Construction of Polygons — ICSE Class 9 Maths Important Questions

Answer: (c) $5$5.

A quadrilateral has five degrees of freedom, so five independent measurements are needed to construct it uniquely.

Answer: (a) Its two diagonals.

The two diagonals of a rhombus bisect each other at right angles, so knowing both diagonals fixes the rhombus uniquely. One side alone or the perimeter alone is not enough.

Answer: (b) $4$4 cm.

In a regular hexagon the side equals the radius of the circumscribing circle, so the radius is $4$4 cm.

Answer: (c) Four sides only.

Four sides give only four measurements and leave the shape 'floppy' (it can flex), so a unique quadrilateral is not fixed. A fifth measurement such as a diagonal or an angle is required.

Answer: (a) Because a square has all sides equal and all angles $90^\circ$90^, the single side length determines every side and angle, so it can be constructed from the side alone. R correctly explains A.

A parallelogram is fixed by two adjacent sides and the included angle between them.

For example: $AB=5$AB=5 cm, $BC=3.5$BC=3.5 cm and $\angle ABC=60^\circ$ABC=60^.

(The opposite sides then equal these, so the parallelogram is uniquely determined. Alternatively, two adjacent sides and one diagonal also suffice.)

1. Draw diagonal $AC=8$AC=8 cm and find its mid-point $O$O (draw its perpendicular bisector).

2. Along the perpendicular at $O$O, mark $OB=3$OB=3 cm and $OD=3$OD=3 cm on opposite sides (since $BD=6$BD=6 cm is bisected at $O$O).

3. Join $A,B,C,D$A,B,C,D in order.

$ABCD$ABCD is the required rhombus, because the diagonals of a rhombus bisect each other at right angles.

This is the SASAS case (three sides and the two included angles).

Steps of construction:

1. Draw $BC=5$BC=5 cm.

2. At $B$B, construct $\angle CBX=105^\circ$CBX=105^. From $B$B along $BX$BX, cut off $BA=4$BA=4 cm.

3. At $C$C, construct $\angle BCY=80^\circ$BCY=80^. From $C$C along $CY$CY, cut off $CD=6.5$CD=6.5 cm.

4. Join $A$A to $D$D.

$ABCD$ABCD is the required quadrilateral.

In a regular hexagon the side equals the circum-radius.

Steps of construction:

1. Draw a circle of radius $3.5$3.5 cm with centre $O$O.

2. Mark any point $A$A on the circle.

3. With the compass still set to $3.5$3.5 cm (the radius), and centre $A$A, cut the circle at $B$B.

4. With centre $B$B and the same radius, cut the circle at $C$C; continue similarly to get $D$D, $E$E and $F$F around the circle.

5. Join $AB$AB, $BC$BC, $CD$CD, $DE$DE, $EF$EF and $FA$FA.

$ABCDEF$ABCDEF is the required regular hexagon, since each side subtends $60^\circ$60^ at the centre and equals the radius.

A quadrilateral has five degrees of freedom, so five independent measurements are needed to fix it.

Here only four sides are given, which is four measurements. Four rods hinged at the corners form a 'floppy' frame: the shape can flex, changing its angles and diagonals while the four side lengths stay the same. Infinitely many different quadrilaterals have these same four sides, so the construction is not unique.

To make it unique, one more independent measurement is required, for example:

• one diagonal (say $AC$AC or $BD$BD), or
• one interior angle (say $\angle B$B).

With four sides and one diagonal (or one angle), the quadrilateral splits into two determinable triangles and is uniquely constructed.

This is the SSSSD case (four sides and one diagonal). The diagonal $AC$AC divides the quadrilateral into $\triangle ABC$ABC and $\triangle ACD$ACD.

Steps of construction:

1. Draw $AC=7$AC=7 cm.

2. Construct $\triangle ABC$ABC: With centre $A$A and radius $AB=4.5$AB=4.5 cm, draw an arc on one side of $AC$AC. With centre $C$C and radius $BC=5.5$BC=5.5 cm, draw another arc to cut the first at $B$B. Join $AB$AB and $CB$CB.

3. Construct $\triangle ACD$ACD: With centre $A$A and radius $DA=6$DA=6 cm, draw an arc on the other side of $AC$AC. With centre $C$C and radius $CD=4$CD=4 cm, draw an arc to cut it at $D$D. Join $AD$AD and $CD$CD.

4. Join the vertices in order $A\to B\to C\to D\to A$A B C D A.

$ABCD$ABCD is the required quadrilateral. It is a general (irregular) quadrilateral in which the diagonal $AC=7$AC=7 cm splits it into two triangles that were each fixed by their three sides (SSS), guaranteeing a unique figure.

Interior angle: For a regular pentagon ($n=5$n=5), each interior angle $=\dfrac{(5-2)\times180^\circ}{5}=\dfrac{540^\circ}{5}=108^\circ$=(5-2)×180^/5=540^/5=108^.

Steps of construction:

1. Draw the first side $AB=4$AB=4 cm.

2. At $B$B, construct an angle of $108^\circ$108^ ($\angle ABC=108^\circ$ABC=108^). Along the new ray cut off $BC=4$BC=4 cm.

3. At $C$C, construct $\angle BCD=108^\circ$BCD=108^. Along the ray cut off $CD=4$CD=4 cm.

4. At $D$D, construct $\angle CDE=108^\circ$CDE=108^. Along the ray cut off $DE=4$DE=4 cm.

5. Join $E$E to $A$A.

If constructed accurately, $EA=4$EA=4 cm and $\angle DEA=\angle EAB=108^\circ$DEA= EAB=108^ automatically, since the exterior angles ($72^\circ$72^ each) sum to $360^\circ$360^.

$ABCDE$ABCDE is the required regular pentagon, with every side $4$4 cm and every interior angle $108^\circ$108^.

(i) Four sides and one diagonal (SSSSD): five independent measurements are given, so a unique quadrilateral can be constructed. The diagonal splits it into two SSS triangles.

(ii) Two adjacent sides and the included angle of a parallelogram: this fixes the parallelogram (opposite sides equal), so a unique parallelogram is constructed.

(iii) Two diagonals of a rhombus: since the diagonals bisect each other at right angles, this data gives a unique rhombus.

(iv) Only four sides are given (four measurements). A quadrilateral needs five, so the frame is flexible and not unique; one more datum (a diagonal or an angle) is needed.