**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 \(\angle AOE=90^{\circ}\), 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. \(\angle AOF=45^{\circ}\)

**(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 \(\angle AOF \).

**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 \(\angle AOE=90^{\circ}\)

**(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 \(\angle AOE\).

**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 \(\bigtriangleup ABC\) 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 \(PQ\perp XY.\)

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

**(v)** Construct \(\angle PAC=30^{\circ}\), meeting XY at B and C respectively.

Therefore, \(\bigtriangleup ABC\) 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, \(\bigtriangleup ABC\) is the required equilateral triangle.

**Q.7: Construct a \( \Delta ABC in which BC = 4.7 cm, \angle B = 60^{\circ} and \angle C = 30^{\circ} \)**

**Sol:**

**Steps of construction:**

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

**(ii)** At B draw \(\angle XBC=60^{\circ}\)

**(iii)** At C draw \(\angle YCB =30^{\circ}\)

Let XB and YC interact at A.

Therefore, \(\bigtriangleup ABC\) 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, \(\bigtriangleup PQR\) 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 \(\angle CBX= 80^{\circ}\), below the line segment BC.

**(iii)** Make \(\angle XBY= 90^{\circ}\)

**(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, \(\bigtriangleup PQR\) 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, \(\bigtriangleup ABC\) is the required triangle.

**Q.11: Construct a \( \Delta ABC \) in which \( \angle B = 30^{\circ} , \angle B = 60^{\circ} \) 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 \(PQ\perp XY\).

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

**(iv)** Through A, draw \(LM\parallel XY\).

**(v)** Construct \(\angle LAB= 30^{\circ}\) and \(\angle MAC= 60^{\circ}\) meeting XY at B and C respectively.

Therefore, \(\bigtriangleup ABC\) 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 \(LQ\parallel NB \; and\; MR\parallel NB\) 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, \(\bigtriangleup PQR\) is the required triangle.

**Q13: Construct a \( \Delta ABC \) in which BC = 4.5 cm , \( \angle B = 60^{\circ} \) and the sum of the other two sides is 8 cm.**

**Sol:**

**Steps of Construction:**

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

**(ii)** Construct \(\angle CBX=60^{\circ}\)

**(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, \(\bigtriangleup ABC\) is the required triangle.

**Q14: Construct a \( \Delta ABC \) in which BC = 5.2 cm, \( \angle B = 30^{\circ} \) and the difference of the other two sides is 3.5 cm.**

**Sol:**

**Steps of Construction:**

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

**(ii)** Construct \(\angle CBX= 30^{\circ}\)

**(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, \(\bigtriangleup ABC\) is the required triangle.

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