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 Meritnation’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 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
AOC = 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 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 YOR = 90°.
5. Draw the bisector of 
YOR = 90° cutting the semi circle at point S. 
6. With S and T as centres draw two arcs intersecting at point A. 
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 YOR = 90°.
5. Draw the bisector of 
YOR = 90° 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 AOY. 
8. 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 YAC = 90°.
5. Draw AB, bisector of XAC.
Thus, YAB = 135°

(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 YOS = 90°.
5. Draw RO, bisector of XOS.
6. Draw AO, bisector of 
ROS. 
AOY = 105° is the required angle. 

(v) 22.5°


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 514:

Question 5:

Construct a ∆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 CBX = 45°.
3. Construct BCY = 75°.
4. The ray BX and CY intersect at A.
Thus, ABC is the required triangle.
When we measure A, we get A = 60°.

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, 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 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. 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 ∆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 CBX = 45°.
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.

Justification:
In APC,
ACP = 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 ∆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 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 ∆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 CBX = 30°.
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, 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 ∆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 BAX = 30°.
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, 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 ∆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 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° and bisect it to get QPX.
3. Construct an angle of 120° and bisect it to get PQY.
4. The ray XP and YQ intersect at A.
5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.
6. Join AB and AC.
Thus, ABC is the required triangle.

Page No 515:

Question 16:

Construct a ∆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° and bisect it to get QPX.
3. Construct an angle of 60° and bisect it to get PQY.
4. The ray XP and YQ intersect at A.
5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.
6. Join AB and AC.
Thus, ABC is the required triangle.

Page No 515:

Question 17:

In each of the following cases, given reasons to show that the construction of ∆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° 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 POS = 90°. 
3. Draw OQ, bisector of POT. Thus, QOS135° is obtained.
4. Draw the bisector of QOS. 
5. 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 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 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 CBX = 90°
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. 
ABC is the required triangle. 

Page No 515:

Question 21:

Construct a ∆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  XY.
3. Along PO, set off OA = 4.5 cm.
4. Draw a line LM  XY.
5. Draw LAB = 45° and MAC = 60°, meeting XY at B and C, respectively.
Thus, ABC is the required triangle.



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