Practical Geometry

Given. ABCD is a quadrilateral.

AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 5.5 cm, AC = 7 cm

Formula used. Area of triangle = × Base × Height

Steps for construction.

Step 1 : Draw a line AB of 5cm

Step 2 : Draw an arc of 7 cm from point A and 6 cm from point B

Step 3 : Make the intersection point C and join AC and BC

Step 4 : Draw an arc of 4cm from point C and 5.5cm from point A

Step 5 : Mark the intersection point D and join AD and DC

Step 6 : Draw perpendicular from B and D on AC and name the intersecting point as E and F respectively

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 7cm × (4.2cm + 3.1cm)

= 25.55cm²

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

25.55cm²

Given. ABCD is a quadrilateral.

AB = 7 cm, BC = 6.5 cm,CD = 6 cm and DA = 4.5 cm,AC = 8 cm

Formula used. Area of triangle = × Base × Height

Steps for construction.

Step 1 : Draw a line AB of 7cm

Step 2 : Draw an arc of 8cm from point A and 6.5cm from point B

Step 3 : Make the intersection point C and join AC and BC

Step 4 : Draw an arc of 6cm from point C and 4.5cm from point A

Step 5 : Mark the intersection point D and join AD and DC

Step 6 : Draw perpendicular from B and D on AC and name the intersecting point as E and F respectively

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 8cm × (3.3cm + 5.4cm)

= 34.8cm²

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

34.8cm²

Given. ABCD is a quadrilateral.

AB = 8 cm, BC = 6.8 cm,CD = 6 cm, AD = 6.4 cm , ∠B = 50°.

Formula used. Area of triangle = × Base × Height

Steps for construction.

Step 1 : Draw a line AB of 8cm

Step 2 : Draw an angle of 50° on point B and extend the line 6.8cm

And make the end point as C

Step 3 : Draw an arc of 6cm from point C and 6.4cm from point A

Step 4 : Mark the intersection point D and join AD and DC

Step 5 : Join A and C to divide quadrilateral in two triangles

Step 6 : Draw perpendicular from B and D on AC and name the intersecting point as E and F respectively

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 6.4 × (6.6cm + 5.3cm)

= 38.08cm²

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

38.08cm²

Given. ABCD is a quadrilateral.

AB = 6 cm, BC = 7 cm,AD = 6 cm, CD = 5 cm , ∠BAC = 45°

Formula used. Area of triangle = × Base × Height

Steps for construction.

Step 1 : Draw a line AB of 6cm

Step 2 : Draw an angle on point A of 45° and extend the ray to AC’

Step 3 : Make an arc of 7cm from point B and mark the point of

intersection of arc and ray AC’ as point C.

Step 4 : Draw an arc of 5cm from point C and 6cm from point A

Step 5 : Mark the intersection point D and join AD and DC

Step 6 : Draw perpendicular from B and D on AC and name the intersecting point as E and F respectively

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 9.8 × (2.5cm + 4.2cm)

= 32.83cm²

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

32.83cm²

Given. ABCD is a quadrilateral.

AB = 5.5 cm, BC = 6.5 cm , BD = 7 cm, AD = 5 cm , ∠BAC = 50°.

Formula used. Area of triangle = × Base × Height

Steps for construction.

Step 1 : Draw a line AB of 5.5cm

Step 2 : Make an angle of 50° on point A and extend the ray to AC’

Step 3 : Make an arc of 6.5cm from point B and mark the point of

intersection of arc and ray AC’ as point C.

Step 4 : Draw an arc of 7cm from point B and 5cm from point A

Step 5 : Mark the intersection point D and join AD and BD

Step 6 : Join C and D to form quadrilateral ABCD

Step 6 : Draw perpendicular from B and D on AC and name the intersecting point as E and F respectively

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 8.5cm × (2.8cm + 4.2cm)

= 29.75cm²

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

29.75cm²

Given. ABCD is a quadrilateral.

AB = 7 cm, BC = 5 cm, CD = 4 cm , ∠ACD = 45°, AC = 6 cm

Formula used. Area of triangle = × Base × Height

Steps for construction.

Step 1 : Draw a line AB of 7cm

Step 2 : Draw an arc of 6cm from point A and 5 cm from point B

Step 3 : Make the intersection point C and join AC and BC

Step 4 : Draw an angle of 45° on point C on AC

Step 5 : Make an arc of 4cm on ray CD’ and mark the intersection

point as D

Step 6 : Join A and D to form quadrilateral ABCD

Step 7 : Draw perpendicular from B and D on AC and name the intersecting point as E and F respectively

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 6cm × (4.9cm + 2.9cm)

= 23.4cm²

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

23.4cm²

Given. ABCD is a quadrilateral.

AB = 5.5 cm, BC = 4.5cm, AC = 6.5 cm

∠CAD = 80° and ∠ACD = 40°

Formula used. Area of triangle = × Base × Height

Steps for construction.

Step 1 : Draw a line AB of 5.5cm

Step 2 : Draw an arc of 6.5 cm from point A and 4.5cm from point B

Step 3 : Mark the intersection point C and join AC and BC

Step 4 : Make an angle of 80° on point A on AC and extend ray AD’

Step 5 : Make an angle of 40° on point C on AC and extend to

intersect at ray AD’ and mark the intersection point as D

Step 6 : Draw perpendicular from B and D on AC and name the intersecting point as E and F respectively

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 6.5cm × (4.7cm + 3.8cm)

= 27.625cm²

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

27.625cm²

Given. ABCD is a quadrilateral.

∠BAD = 100° and ∠DBC = 60.

AB = 5 cm, BC = 4 cm, BD = 7 cm

Formula used. Area of triangle = × Base × Height

Steps for construction.

Step 1 : Draw a line AB of 5cm

Step 2 : Make an angle of 100° on point A and extend the ray to AD’

Step 3 : Make an arc on AD’ of 7cm from point B and mark the

intersection point as D

Step 4 : Make an angle of 60° on point B on BD and extend ray BC’

Step 5 : Mark an arc of 4cm from point B on BC’ and name the

intersection point as C.

Step 6 : Join C and D to form Quadrilateral ABCD

Step 6 : Draw perpendicular from A and C on BD and name the intersecting point as E and F respectively

Area of quadrilateral = Area of ΔABD + Area of ΔCBD

= × AE × BD + × CF × BD

= × BD × (AE + CF)

= × 7cm × (3.5cm + 2.9cm)

= 22.4cm²

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

22.4cm²

Given. ABCD is a quadrilateral.

∠ABC = 100°, ∠ABD = 50° and ∠CAD = 40°, AB = 4 cm, AC = 8 cm

Formula used. Area of triangle = × Base × Height

Steps for construction.

Step 1 : Draw a line AB of 4cm

Step 2 : Draw an angle of 100° on point B and extend ray BC’

Step 3 : Make an arc of 8cm from point A and intersect it on BC’

Mark intersection point of arc and ray BC’ as point C

Step 4 : Draw an another angle of 50° on point B on AB and extend

ray to BD’

Step 5 : Make an angle of 40° on point A on AC and extend ray to

intersect ray BD’ mark the intersection point as D

Step 6 : Join D and C to form quadrilateral ABCD

Step 6 : Draw perpendicular from B and D on AC and name the intersecting point as E and F respectively

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 8cm × (3.1cm + 3.1cm)

= 24.8cm²

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

24.8cm²

Given. ABCD is a quadrilateral.

AB = 6 cm, BC = 6 cm, ∠BAC = 50°, ∠ACD = 30° and ∠CAD = 100°

Formula used. Area of triangle = × Base × Height

Steps for construction.

Step 1 : Draw a line AB of 6cm

Step 2 : Draw an angle of 50° on point A and extend ray AC’

Step 3 : Make an arc of 6cm from point B and intersect it on AC’

Mark intersection point of arc and ray AC’ as point C

Step 4 : Draw an angle of 30° on point C on AC and extend

ray to CD’

Step 5 : Make an angle of 100° on point A on AC and extend ray to

intersect ray CD’ mark the intersection point as D

Step 6 : Join BD

Step 6 : Draw perpendicular from A and C on BD and name the intersecting point as E and F respectively

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × AE × BD + × CF × BD

= × BD × (AE + CF)

= × 10.5cm × (1.4cm + 5.5cm)

= 36.225cm²

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

36.225cm²

Given. PQRS is a trapezium.

PQ = 6.8 cm, RS = 8 cm , QR = 7.2 cm, PR = 8.4 cm

Formula used. Area of trapezium = × height × sum of parallel sides

Area of trapezium = × height × (PQ + SR)

Steps of Construction

Step 1 : Draw a line PQ of 6.8cm

Step 2 : Make an arc of 8.4cm from point P and 7.2cm from point Q

Step 3 : Mark the intersection point as R and join PR and QR

Step 4 : Draw a ray RS’ parallel to PQ from point R

Step 5 : Make arc of 8cm on ray RS’ from point R and name intersection point S

Step 6 : Join point S and P to form trapezium PQRS

Step 7 : Draw a perpendicular on PQ from point R and name the intersection point as E

Area of trapezium = × height × (PQ + SR)

= × RE × (PQ + SR)

= × 6.9cm × (6.8cm + 8cm)

= 51.06cm²

Conclusion. Hence; Trapezium PQRS is constructed and its area is

51.06cm²

Given. PQRS is a trapezium.

PQ = 8 cm, RS = 4.5 cm , QR = 5 cm, PR = 6 cm

Formula used. Area of trapezium = × height × sum of parallel sides Area of trapezium = × height × (PQ + SR)

Steps of Construction

Step 1 : Draw a line PQ of 8cm

Step 2 : Make an arc of 6cm from point P and 5cm from point Q

Step 3 : Mark the intersection point as R and join PR and QR

Step 4 : Draw a ray RS’ parallel to PQ from point R

Step 5 : Make an arc of 4.5cm on ray RS’ from point R and name intersection point as S

Step 6 : Join point S and P to form trapezium PQRS

Step 7 : Draw a perpendicular on PQ from point R and name the intersection point as E

Area of trapezium = × height × (PQ + SR)

= × RE × (PQ + SR)

= × 3.8cm × (4.5cm + 8cm)

= 23.75cm²

Conclusion. Hence; Trapezium PQRS is constructed and its area is 23.75cm²

Given. PQRS is a trapezium.

PQ = 7 cm, QR = 5 cm and RS = 4 cm , ∠Q = 60°

Formula used. Area of trapezium = × height × sum of parallel sides

Area of trapezium = × height × (PQ + SR)

Steps of Construction

Step 1 : Draw a line PQ of 7cm

Step 2 : Make an angle of 60° from point Q and extend ray to QR’

Step 3 : Mark an arc of 5cm from Q on ray QR’ and name intersection point as R

Step 4 : Draw a ray RS’ parallel to PQ from point R

Step 5 : Make arc of 4cm on ray RS’ from point R and name intersection point S

Step 6 : Join point S and P to form trapezium PQRS

Step 7 : Draw a perpendicular on PQ from point R and name the intersection point as E

Area of trapezium = × height × (PQ + SR)

= × RE × (PQ + SR)

= × 4.3cm × (7cm + 4cm)

= 23.65cm²

Conclusion. Hence; Trapezium PQRS is constructed and its area is

23.65cm²

Given. PQRS is a trapezium.

PQ = 6.5 cm, QR = 7 cm, PS = 9 cm , ∠ PQR = 85°

Formula used. Area of trapezium = × height × sum of parallel sides

Area of trapezium = × height × (PQ + SR)

Steps of Construction

Step 1 : Draw a line PQ of 6.5cm

Step 2 : Make an angle of 85° from point Q and extend ray to QR’

Step 3 : Mark an arc of 7cm from Q on ray QR’ and name intersection point as R

Step 4 : Draw a ray RS’ parallel to PQ from point R

Step 5 : Make arc of 9cm on ray RS’ from point P and name intersection point as S

Step 6 : Join point S and P to form trapezium PQRS

Step 7 : Draw a perpendicular on PQ from point R and name the intersection point as E

Area of trapezium = × height × (PQ + SR)

= × RE × (PQ + SR)

= × 7cm × (6.5cm + 11.8cm)

64.05cm²

Conclusion. Hence; Trapezium PQRS is constructed and its area is 64.05cm²

Given. PQRS is a trapezium.

PQ = 7.5 cm, PS = 6.5 cm , ∠QPS = 100° and ∠PQR = 45°.

Formula used. Area of trapezium = × height × sum of parallel sides

Area of trapezium = × height × (PQ + SR)

Steps of Construction

Step 1 : Draw a line PQ of 7.5cm

Step 2 : Make an angle of 45° from point Q and extend ray to QR’

Step 3 : Make an angle of 100° from point P and extend ray to PS’

Step 3 : Mark an arc of 6.5cm from P on ray PS’ and name intersection point as S

Step 4 : Draw a ray SR” parallel to PQ from point S

Step 5 : Mark the intersection point of SR” and QR’ as R

Step 6 : Draw a perpendicular on PQ from point R and name the intersection point as E

Area of trapezium = × height × (PQ + SR)

= × RE × (PQ + SR)

= × 6.4cm × (7.5cm + 2.2cm)

= 31.04cm²

Conclusion. Hence; Trapezium PQRS is constructed and its area is

31.04cm²

Given. PQRS is a trapezium.

PQ = 6 cm, PS = 5 cm , ∠QPS = 60° and ∠PQR = 100°

Formula used. Area of trapezium = × height × sum of parallel sides

Area of trapezium = × height × (PQ + SR)

Steps of Construction

Step 1 : Draw a line PQ of 6cm

Step 2 : Make an angle of 100° from point Q and extend ray to QR’

Step 3 : Make an angle of 60° from point P and extend ray to PS’

Step 4 : Mark an arc of 5cm from P on ray PS’ and name intersection point as S

Step 5 : Draw a ray SR” parallel to PQ from point S

Step 6 : Mark the intersection point of SR” and QR’ as R

Step 7 : Draw a perpendicular on PQ from point S and name the intersection point as A

Area of trapezium = × height × (PQ + SR)

= × SA × (PQ + SR)

= × 4.3cm × (6cm + 4.3cm)

= 22.145cm²

Conclusion. Hence; Trapezium PQRS is constructed and its area is

22.145cm²

Given. PQRS is a trapezium.

PQ = 8 cm, QR = 5 cm, RS = 6 cm and SP = 4 cm

Formula used. Area of trapezium = × height × sum of parallel sides

Area of trapezium = × height × (PQ + SR)

Steps of Construction

Step 1 : Draw a line PQ of 8cm

Step 2 : Mark a point E on PQ such that PE = 6cm and EQ = 2cm

Step 3 : Make arc of 5cm from point Q and 4cm from point E

Step 4 : Mark the intersecting point of both arcs to be R and join ER and RQ

Step 5 : Draw arcs of 6cm and 4cm from point R and P respectively

Step 6 : Mark the intersection point of both arcs as S join SP and RS

Step 8 : Draw a perpendicular on PQ from point R and name the intersection point as F

Area of trapezium = × height × (PQ + SR)

= × RF × (PQ + SR)

= × 3.8cm × (8cm + 6cm)

= 26.6cm²

Conclusion. Hence; Trapezium PQRS is constructed and its area is

26.6cm²

Given. PQRS is a trapezium.

PQ = 4.5 cm, QR = 2.5 cm, RS = 3 cm and SP = 2 cm

Formula used. Area of trapezium = × height × sum of parallel sides

Area of trapezium = × height × (PQ + SR)

Steps of Construction

Step 1 : Draw a line PQ of 4.5cm

Step 2 : Mark a point E on PQ such that PE = 3cm and EQ = 1.5cm

Step 3 : Make arc of 2.5cm from point Q and 2cm from point E

Step 4 : Mark the intersecting point of both arcs to be R and join RE and EQ.

Step 5 : Draw arcs of 3cm and 2cm from point R and P respectively

Step 6 : Mark the intersection point of both arcs as S,join SP and RS

Step 8 : Draw a perpendicular on PQ from point R and name the intersection point as E

Area of trapezium = × height × (PQ + SR)

= × RE × (PQ + SR)

= × 2cm × (4.5cm + 3cm)

= 7.5cm²

Conclusion. Hence; Trapezium PQRS is constructed and its area is

7.5cm²

Given. ABCD is a isosceles trapezium.

AB = 9 cm, DC = 6 cm and AD = BC = 5 cm.

Formula used. Area of trapezium = × height × sum of parallel sides

Area of trapezium = × height × (AB + CD)

Steps of Construction

Step 1 : Draw a line AB of 9cm

Step 2 : Mark a point E on AB such that AE = 6cm and EB = 3cm

Step 3 : Make arc of 5cm from both point E and B

Step 4 : Mark the intersecting point of both arcs to be C and join EC and BC.

Step 5 : Draw arcs of 6cm and 5cm from point C and A respectively

Step 6 : Mark the intersection point of both arcs as D

Step 7 : Join AD and CD to form trapezium ABCD

Step 8 : Draw a perpendicular on AB from point C and name then intersection point as F

Area of trapezium = × height × (AB + CD)

= × CF × (AB + CD)

= × 4.7cm × (9cm + 6cm)

= 35.25cm²

Conclusion. Hence; Trapezium ABCD is constructed and its area is

35.25cm²

Given. ABCD is a isosceles trapezium.

AB = 10 cm, DC = 6 cm and AD = BC = 7 cm

Formula used. Area of trapezium = × height × sum of parallel sides

Area of trapezium = × height × (PQ + SR)

Steps of Construction

Step 1 : Draw a line AB of 10cm

Step 2 : Mark a point E on AB such that AE = 6cm and EB = 4cm

Step 3 : Make arc of 7cm from both point E and B

Step 4 : Mark the intersecting point of both arcs to be C and join EC and EB

Step 5 : Draw arcs of 6cm and 7cm from point C and A respectively

Step 6 : Mark the intersection point of both arcs as D

Step 7 : Join AD and CD to form trapezium ABCD

Step 8 : Draw a perpendicular on AB from point C and name the intersection point as F

Area of trapezium = × height × (AB + CD)

= × RF × (AB + CD)

= = × 6.7cm × (10cm + 6cm)

= 53.6cm²

Conclusion. Hence; Trapezium ABCD is constructed and its area is

53.6cm²

Given. ABCD is a parallelogram.

AB = CD ; BC = AD , AB = 7 cm, BC = 5 cm , ∠ABC = 60°

Formula used. Area of parallelogram = base × height

Steps of construction

Step 1 : Draw a line AB of 7cm

Step 2 : Make an angle of 60° on point B and extend ray to BC’

Step 3 : Make an arc of 5cm on ray BC’ from point B and name the

intersection point as C

Step 4 : Draw arc of 5cm and 7cm from point A and C respectively

Step 5 : Mark the intersection point as D and join AD and CD

Step 6 : Draw a perpendicular on AB from point C and name the intersection point as E

Area of parallelogram = base × height

= AB × CE

= 7cm × 4.3cm

= 30.1cm²

Conclusion. Hence; parallelogram ABCD is constructed and its area

is 30.1cm²

Given. ABCD is a parallelogram.

AB = CD ; BC = AD

AB = 8.5 cm, AD = 6.5 cm

∠DAB = 100°

Formula used. Area of parallelogram = base × height

Steps of construction

Step 1 : Draw a line AB of 8.5cm

Step 2 : Make an angle of 100° on point A and extend ray to AD’

Step 3 : Make an arc of 6.5cm on ray AD’ through point A and name the intersection point as D

Step 4 : Draw arc of 6.5cm and 8.5cm from point D and B Respectively

Step 5 : Mark the intersection point as C and join DC and BC

Step 6 : Draw a perpendicular on AB from point C and name the intersection point as E

Area of parallelogram = base × height

= AB × CE

= 8.5cm × 6.4cm

= 54.4cm²

Conclusion. Hence; parallelogram ABCD is constructed and its area

is 54.4cm²

Given. ABCD is a parallelogram.

AB = CD ; BC = AD

AB = 6 cm, BD = 8 cm and AD = 5 cm

Formula used. Area of parallelogram = base × height

Steps of construction

Step 1 : Draw a line AB of 6cm

Step 2 : Make an arc of 8cm from point B and 5cm from point A

Step 3 : Intersection of Both the arcs gives point D and join AD , AB

Step 4 : Draw arc of 6cm and 5cm from point D and B respectively

Step 5 : Mark the intersection point as C then join DC and BC

Step 6 : Draw a perpendicular on AB from point C and name the intersection point as E

Area of parallelogram = base × height

= AB × CE

= 6cm × 5cm

= 30cm²

Conclusion. Hence; parallelogram ABCD is constructed and its area is 30cm²

Given. ABCD is a parallelogram.

AB = CD ; BC = AD

AB = 5 cm, BC = 4 cm, AC = 7 cm

Formula used. Area of parallelogram = base × height

Steps of construction

Step 1 : Draw a line AB of 5cm

Step 2 : Make an arc of 7cm from point A and 4cm from point B

Step 3 : Intersection of Both the arcs gives point C then join

AB and AC

Step 4 : Draw arc of 4cm and 5cm from point A and C respectively

Step 5 : Mark the intersection point as D and join DC and DA

Step 6 : Draw a perpendicular on AB from point D and name the

intersection point as E

Area of parallelogram = base × height

= AB × DE

= 5cm × 3.9cm

= 19.5cm²

Conclusion. Hence; parallelogram ABCD is constructed and its area

is 19.5cm²

Given. ABCD is a parallelogram.

AC = 10 cm, BD = 8 cm

∠AOB = 100°

Formula used. Area of parallelogram = base × height

Steps of construction

Step 1 : Draw line segment AC of 10cm

Step 2 : Mark O as midpoint of AC

Step 3 : Draw line B’D’ through O which makes ∠AOB’ = 120°

Step 4 : Make Arcs of 4cm form O on both sides of B’D’

Step 5 : Join AB;BC;CD;DA

Step 6 : Draw a perpendicular on AB from point D and name the intersection point as E

Area of parallelogram = base × height

= AB × DE

= 5.7cm × 6.9cm

= 39.33cm²

Conclusion. Hence; parallelogram ABCD is constructed and its area

is 39.33cm²

Given. ABCD is a parallelogram.

AC = 8 cm, BD = 6 cm

∠COD = 90°

Formula used. Area of parallelogram = base × height

Steps of construction

Step 1 : Draw line segment AC of 8cm

Step 2 : Mark O as midpoint of AC

Step 3 : Draw line B’D’ through O which makes ∠COD’ = 120°

Step 4 : Make Arcs of 3cm form O on both sides of B’D’

Step 5 : Join AB;BC;CD;DA

Step 6 : Draw a perpendicular on AB from point D and name the intersection point as E

Area of parallelogram = base × height

= AB × DE

= 5cm × 4.8cm

= 24cm²

Conclusion. Hence; parallelogram ABCD is constructed and its area

is 24cm²

Given. ABCD is a parallelogram.

AB = CD ; BC = AD

AB = 8 cm, AC = 10 cm

∠ABC = 100°

Formula used. Area of parallelogram = base × height

Steps of Construction

Step 1 : Draw AB of length 8cm

Step 2 : Make an angle of 100° on point B and extend the ray BC’

Step 3 : Make an arc of 10cm from point A on ray BC’ and note the

Length of BC

Step 4 : Make an arc of 8cm from point C and another arc of same length of BC from point A

Step 5 : Mark the intersection point as D and join AD and CD

Step 6 : Draw a perpendicular on AB from point D and name the intersection point as E

Area of parallelogram = base × height

= AB × DE

= 8cm × 4.6cm

= 36.8cm²

Conclusion. Hence; parallelogram ABCD is constructed and its area

is 36.8cm²

Given. ABCD is a parallelogram.

AB = CD ; BC = AD , AB = 5.5 cm, BD = 7 cm

∠DAB = 50°

Formula used. Area of parallelogram = base × height

Steps of Construction

Step 1 : Draw AB of length 5.5cm

Step 2 : Make an angle of 50° on point A and extend the ray AD’

Step 3 : Make an arc of 7cm from point B on ray AD’ and also note Down the length of AD

Step 4 : Make an arc of 5.5cm from point D and another arc of same length of AD from point B

Step 5 : Mark the intersection point as C and join AC and CD

Step 6 : Draw a perpendicular on AB from point C and name the

intersection point as E

Area of parallelogram = base × height

= AB × CE

= 5.5cm × 7cm

= 38.5cm²

Conclusion. Hence; parallelogram ABCD is constructed and its area is 38.5cm²