Practical Geometry Class 8th Mathematics Term 1 Tamilnadu Board Solution

Class 8th Mathematics Term 1 Tamilnadu Board Solution
Exercise 4.1
  1. AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 5.5 cm and AC = 7 cm. Draw quadrilateral…
  2. AB = 7 cm, BC = 6.5 cm, AC = 8 cm, CD = 6 cm and DA = 4.5 cm. Draw quadrilateral…
  3. AB = 8 cm, BC = 6.8 cm, CD = 6 cm, AD = 6.4 cm and ∠B = 50°. Draw quadrilateral…
  4. AB = 6 cm, BC = 7 cm, AD = 6 cm, CD = 5 cm, and ∠BAC = 45°. Draw quadrilateral…
  5. AB = 5.5 cm, BC = 6.5 cm, BD = 7 cm, AD = 5 cm and ∠BAC = 50°. Draw…
  6. AB = 7 cm, BC = 5 cm, AC = 6 cm, CD = 4 cm, and ∠ACD = 45°. Draw quadrilateral…
  7. AB = 5.5 cm, BC = 4.5cm, AC = 6.5 cm, ∠CAD = 80°,∠ACD = 40° Draw quadrilateral…
  8. AB = 5 cm, BD = 7 cm, BC = 4 cm, ∠BAD = 100° and ∠DBC = 60. Draw quadrilateral…
  9. AB = 4 cm, AC = 8 cm, ∠ABC = 100°, ∠ABD = 50° ,∠CAD = 40° Draw quadrilateral…
  10. AB = 6 cm, BC = 6 cm, ∠BAC = 50°, ∠ACD = 30° ,∠CAD = 100° Draw quadrilateral…
Exercise 4.2
  1. bar pq is parallel to bar sr PQ = 6.8 cm, QR = 7.2 cm, PR = 8.4 cm and RS = 8…
  2. bar pq is parallel to bar sr PQ = 8 cm,QR = 5 cm, PR = 6 cm,RS = 4.5 cm…
  3. bar pq is parallel to bar sr PQ = 7 cm, ∠Q = 60°,QR = 5 cm RS = 4 cm. Construct…
  4. bar pq is parallel to bar sr PQ = 6.5 cm,QR = 7 cm,∠PQR = 85°,PS = 9cm…
  5. bar pq is parallel to bar sr PQ = 7.5 cm, PS = 6.5 cm, ∠QPS = 100° and ∠PQR =…
  6. bar pq is parallel to bar sr PQ = 6 cm, PS = 5 cm, ∠QPS = 60° and ∠PQR = 100°.…
  7. bar pq is parallel to bar sr PQ = 8 cm, QR = 5 cm,RS = 6 cm,SP = 4 cm Construct…
  8. bar pq is parallel to bar sr PQ = 4.5 cm, QR = 2.5 cm, RS = 3 cm and SP = 2 cm.…
  9. bar ab is parallel to bar dc AB = 9 cm, DC = 6 cm and AD = BC = 5 cm. Construct…
  10. bar ab is parallel to bar dc AB = 10 cm, DC = 6 cm and AD = BC = 7 cm.…
Exercise 4.3
  1. AB = 7 cm, BC = 5 cm and ∠ABC = 60°. Draw parallelogram ABCD with the following…
  2. AB = 8.5 cm, AD = 6.5 cm and ∠DAB = 100°. Draw parallelogram ABCD with the…
  3. AB = 6 cm, BD = 8 cm and AD = 5 cm. Draw parallelogram ABCD with the following…
  4. AB = 5 cm, BC = 4 cm, AC = 7 cm. Draw parallelogram ABCD with the following…
  5. AC = 10 cm, BD = 8 cm and ∠AOB = 100° where bar ac and bar rd intersect at ‘O’.…
  6. AC = 8 cm, BD = 6 cm and ∠COD = 90° where bar ac and bar rd intersect at ‘O’.…
  7. AB = 8 cm, AC = 10 cm and ∠ABC = 100°. Draw parallelogram ABCD with the…
  8. AB = 5.5 cm, ∠DAB = 50° and BD = 7 cm. Draw parallelogram ABCD with the…

Exercise 4.1
Question 1.

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.


Answer:

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



Question 2.

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.


Answer:

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



Question 3.

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°.


Answer:

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



Question 4.

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°.


Answer:

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



Question 5.

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°.


Answer:

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



Question 6.

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°.


Answer:

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



Question 7.

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°


Answer:

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



Question 8.

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.


Answer:

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



Question 9.

Draw quadrilateral ABCD with the following measurements. Find also its area.

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


Answer:

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



Question 10.

Draw quadrilateral ABCD with the following measurements. Find also its area.

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


Answer:

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




Exercise 4.2
Question 1.

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.


Answer:

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



Question 2.

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


Answer:

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



Question 3.

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.


Answer:

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



Question 4.

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


Answer:

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



Question 5.

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°.


Answer:

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



Question 6.

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°.


Answer:

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



Question 7.

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


Answer:

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



Question 8.

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.


Answer:

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