Rs Aggarwal 2020 2021 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 2020 2021 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 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 514:

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:

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:

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:

(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$°$.
5. Draw the bisector of
$\angle$YOR = 90$°$ 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$°$.
5. Draw the bisector of
$\angle$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 $\angle$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$°$.
5. Draw AB, bisector of $\angle$XAC.
Thus, $\angle$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 $\angle$YOS = 90$°$.
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$°$.
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.
Thus, $\angle$BAC is the required angle of measure 22.5o.

#### Page No 514:

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:

Steps of construction:
1. Draw a line segment BC = 4.8 cm.
2. Construct $\angle$CBX = 45$°$.
3. Construct $\angle$BCY = 75$°$.
4. The ray BX and CY intersect at A.
Thus, $△$ABC is the required triangle.
When we measure $\angle$A, we get $\angle$A = 60$°$.

#### Page No 514:

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:

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$°$ and $\angle$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:

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:

Steps of construction:
1. Draw a line segment BC = 4.5 cm.
2. Construct $\angle$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,
$\angle$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:

Steps of construction:

1. Draw a line segment AB = 5.8 cm.
2. Construct $\angle$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:

Steps of construction:
1. Draw a line segment BC = 6 cm.
2. Construct $\angle$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.
Now
BD = AB − AD = AB − AC

#### Page No 515:

Steps of construction:
1. Draw a line segment AB = 5 cm.
2. Construct $\angle$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:

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 $\parallel$ 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, $△$PQR is the required triangle.

#### Page No 515:

Steps of construction:
1. Draw a line segment PQ = 10.4 cm.
2. Construct an angle of 45$°$ and bisect it to get $\angle$QPX.
3. Construct an angle of 120$°$ and bisect it to get $\angle$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:

Steps of construction:
1. Draw a line segment PQ = 11.6 cm.
2. Construct an angle of 45$°$ and bisect it to get $\angle$QPX.
3. Construct an angle of 60$°$ and bisect it to get $\angle$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:

(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:

Steps of Construction:
1. Draw a line TS.
2. Construct $\angle$POS = 90°.
3. Draw OQ, bisector of $\angle$POT. Thus, $\angle$QOS135$°$ is obtained.
4. Draw the bisector of $\angle$QOS.
5. $\angle$ROS thus obtained is equal to 67.5°.

#### Page No 515:

Steps of construction:
1. Draw a line segment AB = 4 cm.
2. Construct $\angle$BAX = 90$°$ and $\angle$ABY = 90$°$.
3. Set off AD = 4 cm and BC = 4 cm.
4. Join DC.
Thus, $\square$ABCD is the required square.

#### Page No 515:

Steps of construction:

1. Draw a line segment BC = 3.5 cm
2. Construct $\angle$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:

2. Take a point O on XY and draw PO $\perp$ XY.
4. Draw a line LM $\parallel$ XY.
5. Draw $\angle$LAB = 45$°$ and $\angle$MAC = 60$°$, meeting XY at B and C, respectively.