## 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:**

- Construction of an Equilateral Triangle
- Construction Of Similar Triangle As Per Given Scale Factor
- Construction Of A Line Segment
- Construction Of The Bisector Of A Given Angle
- Construction Of Perpendicular Bisector Of A Line Segment
- Construction Of An Angle Using Compass And Ruler

**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.