$30^\circ$$, $$\angle{Z}$$ = $$90^\circ$$ and XY + YZ + ZX = 11cm.
Steps of construction: 1. Draw a line segment BC = XY + YZ + ZX = 11cm.
2. Make $$\angle{LBC}$$ = $$\angle{Y}$$ = 30° and $$\angle{MCB}$$ = $$\angle{Z}$$ = 90°.
3. Bisect $$\angle{LBC}$$ and $$\angle{MCB}$$. 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, $$\triangle{XYZ}$$ is the required triangle.

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

Given: In $$\triangle{ABC}$$, base BC = 12 cm, $$\angle{B}$$ = $$90^\circ$$ and AB + BC = 18 cm. 2. At the point B, make an $$\angle{XBC}$$ = $$90^\circ$$.
Then, $$\triangle{ABC}$$ is the required right triangle.