##### Class 8^{th} Mathematics Term 1 Tamilnadu Board Solution

**Exercise 4.1**- AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 5.5 cm and AC = 7 cm. Draw quadrilateral…
- AB = 7 cm, BC = 6.5 cm, AC = 8 cm, CD = 6 cm and DA = 4.5 cm. Draw quadrilateral…
- AB = 8 cm, BC = 6.8 cm, CD = 6 cm, AD = 6.4 cm and ∠B = 50°. Draw quadrilateral…
- AB = 6 cm, BC = 7 cm, AD = 6 cm, CD = 5 cm, and ∠BAC = 45°. Draw quadrilateral…
- AB = 5.5 cm, BC = 6.5 cm, BD = 7 cm, AD = 5 cm and ∠BAC = 50°. Draw…
- AB = 7 cm, BC = 5 cm, AC = 6 cm, CD = 4 cm, and ∠ACD = 45°. Draw quadrilateral…
- AB = 5.5 cm, BC = 4.5cm, AC = 6.5 cm, ∠CAD = 80°,∠ACD = 40° Draw quadrilateral…
- AB = 5 cm, BD = 7 cm, BC = 4 cm, ∠BAD = 100° and ∠DBC = 60. Draw quadrilateral…
- AB = 4 cm, AC = 8 cm, ∠ABC = 100°, ∠ABD = 50° ,∠CAD = 40° Draw quadrilateral…
- AB = 6 cm, BC = 6 cm, ∠BAC = 50°, ∠ACD = 30° ,∠CAD = 100° Draw quadrilateral…

**Exercise 4.2**- bar pq is parallel to bar sr PQ = 6.8 cm, QR = 7.2 cm, PR = 8.4 cm and RS = 8…
- bar pq is parallel to bar sr PQ = 8 cm,QR = 5 cm, PR = 6 cm,RS = 4.5 cm…
- bar pq is parallel to bar sr PQ = 7 cm, ∠Q = 60°,QR = 5 cm RS = 4 cm. Construct…
- bar pq is parallel to bar sr PQ = 6.5 cm,QR = 7 cm,∠PQR = 85°,PS = 9cm…
- bar pq is parallel to bar sr PQ = 7.5 cm, PS = 6.5 cm, ∠QPS = 100° and ∠PQR =…
- bar pq is parallel to bar sr PQ = 6 cm, PS = 5 cm, ∠QPS = 60° and ∠PQR = 100°.…
- bar pq is parallel to bar sr PQ = 8 cm, QR = 5 cm,RS = 6 cm,SP = 4 cm Construct…
- bar pq is parallel to bar sr PQ = 4.5 cm, QR = 2.5 cm, RS = 3 cm and SP = 2 cm.…
- bar ab is parallel to bar dc AB = 9 cm, DC = 6 cm and AD = BC = 5 cm. Construct…
- bar ab is parallel to bar dc AB = 10 cm, DC = 6 cm and AD = BC = 7 cm.…

**Exercise 4.3**- AB = 7 cm, BC = 5 cm and ∠ABC = 60°. Draw parallelogram ABCD with the following…
- AB = 8.5 cm, AD = 6.5 cm and ∠DAB = 100°. Draw parallelogram ABCD with the…
- AB = 6 cm, BD = 8 cm and AD = 5 cm. Draw parallelogram ABCD with the following…
- AB = 5 cm, BC = 4 cm, AC = 7 cm. Draw parallelogram ABCD with the following…
- AC = 10 cm, BD = 8 cm and ∠AOB = 100° where bar ac and bar rd intersect at ‘O’.…
- AC = 8 cm, BD = 6 cm and ∠COD = 90° where bar ac and bar rd intersect at ‘O’.…
- AB = 8 cm, AC = 10 cm and ∠ABC = 100°. Draw parallelogram ABCD with the…
- AB = 5.5 cm, ∠DAB = 50° and BD = 7 cm. Draw parallelogram ABCD with the…

**Exercise 4.1**

- AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 5.5 cm and AC = 7 cm. Draw quadrilateral…
- AB = 7 cm, BC = 6.5 cm, AC = 8 cm, CD = 6 cm and DA = 4.5 cm. Draw quadrilateral…
- AB = 8 cm, BC = 6.8 cm, CD = 6 cm, AD = 6.4 cm and ∠B = 50°. Draw quadrilateral…
- AB = 6 cm, BC = 7 cm, AD = 6 cm, CD = 5 cm, and ∠BAC = 45°. Draw quadrilateral…
- AB = 5.5 cm, BC = 6.5 cm, BD = 7 cm, AD = 5 cm and ∠BAC = 50°. Draw…
- AB = 7 cm, BC = 5 cm, AC = 6 cm, CD = 4 cm, and ∠ACD = 45°. Draw quadrilateral…
- AB = 5.5 cm, BC = 4.5cm, AC = 6.5 cm, ∠CAD = 80°,∠ACD = 40° Draw quadrilateral…
- AB = 5 cm, BD = 7 cm, BC = 4 cm, ∠BAD = 100° and ∠DBC = 60. Draw quadrilateral…
- AB = 4 cm, AC = 8 cm, ∠ABC = 100°, ∠ABD = 50° ,∠CAD = 40° Draw quadrilateral…
- AB = 6 cm, BC = 6 cm, ∠BAC = 50°, ∠ACD = 30° ,∠CAD = 100° Draw quadrilateral…

**Exercise 4.2**

- bar pq is parallel to bar sr PQ = 6.8 cm, QR = 7.2 cm, PR = 8.4 cm and RS = 8…
- bar pq is parallel to bar sr PQ = 8 cm,QR = 5 cm, PR = 6 cm,RS = 4.5 cm…
- bar pq is parallel to bar sr PQ = 7 cm, ∠Q = 60°,QR = 5 cm RS = 4 cm. Construct…
- bar pq is parallel to bar sr PQ = 6.5 cm,QR = 7 cm,∠PQR = 85°,PS = 9cm…
- bar pq is parallel to bar sr PQ = 7.5 cm, PS = 6.5 cm, ∠QPS = 100° and ∠PQR =…
- bar pq is parallel to bar sr PQ = 6 cm, PS = 5 cm, ∠QPS = 60° and ∠PQR = 100°.…
- bar pq is parallel to bar sr PQ = 8 cm, QR = 5 cm,RS = 6 cm,SP = 4 cm Construct…
- bar pq is parallel to bar sr PQ = 4.5 cm, QR = 2.5 cm, RS = 3 cm and SP = 2 cm.…
- bar ab is parallel to bar dc AB = 9 cm, DC = 6 cm and AD = BC = 5 cm. Construct…
- bar ab is parallel to bar dc AB = 10 cm, DC = 6 cm and AD = BC = 7 cm.…

**Exercise 4.3**

- AB = 7 cm, BC = 5 cm and ∠ABC = 60°. Draw parallelogram ABCD with the following…
- AB = 8.5 cm, AD = 6.5 cm and ∠DAB = 100°. Draw parallelogram ABCD with the…
- AB = 6 cm, BD = 8 cm and AD = 5 cm. Draw parallelogram ABCD with the following…
- AB = 5 cm, BC = 4 cm, AC = 7 cm. Draw parallelogram ABCD with the following…
- AC = 10 cm, BD = 8 cm and ∠AOB = 100° where bar ac and bar rd intersect at ‘O’.…
- AC = 8 cm, BD = 6 cm and ∠COD = 90° where bar ac and bar rd intersect at ‘O’.…
- AB = 8 cm, AC = 10 cm and ∠ABC = 100°. Draw parallelogram ABCD with the…
- 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)

= 25.55cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

25.55cm^{2}

**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)

= 34.8cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

34.8cm^{2}

**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)

= 38.08cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

38.08cm^{2}

**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)

= 32.83cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

32.83cm^{2}

**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)

= 29.75cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

29.75cm^{2}

**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)

= 23.4cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

23.4cm^{2}

**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)

= 27.625cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

27.625cm^{2}

**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)

= 22.4cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

22.4cm^{2}

**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)

= 24.8cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

24.8cm^{2}

**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)

= 36.225cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

36.225cm^{2}

**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)

= 25.55cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

25.55cm^{2}

**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)

= 34.8cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

34.8cm^{2}

**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)

= 38.08cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

38.08cm^{2}

**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

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 9.8 × (2.5cm + 4.2cm)

= 32.83cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

32.83cm^{2}

**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

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 8.5cm × (2.8cm + 4.2cm)

= 29.75cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

29.75cm^{2}

**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)

= 23.4cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

23.4cm^{2}

**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

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 6.5cm × (4.7cm + 3.8cm)

= 27.625cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

27.625cm^{2}

**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)

= 22.4cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

22.4cm^{2}

**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

Area of quadrilateral = Area of ΔACB + Area of ΔACD

= × BE × AC + × DF × AC

= × AC × (BE + DF)

= × 8cm × (3.1cm + 3.1cm)

= 24.8cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

24.8cm^{2}

**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)

= 36.225cm^{2}

Conclusion. Hence quadrilateral ABCD is constructed and its Area is

36.225cm^{2}

###### 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)

= 51.06cm^{2}

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

51.06cm^{2}

**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)

= 23.75cm^{2}

Conclusion. Hence; Trapezium PQRS is constructed and its area is 23.75cm^{2}

**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)

= 23.65cm^{2}

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

23.65cm^{2}

**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)

64.05cm^{2}

Conclusion. Hence; Trapezium PQRS is constructed and its area is 64.05cm^{2}

**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)

= 31.04cm^{2}

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

31.04cm^{2}

**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)

= 22.145cm^{2}

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

22.145cm^{2}

**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)

= 26.6cm^{2}

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

26.6cm^{2}

**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

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

= × RE × (PQ + SR)

= × 2cm × (4.5cm + 3cm)

= 7.5cm^{2}

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

7.5cm^{2}

**Question 9.**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.

**Answer:**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

**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)

= 51.06cm^{2}

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

51.06cm^{2}

**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)

= 23.75cm^{2}

Conclusion. Hence; Trapezium PQRS is constructed and its area is 23.75cm^{2}

**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)

= 23.65cm^{2}

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

23.65cm^{2}

**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)

64.05cm^{2}

Conclusion. Hence; Trapezium PQRS is constructed and its area is 64.05cm^{2}

**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)

= 31.04cm^{2}

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

31.04cm^{2}

**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)

= 22.145cm^{2}

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

22.145cm^{2}

**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)

= 26.6cm^{2}

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

26.6cm^{2}

**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

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

= × RE × (PQ + SR)

= × 2cm × (4.5cm + 3cm)

= 7.5cm^{2}

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

7.5cm^{2}

**Question 9.**

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.

**Answer:**

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