Rs Aggarwal 2019 Solutions for Class 9 Math 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 Math Geometrical Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 9 Math Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Page No 514:

#### Question 1:

Draw a line segment *AB *= 5.6 cm and draw its perpendicular bisector. Measure the length of each part.

#### 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:

#### Question 2:

Draw an angle of 80° with the help of a protractor and bisect it. Measure each part of the bisected angle.

#### 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:

#### Question 3:

Construct an angle of 90° using ruler and compasses and bisect it.

#### 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:

#### Question 4:

Construct each of the following angles, using ruler and compasses:

(i) 75°

(ii) 37.5°

(iii) 135°

(iv) 105°

(v) 22.5°

#### 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:

#### Question 5:

Construct a âˆ†*ABC* in which *BC* = 5 cm, *AB* = 3.8 cm and *AC* = 2.6 cm. Bisect the largest angle of this triangle.

#### 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:

#### Question 6:

Construct a ΔABC in which BC = 4.8 cm, ∠B = 45° and ∠C = 75°. Measure ∠A.

#### 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:

#### Question 7:

Construct an equilateral triangle each of whose sides measures 5 cm.

#### 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:

#### Question 8:

Construct an equilateral triangle each of whose altitudes measures 5.4 cm. Measure each of its sides.

#### 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:

#### Question 9:

Construct a right-angled triangle whose hypotenuse measures 5 cm and the length of one of whose sides containing the right angle measures 4.5 cm.

#### 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:

#### Question 10:

Construct a âˆ†*ABC *in which *BC *= 4.5 cm, ∠*B *= 45° and *AB *+ *AC *= 8 cm. Justify your construction.

#### 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:

#### Question 11:

Construct a âˆ†*ABC *in which AB = 5.8 cm, ∠*B *= 60° and *BC *+ *CA *= 8.4 cm. Justify your construction.

#### 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:

#### Question 12:

Construct a âˆ†*ABC *in which *BC *= 6 cm, ∠*B *= 30° and *AB *– *AC *= 3.5 cm. Justify your construction.

#### 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:

#### Question 13:

Construct a âˆ†*ABC *in which base *AB *= 5 cm, ∠*A *= 30° and *AC *– *BC *= 2.5 cm. Justify your construction.

#### 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:

#### Question 14:

Construct a âˆ†*PQR* whose perimeter is 12 cm and the lengths of whose sides are in the ratio 3 : 2 : 4.

#### 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:

#### Question 15:

Construct a triangle whose perimeter is 10.4 cm and the base angles are 45° and 120°.

#### 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:

#### Question 16:

Construct a âˆ†*ABC* whose perimeter is 11.6 cm and the base angles are 45° and 60°

#### 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:

#### Question 17:

In each of the following cases, given reasons to show that the construction of âˆ†*ABC *is not possible:

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

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

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

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

#### 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:

#### Question 18:

Construct an angle of 67.5° by using the ruler and compasses.

#### 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:

#### Question 19:

Construct a square of side 4 cm.

#### 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:

#### Question 20:

Construct a right triangle whose one side is 3.5 cm and the sum of the other side and the hypotenuse is 5.5 cm.

#### 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:

#### Question 21:

Construct a âˆ†*ABC *in which ∠*B* = 45°, ∠*C* = 60° and the perpendicular from the vertex *A* to base *BC *is 4.5 cm.

#### 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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