# Question 1

**Construct a quadrilateral PQRS in which PQ = 3 cm, QR = 5 cm, QS = 5 cm, PS = 4 cm and SR = 4 cm.**

**Solution:**

Steps of construction:

1. Draw SQ = 5cm.

2. With S as centre and SP (= 4 cm) as radius draw an arc.

3. With Q as centre and QP (= 3 cm) as radius draw another arc to cut the arc of step 2 at P.

4. With S as centre and SR (= 4 cm) as radius draw an arc.

5. With Q as centre and QR (= 5 cm) draw another arc to cut the arc of step 4 at R.

6. Join PS, PQ, RS and RQ.

PQRS is the required quadrilateral.

# Question 2

**Construct a parallelogram ABCD in which AB = 3.5 cm, BC = 4 cm and AC = 6.5 cm.**

**Solution:**

Steps of construction:

1. Draw AB = 3.5cm.

2. With A as centre and AC (= 6.5cm) as radius draw an arc above AB.

3. With B as centre and BC (= 4cm) as radius draw another arc to cut the arc of step 2 at C.

4. With A as centre and AD (= 4cm) as radius draw an arc.

5. With C as centre and CD (= 3.5cm) as radius draw another arc to cut the arc of step 4 at D.

6. Join BC, CD and DA.

ABCD is the required parallelogram.

# Question 3

**Is it possible to construct a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 5.5 cm, DA = 6 cm and BD = 9 cm? If not, give reason.**

**Solution:**

The measurements must be such that the sum of any two sides of a traingle is greater than the third side.

AB = 3 cm, BD = 9 cm, and DA = 6 cm.

AB + AD = 3 cm + 6 cm = 9 cm =BD

âˆ´ Î” ABD cannot be constructed.

âˆ´ The quadrilateral ABCD cannot constructed.

# Question 4

**Construct a quadrilateral ABCD in which AB = 5 cm, BC = 4 cm, AD = 3 cm,**

**CD = 6 cm and BD = 5 cm.**

**Solution:**

Steps of construction:

1. Draw AB = 5 cm.

2. With A as centre and AD (= 3 cm) as radius, draw an arc.

3. With B as centre and BD (= 5 cm) as radius draw another arc to cut the arc of step 2 at D.

4. With B as centre and BC (= 4 cm) as radius draw an arc.

5. With D as centre and DC (= 6 cm) as radius draw another arc to cut the arc of step 4 at C.

6. Join AD, DB, CD and BD.

ABCD is the required quadrilateral.

# Question 5

**Is it possible to construct a quadrilateral ABCD in which AB = 3cm, CD = 3 cm, DA = 7.5 cm, AC = 8 cm and BD = 4 cm? If not, given reason.**

**Solution:**

The measurements must be such that the sum of any two sides of a traingle is greater than the third side.

AB = 3 cm, CD = 3 cm, and DA = 7.5 cm, AC = 8 cm and BD = 4 cm.

BD + AB = 4 cm + 3 cm = 7 cm < AD (= 7.5 cm.)

âˆ´ Î” ABD cannot be constructed.

âˆ´ The quadrilateral ABCD cannot constructed.

No, the quadrilateral ABCD cannot be constructed since BD + AB < AD.

# Question 6

**Construct a quadrilateral ABCD in which AB = 7 cm, AD = 6 cm, AC = 7 cm, BD = 7.5 cm and BC = 4cm.**

**Solution:**

Steps of construction:

1. Draw AB = 7 cm.

2. With A as centre and AC (= 7 cm) as radius draw an arc.

3. With B as centre and BC (= 4 cm) as radius draw another arc to cut the arc of step 2 at C.

4. With A as centre and AD (= 6 cm) as radius draw an arc.

5. With B as centre and BD (= 7.5 cm) as radius draw another arc to cut the arc of step 4 at D.

6. Join AC, BC, AD and CD.

ABCD is the required quadrilateral.

# Question 7

**Construct a quadrilateral ABCD in which AB = CD = 3 cm, BC = 2.5 cm, AC = 4 cm and BD = 5 cm.**

**Solution:**

Steps of construction:

1. Draw AB = 3 cm.

2. With A as centre and AC (= 4 cm) as radius draw an arc.

3. With B as centre and BC (= 2.5 cm) draw another arc to cut the arc of step 2 at C.

4. With C as centre and radius = 3cm draw an arc

5. With B as centre and radius = 5cm draw an arc to cut the arc in step 4 at D

6. Join BC, CD and AD.

ABCD is the required quadrilateral.

# Question 8

**Construct a quadrilateral PQRS in which QR = 7.5 cm, RP = PS = 6 cm, RS = 5 cm and QS = 10 cm.**

**Solution:**

Steps of construction:

1. Draw QR = 7.5 cm.

2. With Q as centre and QS (= 10 cm) as radius draw an arc.

3. With R as centre and RS (= 5 cm) as radius draw an arc to cut the arc of step 2 at S.

4. With R and S as centre and radius 6 cm draw two arcs to cut each other at P.

5. Join PQ, PS and SR.

PQRS is the required quadrilateral.

# Question 9

**Construct a quadrilateral ABCD in which BC = 5.5 cm, CD = 4 cm, âˆ A = 70**Â°,

**âˆ B = 110Â° and âˆ D = 85Â°.**

**Solution:**

âˆ CÂ = 360Â° - (âˆ A + âˆ B + âˆ D)

Â Â Â Â Â Â Â Â Â Â Â Â = 360Â° - (70Â° + 110Â° + 85Â° )

Â Â Â Â Â Â Â Â Â Â Â Â = 360Â° - 255Â°Â Â Â Â Â Â Â Â Â Â Â Â = 105Â°

Steps of construction:

1. Draw BC = 5.5 cm.

2. At B, draw an angle BCCâ€™ of measure 105Â° using a protractor.

3. With C as centre draw an arc on CCâ€™ such that CD = 4 cm.

4. At D, draw an angle of measure 85Â° using a protractor.

5. At B, draw an angle BCCâ€™ of measure 110Â° using a protractor to cut DDâ€™ at A.

ABCD is the required quadrilateral.

# Question 10

**Is it possible to construct a quadrilateral ABCD in which AB = 5 cm, BC = 7.5 cm, âˆ A = 80**Â°,

**âˆ B = 140Â° and âˆ C = 145Â°? If not, give reason.**

**Solution:**

No, it is not possible to construct a quadrilateral ABCD with the given measurements.

âˆ A + âˆ B + âˆ C (= 80Â° + 140Â° + 145Â° = 365Â° ) is greater than 360Â° .

The sum of all the four angles is 360Â° , quadrilateral cannot be constructed.

# Question 11

**Construct a quadrilateral ABCD in which AB = 4.5 cm, BC = 3.5 cm, CD = 5 cm, âˆ B = 45**

^{0}and âˆ C = 150^{Â°}.

**Solution:**

Steps of construction:

1. Draw BC = 3.5 cm.

2. At B, draw BM perpendicular to BC.

3. Construct BBâ€™, the bisector of âˆ MBC to get âˆ B'BC = 45Â° .

4. On BBâ€™, mark a point A such that BA = 4.5cm.

5. At C, draw CCâ€™ perpendicular to BC.

6. At C, construct an angle Câ€™DC = 60Â°, to get âˆ C = 150Â°.

7. Mark a point D on CDâ€™ such that CD = 5 cm.

8. Join AD.

ABCD is the required quadrilateral.