Construction of Quadrilaterals – Maharashtra Board Class 8 Solutions for Mathematics

Construction of Quadrilaterals – Maharashtra Board Class 8 Solutions for Mathematics

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise – 52

Solution 1:

Steps of construction:

1. Draw l(DF) = 9 cm, which is the diagonal of □DEFG.
2. Taking D as the centre, draw an arc of radius 3 cm.G will be their point of intersection.
3. Draw seg DG and FG.
4. Taking F as the centre, draw an arc of radius 7 cm.
5. Similarly, taking D as the centre, draw an arc of radius 5 cm and taking F as the centre, draw an arc of radius 6 cm.Draw seg DE and FE.
6. □DEFG is the required quadrilateral.
7. E will be their point of intersection.

Solution 2:

Steps of construction:

1. Draw l(AC) = 5.5 cm which is one of the diagonals of rhombus ABCD.
2. The four sides of a rhombus are congruent.Hence, all the sides of the given rhombus measure 4.5 cm.
3. One side of the rhombus = 4.5 cm
4. Now, taking A as the centre, draw arcs of radius 4.5 cm above and below the seg AC.
5. Similarly, taking C as the centre, draw arcs of radius 4.5 cm above and below the seg AC.
6. The point of intersection of the arcs above the seg AC is the vertex D and the point of intersection of the arcs below the seg AC is the vertex B.
7. ABCD is the required rhombus.

Solution 3:

Steps of construction:

1. PR is one of the diagonals of parallelogram PQRS and PQ is the base. Hence, draw PQ = 3.5 cm.
2. Taking P as the centre, draw an arc of radius 6.5 cm.
3. Taking Q as the centre, draw an arc of radius 4.5 cm to intersect the previous arc at R. Draw seg PR and seg QR.
4. l(PS) = l(QR) = 4.5 cm and l(PQ) = l(SR) = 3.5 cm. Taking P as the centre, draw an arc of radius 4.5 cm and taking R as the centre, draw an arc of radius 3.5 cm to intersect the previous arc at S. Draw seg PS and seg RS.
5. PQRS is the required parallelogram.

Exercise – 53

Solution 1:

Steps of construction:

1. Draw  l(EG) = 8 cm which is one of the diagonals of □DEFG.
2. Taking E as the centre, draw an arc of radius 5.5 cm above EG. Taking G as the centre, draw an arc of radius 5 cm, intersecting the previous arc at D.
3. Taking E as the centre, draw an arc of radius 3.5 cm, below EG. Taking D as the centre, draw an arc of radius 7.5 cm intersecting the previous arc at F. Draw DF, EF and FG.
4. □DEFG is the required quadrilateral.

Solution 2:

Steps of construction:

1. Draw l(OD) = 10 cm.
2. Taking O as the centre, draw an arc of radius 7.5 cm above OD.
3. Taking D as the centre, draw an arc of radius 5 cm, intersecting the previous arc at L. Draw OL and DL.
4. Taking D as the centre, draw an arc of radius 4 cm below OD.
5. Taking L as the centre, draw an arc of radius 6 cm intersecting the previous arc at G. Draw LG, OG and DG.
6. □GOLD is the required quadrilateral.

Solution 3:

Steps of construction:

1. Draw l(PA) = 8.5 cm.
2. Taking P as the centre, draw an arc of radius 4.5 cm above PA.
3. Taking A as the centre, draw an arc of radius 6.5 cm, intersecting the previous arc at L. Draw seg PL and seg AL.
4. Taking P as the centre, draw an arc of radius 5.5 cm below PA.
5. Taking L as the centre, draw an arc of radius 7.5 cm, intersecting the previous arc at Y. Join P and Y, A and Y.
6. □PLAY is the required quadrilateral.

Exercise – 54

Solution 1:

Steps of construction:

1. Draw DE = 4.5 cm.
2. Since m∠D = 65°, measure an angle of 65° at vertex D by drawing ray DX.
3. Since m∠E = 100°, measure an angle of 100° at vertex E by drawing ray EY.
4. l(EF) = 6.5 cm, hence, taking E as the centre and radius equal to 6.5 cm, draw an arc of circle intersecting the ray EY at F.
5. m∠F = 60°, hence, measure an angle of 60° at vertex F and draw ray FZ, intersecting the ray DX at G.
6. □DEFG is the required quadrilateral.

Solution 2:

Steps of construction:

1. Draw MT = 4 cm.
2. Since m∠M = 50°, measure an angle of 50° at vertex M by drawing ray MX.
3. Since m∠T = 110°, measure an angle of 110° at vertex T by drawing ray TY.
4. l(TY) = 5 cm, hence, taking T as the centre and radius equal to 5 cm, draw an arc of circle intersecting the ray TY at S.
5. m∠S = 70°, hence, measure an angle of 70° at vertex S and draw ray SZ, intersecting the ray MX at N.
6.  □MTSN is the required quadrilateral.

Solution 3:

Steps of construction:

1. Draw XY = 5.5 cm.
2. Since m∠X = 70°, measure an angle of 70° at vertex X by drawing ray XA.
3. l(XW) = 3 cm, hence, taking X as the centre and radius equal to 3 cm, draw an arc of circle intersecting the ray XA at W.
4. Since m∠Y = 95°, measure an angle of 95° at vertex Y by drawing ray YB.
5. Sum of the measures of all angles of a quadrilateral = 360°360° = m∠W + 250° Measure an angle of 110° at vertex W by drawing ray WC, so that m∠Z equals to 85°. The point of intersection of WC and YB is the vertex Z.
6. □XYZW is the required quadrilateral.
7. m∠W = (360 – 250) ° = 110°
8. m∠W + m∠X + m∠Y + m∠Z = m∠W + 70° + 95° + 85°

Exercise – 55

Solution 1:

Steps of construction:

1. Draw YR = 5 cm.
2. Since m∠Y = 60°, measure an angle of 60° at vertex Y by drawing ray YA.
3. Since m∠R = 120°, measure an angle of 120° at vertex R by drawing ray RB.
4. Taking Y as the centre and radius equal to 3.5 cm, draw an arc of circle intersecting the ray YA at T.
5. Taking R as the centre and radius equal to 4.5 cm, draw an arc of circle intersecting the ray RB at E.
6. □TYRE is the required quadrilateral.

Solution 2:

Steps of construction:

1. Draw IN = 5.5 cm.
2. Since m∠I = 115°, measure an angle of 115° at vertex I by drawing ray IA.
3. Since m∠N = 75°, measure an angle of 75° at vertex N by drawing ray NB.
4. Taking I as the centre and radius equal to 3.5 cm, draw an arc of circle intersecting the ray IA at K.
5. Taking N as the centre and radius equal to 5 cm, draw an arc of circle intersecting the ray NB at G.
6. □KING is the required quadrilateral.

Solution 3:

Steps of construction:

1. Draw RD = 6 cm.
2. Since m∠D = 100°, measure an angle of 100° at vertex D by drawing ray DB.
3. Taking D as the centre and radius equal to 4 cm, draw an arc of circle intersecting the ray DB at T.
4. Since m∠T = 90°, measure an angle of 90° at vertex T by drawing ray TC.
5. Taking T as the centre and radius equal to 4 cm, draw an arc of circle intersecting the ray TC at G.
6.  □RDTG is the required quadrilateral.