Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 5.5 cm and AC = 7 cm.
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
Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 7 cm, BC = 6.5 cm, AC = 8 cm, CD = 6 cm and DA = 4.5 cm.
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
Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 8 cm, BC = 6.8 cm, CD = 6 cm, AD = 6.4 cm and ∠B = 50°.
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
Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 6 cm, BC = 7 cm, AD = 6 cm, CD = 5 cm, and ∠BAC = 45°.
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
Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 5.5 cm, BC = 6.5 cm, BD = 7 cm, AD = 5 cm and ∠BAC = 50°.
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
Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 7 cm, BC = 5 cm, AC = 6 cm, CD = 4 cm, and ∠ACD = 45°.
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
Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 5.5 cm, BC = 4.5cm, AC = 6.5 cm, ∠CAD = 80°,∠ACD = 40°
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
Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 5 cm, BD = 7 cm, BC = 4 cm, ∠BAD = 100° and ∠DBC = 60.
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
Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 4 cm, AC = 8 cm, ∠ABC = 100°, ∠ABD = 50° ,∠CAD = 40°
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
Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 6 cm, BC = 6 cm, ∠BAC = 50°, ∠ACD = 30° ,∠CAD = 100°
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
Construct trapezium PQRS with the following measurements. Find also its area.
is parallel to PQ = 6.8 cm, QR = 7.2 cm, PR = 8.4 cm and RS = 8 cm.
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
Construct trapezium PQRS with the following measurements. Find also its area.
is parallel to PQ = 8 cm,QR = 5 cm, PR = 6 cm,RS = 4.5 cm
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
Construct trapezium PQRS with the following measurements. Find also its area.
is parallel to PQ = 7 cm, ∠Q = 60°,QR = 5 cm RS = 4 cm.
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
Construct trapezium PQRS with the following measurements. Find also its area.
is parallel to PQ = 6.5 cm,QR = 7 cm,∠PQR = 85°,PS = 9cm
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
Construct trapezium PQRS with the following measurements. Find also its area.
is parallel to PQ = 7.5 cm, PS = 6.5 cm, ∠QPS = 100° and ∠PQR = 45°.
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
Construct trapezium PQRS with the following measurements. Find also its area.
is parallel to PQ = 6 cm, PS = 5 cm, ∠QPS = 60° and ∠PQR = 100°.
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
Construct trapezium PQRS with the following measurements. Find also its area.
is parallel to PQ = 8 cm, QR = 5 cm,RS = 6 cm,SP = 4 cm
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
Construct trapezium PQRS with the following measurements. Find also its area.
is parallel to PQ = 4.5 cm, QR = 2.5 cm, RS = 3 cm and SP = 2 cm.
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
Construct isosceles trapezium ABCD with the following measurements and find its area.
is parallel to AB = 9 cm, DC = 6 cm and AD = BC = 5 cm.
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
Construct isosceles trapezium ABCD with the following measurements and find its area.
is parallel to AB = 10 cm, DC = 6 cm and AD = BC = 7 cm.
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
Draw parallelogram ABCD with the following measurements and calculate its area.
AB = 7 cm, BC = 5 cm and ∠ABC = 60°.
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
Draw parallelogram ABCD with the following measurements and calculate its area.
AB = 8.5 cm, AD = 6.5 cm and ∠DAB = 100°.
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
Draw parallelogram ABCD with the following measurements and calculate its area.
AB = 6 cm, BD = 8 cm and AD = 5 cm.
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
Draw parallelogram ABCD with the following measurements and calculate its area.
AB = 5 cm, BC = 4 cm, AC = 7 cm.
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
Draw parallelogram ABCD with the following measurements and calculate its area.
AC = 10 cm, BD = 8 cm and ∠AOB = 100° where and intersect at ‘O’.
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
Draw parallelogram ABCD with the following measurements and calculate its area.
AC = 8 cm, BD = 6 cm and ∠COD = 90° where and intersect at ‘O’.
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
Draw parallelogram ABCD with the following measurements and calculate its area.
AB = 8 cm, AC = 10 cm and ∠ABC = 100°.
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
Draw parallelogram ABCD with the following measurements and calculate its area.
AB = 5.5 cm, ∠DAB = 50° and BD = 7 cm.
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