Rs Aggarwal 2021 2022 Solutions for Class 9 Maths Chapter 13 Geometrical Constructions are provided here with simple step-by-step explanations. These solutions for Geometrical Constructions are extremely popular among Class 9 students for Maths Geometrical Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 9 Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 514:

#### Answer:

**Steps of Construction**

1. Draw a line AB = 5.6 cm.

2. With A as centre and radius more than half of AB, draw one above and other below line AB.

3. Similarly, with B as centre draw two arcs cutting the previous drawn arcs and name the points obtained as M and N respectively.

4. Join MN intersecting AB at point O.

MN is the required perpendicular bisector.

AO = OB = 2.8 cm

#### Page No 514:

#### Answer:

**Steps of construction**

1. Draw $\angle $AOB = 80° using protractor.

2. With O as centre and a convenient radius, draw an arc cutting AO at N and OB at M.

3. With N as centre and a convenient radius, draw an arc.

4. Similarly, with M as centre and same radius, cut the previous drawn arc and name it as point C.

5. Join OC.

OC is the required angle bisector.

On measuring we get

$\angle $AOC = $\angle $BOC = 40°

#### Page No 514:

#### Answer:

Steps of construction:

1. Draw a line segment *AB*.

2. With *A* as the centre and and a small radius, draw an arc cutting *AB* *at M*.

3. With *M* as the centre and the same radius as above, draw an arc cutting the previously drawn arc at *N*.

4. With *N* as the centre and the same radius as above, draw an arc cutting the previously drawn arc at *P*.

5. Again, with *N* as the centre and a radius more than half of *PN,* draw an arc.

6. With *P* as the centre and the same radius as above, draw an arc cutting the previously drawm arc at *Q*.

7. Join *AQ* cutting the arc at *O* and produced it to C.

8. With *O* as the centre and a radius more than half of *OM*, draw an arc.

9. With *M* as the centre and the same radius as above, draw another arc cutting the previously drawn arc at point *R*.

10. Join *AR*.

Thus, *AR *bisects $\angle $*BAC.*

#### Page No 514:

#### Answer:

(i) 75°

Steps of construction

1. Draw a line XY.

2. Take a point O on XY.

3. With O as centre, draw a semi circle, cutting *XY* at *P* and *Q*.

4. Construct $\angle $YOR* = *90$\xb0$.

5. Draw the bisector of $\angle $YOR* = *90$\xb0$ cutting the semi circle at point S.

6. With S and T as centres draw two arcs intersecting at point A.

$\angle $AOY = 75°.

(ii) 37.5°

**Steps of construction**

1. Draw a line XY.

2. Take a point O on XY.

3. With O as centre, draw a semi circle, cutting *XY* at *P* and *Q*.

4. Construct $\angle $YOR* = *90$\xb0$.

5. Draw the bisector of $\angle $YOR* = *90$\xb0$ cutting the semi circle at point S.

6. With S and T as centres draw two arcs intersecting at point A.

7. Draw the angle bisector of $\angle $AOY.

8. $\angle $BOY is the required angle of 37.5°.

(iii) 135°

Steps of construction:

1. Draw a line *XY*.

2. Take a point *A* on *XY*.

3. With *A* as centre, draw a semi circle, cutting *XY* at *P* and *Q*.

4. Construct $\angle $*YAC = *90$\xb0$.

5. Draw *AB,* bisector of $\angle $*XAC*.

Thus, $\angle $*YAB = *135$\xb0$

(iv) 105°

Steps of construction

1. Draw a line *XY*.

2. Take a point O on *XY*.

3. With O as centre, draw a semi circle, cutting *XY* at *P* and *Q*.

4. Construct $\angle $YOS* = *90$\xb0$.

5. Draw RO*,* bisector of $\angle $XOS.

6. Draw AO, bisector of $\angle $ROS.

$\angle $AOY = 105° is the required angle.

(v) 22.5°

Steps of construction:

1. Draw a ray *AB*.

2. Draw an angle $\angle $*BAE* = 45$\xb0$.

3. With *A* as the centre and a small radius, draw an arc cutting *AB* *at P *and *AE *at* Q*.

4. With *P* as the centre and a radius more than half of *PQ*, draw an arc.

5. With *Q* as the centre and the same radius as above, draw another arc cutting the previously drawn arc at *D*.

6. Join *AD*.

Thus, $\angle $*BAC *is the required angle of measure 22.5^{o}.

#### Page No 514:

#### Answer:

Steps of construction:

1. Draw a line segment *BC* = 5 cm.

2. With *B* as the centre and a radius equal to 3.8 cm, draw an arc.

3. With* C* as the centre and a radius equal to 2.6 cm, draw an arc cutting the previously drawn arc at *A*.

4. Join *AB* and *AC*.

Thus, *ABC *is the required triangle.

Take the largest angle and draw the angle bisector.

#### Page No 514:

#### Answer:

Steps of construction:

1. Draw a line segment *BC* = 4.8 cm.

2. Construct $\angle $*CBX* = 45$\xb0$.

3. Construct $\angle $*BCY* = 75$\xb0$.

4. The ray* BX* and *CY* intersect at *A.*

Thus, $\u25b3$*ABC *is the required triangle.

When we measure $\angle $*A, *we get* *$\angle $*A* = 60$\xb0$.

#### Page No 514:

#### Answer:

Steps of construction:

1. Draw a line segment *AB* = 5 cm.

2. With *A* as the centre and a radius equal to *AB*, draw an arc.

3. With *B* as the centre and the same radius as above, draw another arc cutting the previously drawn arc at *C*.

4. Join *AC* and *BC*.

Thus, $\u25b3$*ABC *is the required triangle.

#### Page No 514:

#### Answer:

Steps of construction:

1. Draw a line* XY*.

2. Mark any point *P*.

3. From *P* draw *PQ *$\perp $ *XY.*

4. From *P*, set off *PA* = 5.4 cm, cutting* PQ* at *A*.

5. Construct $\angle $*PAB* = 30$\xb0$ and $\angle $*PAC = *30$\xb0$, meeting *XY* at *B *and *C,* respectively.

Thus, $\u25b3$*ABC *is the required triangle. Measure of each side is 6 cm.

#### Page No 514:

#### Answer:

**Steps of construction:**

1. Draw a line segment *BC* = 5 cm.

2. Find the midpoint *O* of *BC*.

3. With *O* as the centre and radius *OB*, draw a semicircle on *BC*.

4. With *B* as the centre and radius equal to 4.5 cm, draw an arc cutting the semicircle at *A*.

4. Join *AB *and* AC**.*

Thus, ABC* *is the required triangle.

#### Page No 514:

#### Answer:

Steps of construction:

1. Draw a line segment *BC* = 4.5 cm.

2. Construct $\angle $*CBX* = 45$\xb0$.

3. Set off *BP* = 8 cm.

4. Join *PC*.

5. Draw the right bisector of *PC*, meeting *BP* at *A*.

6. Join *AC*.

Thus, $\u25b3$*ABC *is the required triangle.

Justification:

In $\u25b3$APC,

$\angle $ACP = $\angle $APC (By construction)

⇒ AC = AP (Sides opposite to equal angles are equal)

Now

AB = BP − AP = BP − AC

⇒ AB + AC = BP

#### Page No 514:

#### Answer:

**Steps of construction:**

1. Draw a line segment *AB* = 5.8 cm.

2. Construct *ABX* = 60.

3. Set off *BP* = 8.4 cm.

4. Join *PA.*

5. Draw the right bisector of *PA*, meeting *BP* at *C*.

6. Join *AC*.

Thus, *ABC *is the required triangle.

**Justification:**

In $\u25b3$APC,

∠CAP = ∠CPA (By construction)

⇒ CP = AC (Sides opposite to equal angles are equal)

Now

BC = PB − PC = PB − AC

⇒ BC + AC = PB

#### Page No 515:

#### Answer:

Steps of construction:

1. Draw a line segment *BC* = 6 cm.

2. Construct $\angle $*CBX* = 30$\xb0$.

3. Set off *BD* = 3.5 cm.

4. Join *DC*.

5. Draw the right bisector of *DC*, meeting *BD* produced at *A*.

6. Join *AC*.

Thus, $\u25b3$*ABC *is the required triangle.

Justification:

Point A lies on the perpendicular bisector of DC.

⇒ AD = AC

Now

BD = AB − AD = AB − AC

#### Page No 515:

#### Answer:

**Steps of construction:**

1. Draw a line segment *AB* = 5 cm.

2. Construct $\angle $*BAX *= 30$\xb0$.

3. Set off *AD* = 2.5 cm.

4. Join *DB*.

5. Draw the right bisector of *DB*, meeting *DB* produced at *C*.

6. Join *CB*.

Thus, $\u25b3$*ABC *is the required triangle.

**Justification:**

Point C lies on the perpendicular bisector of DB.

⇒ CD = BC

Now

AD = AC − DC = AC − BC

#### Page No 515:

#### Answer:

1. Draw a line segment *XY* = 12 cm.

2. In the downward direction, construct an acute angle with *XY* at *X*.

3. From *X*, set off (3 + 2 + 4) = 9 arcs of equal distances along *XZ*.

4. Mark points *L, M *and *N* such that X*L* = 3 units,* LM* = 2 units and *MN* = 4 units.

5. Join *NY*.

6. Through L and M, draw *LQ $\parallel $ NY* and *MR $\parallel $ NY* cutting* XY *at* Q *and *R, *respectively.

7. With *Q* as the centre and radius *QX*, draw an arc.

8. With *R* as the centre and radius *RY*, draw an arc, cutting the previously drawn arc at *P.*

9. Join *PQ* and *PR.*

Thus, $\u25b3$*PQR *is the required triangle.

#### Page No 515:

#### Answer:

Steps of construction:

1. Draw a line segment *PQ* = 10.4 cm.

2. Construct an angle of 45$\xb0$ and bisect it to get $\angle $*QPX*.

3. Construct an angle of 120$\xb0$ and bisect it to get $\angle $*PQY.*

4. The ray* XP* and *YQ* intersect at *A.*

5. Draw the right bisectors of A*P* and *AQ*, cutting *PQ* at *B* and *C*, respectively.

6. Join *AB* and *AC*.

Thus, $\u25b3$*ABC *is the required triangle.

#### Page No 515:

#### Answer:

Steps of construction:

1. Draw a line segment *PQ* = 11.6 cm.

2. Construct an angle of 45$\xb0$ and bisect it to get $\angle $*QPX*.

3. Construct an angle of 60$\xb0$ and bisect it to get $\angle $*PQY.*

4. The ray* XP* and *YQ* intersect at *A.*

5. Draw the right bisectors of A*P* and *AQ*, cutting *PQ* at *B* and *C*, respectively.

6. Join *AB* and *AC*.

Thus, $\u25b3$*ABC *is the required triangle.

#### Page No 515:

#### Answer:

(i) *AB *= 6 cm, ∠*A* = 40° and (*BC* + *AC*) = 5.8 cm.

Here BC + AC is not greater than AB so, this triangle is not possible.

(ii) *AB *= 7 cm, ∠*A* = 50° and (*BC *– *AC*) = 8 cm.

Here *BC *– *AC *is not less than AB so this triangle is not possible.

(iii) *BC *= 5 cm, ∠*B* = 80°, ∠*C* = 50° and ∠*A* = 60°.

Sum of the angles of the given triangle is not equal to 180$\xb0$ so, given triangle is not possible.

(iv) *AB *= 4 cm, *BC *= 3 cm and *AC *= 7 cm.

Sum of two sides of this triangle not greater than third side so given triangle not possible.

#### Page No 515:

#### Answer:

**Steps of Construction:**

1. Draw a line TS.

2. Construct $\angle $POS = 90°.

3. Draw OQ*,* bisector of $\angle $POT. Thus, $\angle $QOS* = *135$\xb0$ is obtained.

4. Draw the bisector of $\angle $QOS.

5. $\angle $ROS thus obtained is equal to 67.5°.

#### Page No 515:

#### Answer:

Steps of construction:

1. Draw a line segment *AB* = 4 cm.

2. Construct $\angle $*BAX* = 90$\xb0$ and $\angle $*ABY* = 90$\xb0$.

3. Set off *AD* = 4 cm and* BC* = 4 cm.

4. Join *DC*.

Thus, $\square $*ABCD *is the required square.

#### Page No 515:

#### Answer:

Steps of construction:

1. Draw a line segment BC = 3.5 cm

2. Construct $\angle $CBX = 90$\xb0$.

3. With B as centre and 5.5 cm radius cut an arc on BX and name it as D.

4. Join CD.

5. Construct the perpendicular bisector of CD intersecting BD at point A.

6. Join AC.

$\u25b3$ABC is the required triangle.

#### Page No 515:

#### Answer:

Steps of construction:

1. Draw a line segment XY.

2. Take a point O on XY and draw PO* *$\perp $ XY.

3. Along PO, set off OA = 4.5 cm.

4. Draw a line LM* *$\parallel $ XY*.*

5. Draw $\angle $LAB* = *45$\xb0$ and $\angle $MAC* = *60$\xb0$, meeting XY at B and C*, *respectively.

Thus, ABC is the required triangle.

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