Class 8 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 8: Quadrilateral : Constructions and Types

Rough diagram

Steps of Construction:

1. Draw line OR = 4.4 cm.
2. Draw a ray using vertex O and making an angle of 105˚.
3. Draw a ray at R making an angle of 90˚.
4. Taking O as the centre and radius 5.8 cm, draw an arc to cut on the ray at point M.
5. Draw an angle of 58˚ at point M, intersecting the ray drawn from R. Name the point of intersection as E.
6. □MORE is the required quadrilateral.

Rough diagram

Steps of construction:

1. Draw seg DE = 4.5 cm
2. Draw an arc with centre D and radius 7.2 cm.
3. With centre E and radius 6.5 cm, draw another arc cutting the first arc and name the point of intersection as F.
4. Draw seg DF and EF.
5. With centre E and radius 7.8 cm, draw an arc on the opposite side of F.
6. With centre D and radius 5.5 cm, draw an arc cutting the previous arc at point G.
7. Draw seg EG and DG.
8. □ DEFG is required quadrilateral.

Rough Diagram

Steps of Construction:

1. Draw seg BC = 4.8 cm.
2. Draw a ray at point B making an angle of 50˚.
3. Draw a ray at point C making an angle of 140˚
4. Taking B as centre and radius 6.4 cm, draw an arc to cut the ray (drawn from B) at point A.
5. Draw a ray at point A making an angle of 70˚ to cut the ray drawn from C and name the point of intersection as D.
6. □ ABCD is required quadrilateral.

Rough diagram

Steps of construction:

1. Draw seg MO = 7.5 cm
2. Taking O as centre and radius 6 cm, draw an arc on one side of OM.
3. With centre M and radius 6 cm, draw an arc to cut the previous arc and name the point of intersection as L.
4. Again taking O as centre and radius 4.5 cm, draw an arc on the other side of OM.
5. With centre M and radius 4.5 cm, draw an arc cutting the previous arc (drawn in step 4) at point N.
6. □ LMNO is required quadrilateral.

Rough Diagram:

Steps of Constructions:

1. Draw seg AB = 6 cm

2. Draw a ray AX from A making an angle of 90^o.

3. Draw another ray BY from B making an angle of 90^o.

4. With centre A and radius 4.5 cm, draw an arc cutting AX at D.

5. With centre B and radius 4.5 cm, drawn an arc cutting BY at C.

6. Draw seg DC.

7. ABCD is the required rectangle.

Rough diagram:

Steps of construction:

1. Draw seg WX = 5.2 cm.
2. Draw two rays WP and XQ from points W and X respectively making an angle of 90o each.
3. Draw two arcs each with centres W and X and radius 5.2 cm intersecting the rays WP and XQ at Z and Y respectively.
4. Join Y and Z.
5. □ WXYZ is the required quadrilateral.

Rough diagram:

Steps of construction:

1. Draw seg KL = 4 cm. 
2. Draw a ray KX at vertex K making an angle of 75˚ and a ray LY at L making an angle of 105˚.
3. Taking K and L as centers with same radius 4 cm, draw two arcs to cut the rays KX and LY at N and M respectively.
4. Join N and M.
5. □ KLMN is a required quadrilateral.

From the diagram, ∆DAB is a right angled triangle.

Using Pythagoras theorem, we have

DA² + AB² = DB²

∴ 24² + AB² = 26²

∴ 576 + AB² = 676

∴ AB² = 676 - 576 = 100

∴ AB = 10 cm

Hence, the other side of rectangle is 10 cm.

The diagonals of a rhombus are perpendicular bisector of each other.

As AC = 16 cm and BD = 12 cm,

OA = OC =  = 8 cm

OB = OD =  = 6 cm

∆AOB is a right angled triangle at O.

Using Pythagoras theorem, we have

AO² + BO² = AB²

∴ AB² = 8² + 6² = 64 + 36 = 100

∴ AB = 10 cm

∴ The side of a rhombus is 10 cm and perimeter = 4 × 10 = 40 cm.

Let us consider, square ABCD

Here, AB = BC = CD = AD = 8 cm

The diagonals of a square are perpendicular bisector of each other.

In right angled ∆ABC,

AC² = AB² + BC²

∴ AC² = 8² + 8² = 64 + 64 = 128

∴ AC =  cm

Hence, the diagonal of a square is  cm.

Let ABCD be a rhombus.

The opposite angles of a rhombus are equal.

∴ ∠B = ∠D = 50˚

The adjacent angles of a rhombus are supplementary.

∴ ∠B + ∠C = 180˚

∴ 50˚ + ∠ C = 180˚

∴ ∠ C = 180˚ - 50˚ = 130˚

∠ C = ∠ A = 130˚

The remaining angles of a rhombus are 50˚, 130˚, 130˚.

Let the parallelogram be PQRS where ∠P = (3x - 2)° and ∠R = (50 - x)° are one pair of opposite angles.

The other pair of opposite angles be ∠Q and ∠S.

By the property of parallelogram, opposite angles of a parallelogram are congruent, we have

∠P = ∠R

∴ 3x - 2 = 50 - x

∴ 3x + x = 50 + 2

∴ 4x = 52

∴ x = 13

∴ ∠P = (50 - x)° = (50 - 13)° = 37° 

∴ ∠R = 37° 

As adjacent angles of a parallelogram are supplementary.

∴ ∠P + ∠Q = 180°

∴ 37° + ∠Q = 180°

∴ ∠Q = 180° - 37° = 143°

∴ ∠S = 143°

∴ The measures of the angles of the parallelogram are 37°, 143°, 37° and 143°.

In parallelogram XYZW,

(1)

Side XY = side WZ = 4.5 cm … Opposite sides of parallelogram

(2)

Side XW = side YZ = 8.2 cm … Opposite sides of parallelogram

(3)

Length OZ = Length OX = 2.5 cm … Diagonals bisect each other in parallelogram

(4)

Since, Diagonals bisect each other in parallelogram

∴ Length WY = 2 × length of WO = 2 × 3.3 = 6.6 cm

(5)

∠ WXY = ∠ WZY = 120˚ … Opposite angles are congruent in parallelogram

∠ XWY + ∠ WZY = 180˚ … Adjacent angles are supplementary

∴ ∠ XWZ + 120˚ = 180˚

∴ ∠ XWZ = 180˚ - 120˚

∴ ∠ XWZ = 60˚ 

Rough diagram:

Steps of Construction:

1. Draw seg BC = 7 cm.

2. Draw ∠B = 40˚ and B and C are supplementary angles of parallelogram.

∴ ∠B + ∠C = 180˚

∴ 40˚ + ∠C = 180˚

∴ ∠C = 180˚- 40˚

∴ ∠C = 140˚

3. From vertex C, draw a ray of an angle 140˚.

4. With center B and C with radius 3 cm. Cut the ray and name the points A and D respectively.

5. Join points A and D.

6. □ ABCD is a parallelogram.

Let □ PQRS be the quadrilateral and given that their angles are in the ratio 1:2:3:4.

Let the angles be x, 2x, 3x, 4x.

The sum of all the angles in a quadrilateral is 360˚.

∴ x + 2x + 3x + 4x = 360˚

∴ 10x = 360˚

∴ x = 36˚

Hence, ∠ P = 36˚, ∠ Q = 2x = 72˚, ∠ R = 3x = 108˚, ∠S = 4x = 144˚

Angles P and S are supplementary, angles Q and R are supplementary.

Hence, side PQ || side RS.

Only one pair of side is parallel hence, □PQRS is a trapezium.

Rough Diagram:

Steps of Constructions:

1. Draw seg AC which is diagonal of a quadrilateral BARC.
2. Taking centre A and C with radius 4.2 cm. Draw two arcs on one side of seg AC. Point B is the intersection point. Join B, C and B, A.
3. Taking centre A and C with radius 5.6 cm. Draw two arcs on the other side of seg AC. Point R is the intersection point.
4. Join R, C and A, R.
5. □BARC is the required quadrilateral.

Rough diagram:

Steps of Construction:

1. Draw seg QR = 5.6 cm.

2. Draw ray at vertex Q of an angle 110˚ = ∠ PQR.

3. Draw ray at vertex R of an angle 70˚ = ∠ SRQ.

4. With centre Q of radius 3.5 cm. Draw an arc to cut the ray and name the point of intersections as P.

5. With centre R of radius 3.5 cm. Draw an arc to cut the ray and name the point of intersections as S.

6. □ PQRS is a required quadrilateral.

PQ = 3.5 cm this information is unnecessary because opposite sides of a parallelogram are congruent.