Rs Aggarwal 2017 Solutions for Class 9 Math Chapter 12 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 2017 Book of Class 9 Math Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Page No 474:

#### Question 1:

Draw a line segment *AB* = 5 cm and draw its perpendicular bisector.

#### Answer:

Steps of construction:

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

2. With *A* as the centre and a radius equal to more than half of *AB, *draw two arcs: one above *AB* and the other below *AB*.

3. With *B* as the centre and the same radius same as above, draw two arcs cutting the previously drawn arcs at *M* and *N*, respectively.

4. Join *MN* intersecting *AB* at *P*.

Thus, *MN* perpendicularly bisects *AB* at *P*.

#### Page No 474:

#### Question 2:

Draw an angle of 45° using scale and compasses only. Draw the bisector of this angle.

#### Answer:

Steps of construction:

1. Draw a line segment *AB*.

2. Draw an angle $\angle $*BAC* = 90$\xb0$.

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

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

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

6. Join *AN* cutting the arc at *Z* and produced it to* X *to make the ray* AX*.

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

8. With *Z* as the centre and the same radius as above, draw another arc cutting the previously drawn arc at *M* .

9. Join *AM*.

Thus, *AM *bisects $\angle $*BAX.*

#### Page No 474:

#### Question 3:

Construct an angle of 90° and draw its bisector.

#### 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 474:

#### Question 4:

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 474:

#### Question 5:

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

#### 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.

#### Page No 474:

#### Question 6:

Construct a âˆ†*ABC* in which *BC* = 5 cm, *AB* = 3.8 cm and *AC* = 2.6 cm.

#### 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.

#### Page No 474:

#### Question 7:

Construct a âˆ†*ABC* in which *BC* = 4.7 cm, ∠*B* = 60° and ∠*C* = 30°. Measure ∠*A*.

#### Answer:

Steps of construction:

1. Draw a line segment *BC* = 4.7 cm.

2. Construct $\angle $*CBX* = 60$\xb0$.

3. Construct $\angle $*BCY* = 30$\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 $*A* = 90$\xb0$.

#### Page No 474:

#### Question 8:

Construct an isosceles âˆ†*PQR* whose base measures 5 cm and each of equal sides measures 4.5 cm.

#### Answer:

Steps of construction:

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

2. With *P* as the centre and radius equal to 4.5 cm, draw an arc.

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

4. Join *PR* and *QR*.

Thus, *PQR *is the required triangle.

#### Page No 474:

#### Question 9:

Construct an isosceles triangle whose base is 4.8 cm and whose vertical angle is 80°.

#### Answer:

Steps of construction:

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

2. Construct $\angle $*CBX = *80$\xb0$.

4. Make $\angle $*XBY = *90$\xb0$.

5. Draw the right bisector *PQ* of* BC* intersecting *BY* at *O*.

6. With O as the centre and radius *OB*, draw a arc intersecting *PQ* at *A*.

7. Join *AC *and *AB**.*

Thus, *ABC** *is the required triangle.

#### Page No 474:

#### Question 10:

Construct a right-angled triangle whose hypotenuse measures 5.3 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 *AB* = 5.3 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 474:

#### Question 11:

Construct a âˆ†*ABC* in which ∠*B* = 30°, ∠*C* = 60° and the length of the perpendicular from the vertex *A* to the base *BC* is 4.8 cm.

#### Answer:

Steps of construction:

1. Draw a line segment *XY*.

2. Take a point *D* on *XY* and draw *PD *$\perp $ XY.

3. Along *PD*, set off *DA* = 4.8 cm.

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

5. Draw $\angle $*LAB = *30$\xb0$ and $\angle $*MAC = *60$\xb0$, meeting *XY* at *B* and *C, *respectively.

Thus, *ABC* is the required triangle.

#### Page No 474:

#### Question 12:

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 474:

#### Question 13:

Construct a âˆ†*ABC* in which *BC* = 4.5 cm, ∠*B* = 60° and the sum of the other two sides is 8 cm.

#### Answer:

Steps of construction:

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

2. Construct $\angle $*CBX* = 60$\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.

#### Page No 474:

#### Question 14:

Construct a âˆ†*ABC* in which *BC* = 5.2 cm, ∠*B* = 30° and the difference of the other two sides is 3.5 cm.

#### Answer:

Steps of construction:

1. Draw a line segment *BC* = 5.2 cm.

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

3. Set off *BP* = 3.5 cm.

4. Join *PC*.

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

6. Join *AC*.

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

#### Page No 475:

#### Question 1:

Which of the following angles can be constructed using ruler and compass only?

(a) 25°

(b) 50°

(c) 22.5°

(d) 42.5°

#### Answer:

(c) 22.5°

We can draw 45° with the help of a ruler and a compass and bisect it to get 22.5°.

#### Page No 475:

#### Question 2:

Which of the following angles can be constructed using ruler and compass only?

(a) 65°

(b) 72°

(c) 80°

(d) 67.5°

#### Answer:

(d) 67.5°

It is because we can draw (90°+ 45° = 135°) with the help of a ruler and a compass and bisect it to get 67.5°.

#### Page No 475:

#### Question 3:

Which of the following angles cannot be constructed using ruler and compass only?

(a) 40°

(b) 120°

(c) 135°

(d) 37.5°

#### Answer:

(a) 40°

It is because we cannot draw an angle of 80° using a compass and a ruler (half of 80° is 40°).

#### Page No 475:

#### Question 4:

Which of the following angles cannot be constructed using ruler and compass only?

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

(b) 15°

(c) $52\frac{1}{2}\xb0$

(d) $32\frac{1}{2}\xb0$

#### Answer:

(d) $32\frac{1}{2}\xb0$

It is because we can not draw 65° using a compass and a ruler (half of 65° is $32\frac{1}{2}\xb0$).

#### Page No 475:

#### Question 5:

The construction of a âˆ†*ABC* in which *AB* = 6 cm, ∠*A* = 45° is possible when (*BC* + *AC*) is

(a) 7 cm

(b) 5.8 cm

(c) 5 cm

(d) 4.9 cm

#### Answer:

(a) 7 cm

We know that the sum of any two sides of a triangle is always greater than the third side.

i.e., *BC* + *AC > AB*

Here*, BC* + *AC > *6 cm and 7 > 6

#### Page No 475:

#### Question 6:

The construction of a âˆ†*PQR* in which *QR* = 5.4 cm and ∠*Q* = 60° is not possible when (*PQ* + *QR*) is

(a) 6 cm

(b) 6.5 cm

(c) 5 cm

(d) 7 cm

#### Answer:

(c) 5 cm

Given: QR = 5.4 cm

Hence, *PQ* + *QR *can not be less than 5.4.

#### Page No 475:

#### Question 7:

The construction of a âˆ†*ABC* in which *AB* = 7 cm, ∠*A* = 75° is possible when (*BC* − *AC*) is equal to

(a) 7.5 cm

(b) 7 cm

(c) 8 cm

(d) 6.5 cm

#### Answer:

*(d) 6.5 cm
It is because the difference of any two sides of a triangle is always less than the third side.*

*i.e., BC $-$ AC < AB*

Here,

*BC*$-$

*AC <*7 cm and 6.5 < 7.

#### Page No 476:

#### Question 8:

The construction of a âˆ†*ABC* in which *BC* = 6 cm and ∠*B* = 50° is not possible when (*AB* − *AC*) is equal to

(a) 5.6 cm

(b) 5 cm

(c) 6 cm

(d) 4.8 cm

#### Answer:

(c) 6 cm

It is because the difference of any two sides of a triangle is always less than the third side.

i.e., *AB* $-$*AC < BC
So, AB* $-$

*AC <*6

Thus, the construction of a âˆ†

*ABC*in which

*BC*= 6 cm and ∠

*B*= 50° is not possible when (

*AB*−

*AC*) is equal to 6 cm.

#### Page No 476:

#### Question 9:

Is it possible to construct a triangle whose sides measure 7 cm, 5 cm and 12 cm?

(a) Yes

(b) No

#### Answer:

(b) No

It is because the sum of any two sides of a triangle is always greater than the third side.

Here, the sum of two sides (i.e., 7+5) is equal to the third side (12 cm).

#### Page No 476:

#### Question 10:

Is it possible to construct a triangle whose sides measure 6 cm, 5 cm and 10 cm?

(a) Yes

(b) No

#### Answer:

(a) Yes

It is because the sum of any two sides of a triangle is always greater than the third side.

#### Page No 476:

#### Question 11:

Is it possible to construct a âˆ†*ABC* in which *BC* = 5 cm, ∠*B* = 120° and ∠*C* = 60°?

(a) Yes

(b) No

#### Answer:

(b) No

It is because the sum of two angles of a triangle is always less than 180°.

Here, the sum of $\angle B\mathrm{and}\angle C$ is 180°.

#### Page No 476:

#### Question 12:

Is it possible to construct a âˆ†*ABC* in which ∠*A* = 60°, ∠*B* = 70° and ∠*C* = 60°?

(a) Yes

(b) No

#### Answer:

(b) No

It is because, the sum of the angles of a triangle cannot be greater than 180°.

Here, the sum of $\angle A,\angle B\mathrm{and}\angle C$ is 190°, which is greater than 180°.

#### Page No 476:

#### Question 13:

Is it possible to construct an angle of 35° using ruler and compass only?

(a) Yes

(b) No

#### Answer:

(b) No

It is because we cannot draw an angle of 70° using a compass and a ruler (here, half of 70° is 35°).

#### Page No 476:

#### Question 14:

Is it possible to construct an angle of 67.5° using ruler and compass only?

(a) Yes

(b) No

#### Answer:

(a) Yes

We can draw an angle of 135° using a compass and a ruler and bisect it to get 67.5°.

#### Page No 476:

#### Question 1:

Is it possible to construct a âˆ†*ABC* in which *BC* = 5 cm, ∠*B* = 60°, ∠*C* = 60°?

(a) Yes

(b) No

#### Answer:

(a) Yes

It is because the sum of two angles of a triangle is always less than 180°.

Here, the sum of $\angle B\mathrm{and}\angle C$ is 120°, which is less than 180°.

#### Page No 477:

#### Question 2:

Is it possible to construct a âˆ†*ABC* in which *AB* = 5 cm, *BC* = 5 cm and *AC* = 10 cm?

(a) Yes

(b) No

#### Answer:

(b) No

It is because the sum of any two sides of a triangle is always greater than the third side.

Here, the sum of two sides (i.e., 5+5 = 10 cm) is equal to the third side (10 cm).

#### Page No 477:

#### Question 3:

Is it possible to construct an angle of 75° using ruler and compass only?

(a) Yes

(b) No

#### Answer:

(a) Yes

We can draw 60° and 15° using a compass and a ruler.

Therefore, we can draw an angle of 75^{o}.

#### Page No 477:

#### Question 4:

Construct a âˆ†*ABC* whose perimeter is 12 cm and whose base angle are 45° and 60°.

#### Answer:

Steps of construction:

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

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

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

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

5. Draw the right bisectors of *AX* and *AY*, cutting *XY* at *B* and *C*, respectively.

6. Join *AB* and *AC*.

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

#### Page No 477:

#### Question 5:

Construct a âˆ†*ABC* whose perimeter is 15 cm and sides are in the ratio 3 : 4 : 5.

#### Answer:

Steps of construction:

1. Draw a line segment *XY* = 15 cm.

2. Construct an acute angle with *XY* in the downward direction.

3. With *X*, set off (3 + 4 + 5) = 12 equal distances along *XP*.

4. Mark points *L, M *and *N* such that *XL* = 3 units, *LM* = 4 units and *MN* = 5 units.

5. Join *YN*.

6. Through *L* and *M*, draw *LB $\parallel $ YN* and *MC $\parallel $ YN*, cutting* XY *at* B *and *C, *respectively.

7. With *B* as the centre and radius *XB,* draw an arc.

8. With *C* as the centre and radius CY, draw an arc cutting the previously drawn arc at *A.*

9. Join *AB* and *AC.*

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

#### Page No 477:

#### Question 6:

Construct an isosceles triangle whose base is 6 cm and whose vertical angle 18 75°.

#### Answer:

Steps of construction:

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

2. Construct $\angle $*CBX = *75$\xb0$.

4. Make $\angle $*XBY = *90$\xb0$.

5. Draw the right bisector *PQ* of* BC*, intersecting *BY* at *O*.

6. With *O* as the centre and radius *OB*, draw a circle intersecting *PQ* at *A*.

7. Join *AC *and *AB**.*

Thus, *ABC** *is the required triangle.

#### Page No 477:

#### Question 7:

Draw a right-angled triangle having hypotenuse = 6 cm and one of the sides containing the right angle having length 4.5 cm.

#### Answer:

Steps of construction:

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

2. Find the mid point *O* of *BC*.

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

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

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

Thus, $\u2206$ABC* *is the required triangle.

#### Page No 477:

#### Question 8:

Construct a âˆ†*ABC* in which ∠*B* = 60°, ∠*C* = 45° and the length of perpendicular from vertex *A* to base *BC* as 5 cm.

#### Answer:

Steps of construction:

1. Draw a line *XY*.

2. Take a point *D* on *XY* and draw *PD *$\perp $ XY.

3. Along *PD*, set off *DA* = 5 cm.

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

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

Thus, *ABC** *is the required triangle.

#### Page No 477:

#### Question 9:

Construct an angle of $22\frac{1}{2}\xb0$.

#### Answer:

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 477:

#### Question 10:

Construct an angle of 135°.

#### Answer:

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$

#### Page No 477:

#### Question 11:

Construct an equilateral triangle of side 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 477:

#### Question 12:

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 477:

#### Question 13:

Draw a line segment *AB* of length 5.2 cm and construct the perpendicularr bisector of *AB*.

#### Answer:

Steps of construction:

1. Draw a line segment *AB* = 5.2 cm.

2. With *A* as the centre and a radius more than half of *AB, *draw two arcs: one above *AB* and the other below *AB*.

3. With *B* as the centre and the same radius as above, draw two arcs, cutting the previously drawn arcs at points *C* and *D. *

4. Join *CD*, intersecting *AB* at *P*.

Thus, *CD* bisects* AB* perpendicularly at point *P*.

#### Page No 477:

#### Question 14:

Construct an angle of 60°and bisect it.

#### Answer:

Steps of construction:-

1. Draw a line segment *AB*.

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

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

4. Join *AQ*.

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

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

7. Join *AD*.

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

#### Page No 477:

#### Question 15:

Construct a âˆ†*ABC* in which base *BC* = 5.2 cm, ∠*B* = 60° and (*AB* + *AC*) = 7.6 cm.

#### Answer:

Steps of construction:-

1. Draw a line segment *BC* = 5.2 cm.

2. Construct $\angle $*CBX* = 60$\xb0$.

3. Set off *BP* = 7.6 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.

#### Page No 477:

#### Question 16:

Construct a square each of whose sides measures 3.2 cm.

#### Answer:

Steps of construction:-

1. Draw a line segment *AB* = 3.2 cm.

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

3. Set off *AD* = 3.2 cm and* BC* = 3.2 cm.

4. Join *DC*.

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

#### Page No 477:

#### Question 17:

Construct a âˆ†*ABC* in which base *BC* = 4.8 cm, ∠*B* = 45° and (*AB* − *AC*) = 2.5 cm.

#### Answer:

Steps of construction:-

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

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

3. Set off *BP* = 2.5 cm.

4. Join *PC*.

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

6. Join *AC*.

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

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