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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

Δ ABD cannot be constructed.

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.)

Δ ABD cannot be constructed.

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, A = 70°, B = 110° and D = 85°.


Solution:
C  = 360° - ( A + B + D)
            
= 360° - (70° + 110° + 85° )
             = 360° - 255°             = 105°

Steps of construction:

1. Draw BC = 5.5 cm.
2. At B, draw an angle BCC
of measure 105° using a protractor.
3. With C as centre draw an arc on CC
such that CD = 4 cm.
4. At D, draw an angle of measure 85
° using a protractor.
5. At B, draw an angle BCC
of measure 110° using a protractor to cut DD 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, A = 80°, B = 140° and C = 145°? If not, give reason.


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

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

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

Question 11


Construct a quadrilateral ABCD in which AB = 4.5 cm, BC = 3.5 cm, CD = 5 cm, B = 450 and C = 150°.

 


Solution:

Steps of construction:

1. Draw BC = 3.5 cm.
2. At B, draw BM perpendicular to BC.
3. Construct BB
, the bisector of MBC to get B'BC = 45° .
4. On BB
, mark a point A such that BA = 4.5cm.
5. At C, draw CC
perpendicular to BC.
6. At C, construct an angle C
DC = 60°, to get C = 150°.
7. Mark a point D on CD
such that CD = 5 cm.
8. Join AD.

ABCD is the required quadrilateral.





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