Here you can get solutions of RS Aggarwal Solutions Class 9 Chapter 12 Geometrical Constructions. These Solutions are part of RS Aggarwal Solutions Class 9. we have given RS Aggarwal Solutions Class 9 Chapter 12 Geometrical Constructions download pdf.

## RS Aggarwal Solutions Class 9 Chapter 12 Geometrical Constructions

**Exercise 12A**

**Question 1:****Steps of Construction:**

(i) Draw a line segment AB = 5 cm

(ii) With A as centre and radius equal to more than half of AB, draw two arcs, one above AB and the other below AB.

(iii) With B as a centre and the same radius draw two arcs which cuts the previously drawn arcs at C and D.

(iv) Join CD , intersecting AB at point P.

∴ CD is the perpendicular bisector of AB at the point P.

**Question 2:****Step of Construction:**

(i) Draw a line segment OA.

(ii) AT A, draw ∠AOE=90^{0}, using ruler and compass.

(iii) With B as centre and radius more than half of BD, draw an arc.

(iv) With D as centre and same radius draw another arc which cuts the previous arc at F.

(v) Join OF. ∴ ∠AOF=45^{0}.

(vi) Now with centre B and radius more than half of BC, draw an arc.

(vii) With centre C and same radius draw another arc which cuts the previously drawn arc at X.

(viii) Join OX. ∴ OX is the bisector of ∠AOF.

**Read More:**

**Question 3:****Step of Construction:**

(i) Draw a line segment OA.

(ii) With O as centre and any suitable radius draw an arc, cutting OA at B.

(iii) With B as centre and the same radius cut the previously drawn arc at C.

(iv) With C as centre and the same radius cut the arc at D.

(v) With C as centre and the radius more than half CD draw an arc.

(vi) With D as centre and the same radius draw another arc which cuts the previous arc at E.

(vii) Join E Now, ∠AOE =90^{0}

(viii) Now with B as centre and radius more than half of CB draw an arc.

(iv) With C as centre and same radius draw an arc which cuts the previous at F.

(x) Join OF.

(xi) ∴ F is the bisector of right ∠AOE.

**Question 4:****Step of construction:**

(i) Draw a line segment BC=5cm.

(ii) With B as centre and radius equal to BC draw an arc.

(iii) With C as centre and the same radius draw another arc which cuts the previous arc at A.

(iv) Join AB and AC.

Then ∆ABC is the required equilateral triangle.

**Question 5:**

**Question 6:**

**Question 7:**

**Question 8:**

**Question 9:**

**Question 10:**

**Question 11:**

**Question 12:**

**Question 13:****Steps of Construction:**

(i) Draw BC = 4.5 cm.

(ii) Construct ∠CBX = 60^{0}

(iii) Along BX set off BP =8cm.

(iv) Join CP.

(v) Draw the perpendicular bisector of CP to intersecting BP at A.

(vi) Join AC. ∴ ∆ABC is the required triangle.

**Question 14:****Steps of Construction:**

(i) Draw BC = 5.2 cm.

(ii) Construct ∠CBX = 30^{0}

(iii) Set off BP = 3.5 cm.

(iv) Join PC.

(v) Draw the right bisector of PC, meeting BP produced at A.

(vi) Join AC. ∴ ∆ABC is the required triangle.

**Complete RS Aggarwal Solutions Class 9**

If You have any query regarding this chapter, please comment on below section our team will answer you. We Tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share **alarity.in** to your friends.

**Best of Luck For Your Future!!**

## No comments: