# RS Aggarwal Class 9 Solutions Geometrical Constructions

## 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 450 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$\angle AOE=90^{\circ}$, 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. AOF=45$\angle AOF=45^{\circ}$

(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 AOF$\angle AOF$.

Q.3: Draw an angle of 900 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$\angle AOE=90^{\circ}$

(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 AOE$\angle AOE$.

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 ABC$\bigtriangleup ABC$ 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 PQXY.$PQ\perp XY.$

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

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

Therefore, ABC$\bigtriangleup ABC$ 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, ABC$\bigtriangleup ABC$ is the required equilateral triangle.

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

Sol:

Steps of construction:

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

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

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

Let XB and YC interact at A.

Therefore, ABC$\bigtriangleup ABC$ 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, PQR$\bigtriangleup PQR$ is the required isosceles triangle.

Q.9: Construct an isosceles triangle whose base is 4.8 cm and whose vertical angle is 800

Sol:

Steps of construction:

(i) Draw a line segment BC = 4.8 cm

(ii) Make CBX=80$\angle CBX= 80^{\circ}$, below the line segment BC.

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

(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, PQR$\bigtriangleup PQR$ 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, ABC$\bigtriangleup ABC$ is the required triangle.

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

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

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

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

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

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

Sol:

Steps of Construction:

(i) Draw BC = 4.5 cm

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

(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, ABC$\bigtriangleup ABC$ is the required triangle.

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

Sol:

Steps of Construction:

(i) Draw BC= 5.2 cm

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

(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, ABC$\bigtriangleup ABC$ is the required triangle.

#### Practise This Question

34 gm of H2O is decomposed to _______ gm of hydrogen and ____ gm of oxygen.