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 Meritnation’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 BAC = 90°.
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 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 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, 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 XY.
4. From P, set off PA = 5.4 cm, cutting PQ at A.
5. Construct PAB = 30° and PAC = 30°, meeting XY at B and C, respectively.

Thus, ABC is the required triangle.

Page No 474:

Question 6:

Construct a ∆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 ∆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 CBX = 60°.
3. Construct BCY = 30°.
4. The ray BX and CY intersect at A.
Thus, ABC is the required triangle.
When we measure A, we get A = 90°.

Page No 474:

Question 8:

Construct an isosceles ∆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 CBX = 80°.
4. Make XBY = 90°.
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 ∆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 XY.
3. Along PD, set off DA = 4.8 cm.
4. Draw a line LM XY.
5. Draw LAB = 30° and MAC = 60°, meeting XY at B and C, respectively.
Thus, ABC is the required triangle.

Page No 474:

Question 12:

Construct a ∆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 XL = 3 units, LM = 2 units and MN = 4 units.
5. Join NY.
6. Through L and M, draw LQ  NY and MR  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, PQR is the required triangle.

Page No 474:

Question 13:

Construct a ∆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 CBX = 60°.
3. Set off BP = 8 cm.
4. Join PC.
5. Draw the right bisector of PC, meeting BP at A.
6. Join AC.
Thus, ABC is the required triangle.

Page No 474:

Question 14:

Construct a ∆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 CBX = 30°.
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, 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) 12
(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) 2212°
(b) 15°
(c) 5212°
(d) 3212°

Answer:

(d) 3212°
It is because we can not draw 65° using a compass and a ruler (half of 65° is 3212°).

Page No 475:

Question 5:

The construction of a ∆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 ∆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 ∆ABC in which AB = 7 cm, ∠A = 75° is possible when (BCAC) 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 ∆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 ∆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 ∆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 B and 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 A, B and 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 ∆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 B and C is
120°, which is less than 180°.



Page No 477:

Question 2:

Is it possible to construct a ∆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 75o.

Page No 477:

Question 4:

Construct a ∆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° and bisect it to get YXP.
3. Construct an angle of 60° and bisect it to get 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, ABC is the required triangle.

Page No 477:

Question 5:

Construct a ∆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 YN and MC 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, 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 CBX = 75°.
4. Make XBY = 90°.  
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, ABC is the required triangle.

Page No 477:

Question 8:

Construct a ∆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 XY.
3. Along PD, set off DA = 5 cm.
4. Draw a line LM XY.
5. Draw LAB = 60° and MAC = 45° meeting XY at B and C, respectively.
Thus, ABC is the required triangle.

Page No 477:

Question 9:

Construct an angle of 2212°.

Answer:



Steps of construction:
1. Draw a ray AB.
2. Draw an angle BAE = 45°.
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, BAC is the required angle of measure 22.5o.

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 YAC = 90°.
5. Draw AB, bisector of XAC.
Thus, YAB = 135°

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, 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 BAX = 90° and ABY = 90°.
3. Set off AD = 4 cm and BC = 4 cm.
4. Join DC.
Thus, 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 BAC.

Page No 477:

Question 15:

Construct a ∆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 CBX = 60°.
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, 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 BAX = 90° and ABY = 90°.
3. Set off AD = 3.2 cm and BC = 3.2 cm.
4. Join DC.
Thus, ABCD is the required square.

Page No 477:

Question 17:

Construct a ∆ABC in which base BC = 4.8 cm, ∠B = 45° and (ABAC) = 2.5 cm.

Answer:



Steps of construction:-
1. Draw a line segment BC = 4.8 cm.
2. Construct CBX = 45°.
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, ABC is the required triangle.



View NCERT Solutions for all chapters of Class 9