## RS Aggarwal Class 9 Solutions Chapter 12

**Q.1: Draw a line segment AB = 5 cm and draw its perpendicular bisector.**

**Sol:**

**Steps of Construction:**

**(i)** Draw a line segment AB= 5cm

**(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 cut the previously drawn arcs at C and D.

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

Therefore, CD is the perpendicular bisector of AB at the point P.

**Q.2: Draw an angle of 45 ^{0} using scale and compass only. Draw the bisector of this angle.**

**Sol:**

**Steps of Construction:**

**(i)** Draw a line segment OA.

**(ii)** At A, draw ∠AOE=90∘, using rulers 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∘

**(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.

Therefore, OX is the bisector of ∠AOF.

**Q.3: Draw an angle of 90 ^{0 }and draw its bisector.**

**Sol:**

**Steps of construction:**

**(i)** Draw a line segment OA.

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

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

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

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

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

**(vii)** Join E, So ∠AOE=90∘

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

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

**(x)** Join OF. Therefore, it is the bisector of the right ∠AOE.

**Q.4: Construct an equilateral triangle each of whose sides measures 5 cm.**

**Sol:**

**Steps 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 the 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.

**Q.5: Construct an equilateral triangle each of whose altitudes measures 5.4 cm**

**Sol:**

**Steps of construction:**

**(i)** Draw a line XY.

**(ii)** Mark any point P on it.

**(iii)** From P, draw PQ⊥XY.

**(iv)** From P, set off PA= 5.4 cm cutting PQ at A.

**(v)** Construct ∠PAC=30∘, meeting XY at B and C respectively.

Therefore, △ABC is required equilateral triangle.

**Q.6: Construct a triangle ABC in which BC = 5 cm, AB = 38 cm and AC = 26 cm.**

**Sol:**

**Steps of construction:**

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

**(ii)** With centre B and radius equal to 3.8 cm draw an arc.

**(iii)** With centre B and radius equal to 3.8 cm draw another arc which cuts the previously drawn arc at A.

**(iv)** Join AB and AC

Therefore, △ABC is the required equilateral triangle.

**Q.7: Construct a ΔABCinwhichBC=4.7cm,∠B=60∘and∠C=30∘**

**Sol:**

**Steps of construction:**

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

**(ii)** At B draw ∠XBC=60∘

**(iii)** At C draw ∠YCB=30∘

Let XB and YC interact at A.

Therefore, △ABC is the required triangle.

**Q.8: Construct an isosceles triangle PQR whose base measures 5 cm and each of equal sides measures 4.5 cm.**

**Sol:**

**Steps of construction:**

**(i)** Draw a line segment QR= 5cm which is the base.

**(ii)** With centre Q and radius equal to 4.5 cm, draw an arc.

**(iii)** With centre P and the same radius is drawn another arc which cuts the previous arc at P.

**(iv)** Join PQ and PR.

Therefore, △PQR is the required isosceles triangle.

**Q.9: Construct an isosceles triangle whose base is 4.8 cm and whose vertical angle is 80 ^{0}**

**Sol:**

**Steps of construction:**

**(i) **Draw a line segment BC = 4.8 cm

**(ii)** Make ∠CBX=80∘, below the line segment BC.

**(iii)** Make ∠XBY=90∘

**(iv)** Draw the right bisector PQ of BC, intersecting BY at O.

**(v)** With O as centre and radius OB, draw a circle intersecting PQ at A.

**(vi)** Join AB and AC.

Therefore, △PQR is the required isosceles triangle in which AB=AC.

**Q.10: Construct a right angled triangle whose hypotenuse measures 5.3 am and the length of one of whose sides containing the right angle measures 4.5 cm.**

**Sol:**

**Steps of construction:**

**(i)** Draw a line segment BC= 5.3 cm.

**(ii)** Find the mid-point O of BC.

**(iii)** With O as a centre and radius OB, draw a semicircle on BC.

**(iv)** With B as centre and radius equal to 4.5 cm draw an arc cutting the semicircle at A.

**(v)** Join AB and AC.

Therefore, △ABC is the required triangle.

**Q.11: Construct a ΔABC in which ∠B=30∘,∠B=60∘ and the length of the perpendicular form the vertex A to the base BC is 4.8 cm.**

**Sol:**

**Steps of construction:**

**(i)** Draw a line segment XY.

**(ii)** Take any point P on XY and draw PQ⊥XY.

**(iii)** Along PQ, set off PA= 4.8 cm.

**(iv)** Through A, draw LM∥XY.

**(v)** Construct ∠LAB=30∘ and ∠MAC=60∘ meeting XY at B and C respectively.

Therefore, △ABC is the required triangle.

**Q12: Construct a triangle PQR whose perimeter is 12 am and lengths of whose sides are in the ratio 3:2:4.**

**Sol:**

**Steps of construction:**

**(i) **Draw a line segment AB = 12 cm.

**(ii)** Draw a ray AX, making an acute angle with AB and draw in the downward direction.

**(iii)** From A set of (3+2+4) = 9 equal distances along AX.

**(iv)** Mark points L,M,N on AX such as that AL = 3 units, Lm = 2 units and MN = 4 units.

**(v)** Join NB.

**(vi)** Through L and M, draw LQ∥NBandMR∥NB cutting AB at Q and R respectively.

**(vii)** With Q as centre and radius AQ, draw an arc.

**(viii)** With R as centre and radius RB, draw another arc, cutting the previous arc at P.

**(ix)** Join PQ and PR.

Therefore, △PQR is the required triangle.

**Q13: Construct a ΔABC in which BC = 4.5 cm , ∠B=60∘ and the sum of the other two sides is 8 cm.**

**Sol:**

**Steps of Construction:**

**(i)** Draw BC = 4.5 cm

**(ii)** Construct ∠CBX=60∘

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

**(iv)** Join CP.

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

**(vi)** Join AC.

Therefore, △ABC is the required triangle.

**Q14: Construct a ΔABC in which BC = 5.2 cm, ∠B=30∘ and the difference of the other two sides is 3.5 cm.**

**Sol:**

**Steps of Construction:**

**(i)** Draw BC= 5.2 cm

**(ii)** Construct ∠CBX=30∘

**(iii)** Along BX set off BP = 3.5 cm

**(iv)** Join PC.

**(v)** Draw the perpendicular bisector of PC meeting BP produced at A.

**(vi)** Join AC.

Therefore, △ABC is the required triangle.