Constructions (Circles) — ICSE Class 10 Maths Important Questions

Answer: (b) Perpendicular bisectors of the sides.

The circumcentre is equidistant from the vertices and is located at the intersection of the perpendicular bisectors of the sides.

Answer: (b) Internal bisectors of the angles.

The incentre is equidistant from the sides, so it lies where the internal angle bisectors meet.

Answer: (c) $2$2.

Exactly two tangents can be drawn to a circle from an external point (from a point on the circle only one, and from an interior point none).

Answer: (c) $120^{\circ}$120^.

In quadrilateral $OAPB$OAPB, $\angle OAP=\angle OBP=90^{\circ}$OAP= OBP=90^, so $\angle AOB=360^{\circ}-90^{\circ}-90^{\circ}-60^{\circ}=120^{\circ}$AOB=360^-90^-90^-60^=120^.

Answer: (a) The construction draws the perpendicular to $OP$OP at $P$P precisely because the tangent must be perpendicular to the radius at the point of contact; hence R correctly explains A.

Steps of construction:

1. Draw a circle with centre $O$O and radius $4\,\text{cm}$4\,cm; mark any point $P$P on it.
2. Join $OP$OP (the radius to the point of contact).
3. At $P$P, construct a line perpendicular to $OP$OP (using compasses to erect the perpendicular).

This perpendicular line is the required tangent, since a tangent is perpendicular to the radius at the point of contact.

In a right-angled triangle the hypotenuse subtends a right angle at the opposite vertex, so the hypotenuse is a diameter of the circumcircle (angle in a semicircle).

Hence the circumcentre is the midpoint of the hypotenuse $AC$AC, and the circumradius is
$R=\frac{AC}{2}=\frac{6}{2}=3\,\text{cm}.$R=AC/2=6/2=3\,cm.

Steps of construction:

1. Draw $BC=6\,\text{cm}$BC=6\,cm. With centre $B$B and radius $5\,\text{cm}$5\,cm, and centre $C$C and radius $7\,\text{cm}$7\,cm, draw arcs meeting at $A$A. Join $AB$AB and $AC$AC to complete $\triangle ABC$ABC.
2. Construct the perpendicular bisector of $BC$BC (equal arcs from $B$B and $C$C, joined).
3. Construct the perpendicular bisector of $AC$AC in the same way.
4. Let the two perpendicular bisectors meet at $O$O; this is the circumcentre.
5. With centre $O$O and radius $OA$OA (=,$OB=OC$OB=OC) draw the circle; it passes through $A$A, $B$B, $C$C and is the required circumcircle.

Steps of construction:

1. Draw $QR=6\,\text{cm}$QR=6\,cm. At $Q$Q construct $\angle Q=60^{\circ}$Q=60^ and cut off $QP=5\,\text{cm}$QP=5\,cm; join $PR$PR to complete $\triangle PQR$PQR.
2. Construct the bisector of $\angle Q$Q.
3. Construct the bisector of $\angle R$R. Let the two bisectors meet at $I$I; this is the incentre.
4. From $I$I, drop a perpendicular $IM$IM to side $QR$QR.
5. With centre $I$I and radius $IM$IM, draw the circle; it touches all three sides and is the required incircle.

Steps of construction:

1. Draw a circle with centre $O$O and radius $3\,\text{cm}$3\,cm. Mark $P$P with $OP=7\,\text{cm}$OP=7\,cm and join $OP$OP.
2. Construct the perpendicular bisector of $OP$OP; let it meet $OP$OP at its midpoint $M$M.
3. With centre $M$M and radius $MO\ (=MP)$MO (=MP), draw a circle; it cuts the given circle at points $A$A and $B$B.
4. Join $PA$PA and $PB$PB. These are the required tangents, because $\angle OAP=\angle OBP=90^{\circ}$OAP= OBP=90^ (angles in the semicircle on diameter $OP$OP), so $PA\perp OA$PA OA and $PB\perp OB$PB OB.

The length of each tangent $=\sqrt{OP^2-OA^2}=\sqrt{7^2-3^2}=\sqrt{40}=2\sqrt{10}\approx6.32\,\text{cm}.$=√OP^2-OA^2=√7^2-3^2=√40=2√106.32\,cm.

Steps of construction:

1. Draw a circle with centre $O$O and radius $4\,\text{cm}$4\,cm (the side of a regular hexagon equals its circumradius).
2. Starting from any point $A$A on the circle, step off chords of $4\,\text{cm}$4\,cm six times to get $B$B, $C$C, $D$D, $E$E and $F$F, and join them to form the hexagon. This circle is the circumcircle: $R=4\,\text{cm}$R=4\,cm.
3. For the incircle, draw the perpendicular bisectors of two adjacent sides (or the bisectors of two interior angles); they meet at $O$O. Drop a perpendicular $OM$OM to any side.
4. With centre $O$O and radius $OM$OM, draw the incircle; it touches all six sides. $r=OM\approx3.5\,\text{cm}$r=OM3.5\,cm (exactly $4\times\dfrac{\sqrt3}{2}\approx3.46\,\text{cm}$4×3/23.46\,cm).

Steps of construction:

1. Draw $BC=7\,\text{cm}$BC=7\,cm. At $B$B construct $45^{\circ}$45^ and at $C$C construct $60^{\circ}$60^; the arms meet at $A$A, completing $\triangle ABC$ABC (with $\angle A=75^{\circ}$A=75^).
2. Construct the perpendicular bisectors of $BC$BC and $AB$AB; they meet at $O$O, the circumcentre.
3. With centre $O$O and radius $OB$OB draw the circumcircle through $A$A, $B$B, $C$C.
4. Measure $OB$OB; on an accurate drawing the circumradius is about $3.6\,\text{cm}$3.6\,cm.

Reason for equidistance: Any point on the perpendicular bisector of a segment is equidistant from its two endpoints. $O$O lies on the perpendicular bisector of $BC$BC, so $OB=OC$OB=OC; it also lies on the perpendicular bisector of $AB$AB, so $OA=OB$OA=OB. Therefore $OA=OB=OC$OA=OB=OC, i.e. $O$O is equidistant from all three vertices, which is why a single circle passes through them.

Steps of construction:

1. Draw a circle with centre $O$O and radius $4\,\text{cm}$4\,cm. Draw radii $OA$OA and $OB$OB with $\angle AOB=120^{\circ}$AOB=120^.
2. At $A$A, construct the perpendicular to $OA$OA; at $B$B, construct the perpendicular to $OB$OB. These are the tangents at $A$A and $B$B.
3. Let the two tangents meet at $P$P.

Finding $\angle APB$APB: In quadrilateral $OAPB$OAPB, $\angle OAP=\angle OBP=90^{\circ}$OAP= OBP=90^, so
$\angle APB=360^{\circ}-90^{\circ}-90^{\circ}-120^{\circ}=60^{\circ}.$APB=360^-90^-90^-120^=60^.

Finding $OP$OP: $OP$OP bisects $\angle AOB$AOB, so $\angle AOP=60^{\circ}$AOP=60^. In right triangle $OAP$OAP (right-angled at $A$A),
$\cos 60^{\circ}=\frac{OA}{OP}\ \Rightarrow\ OP=\frac{OA}{\cos 60^{\circ}}=\frac{4}{\tfrac12}=8\,\text{cm}.$60^=OA/OP OP=OA 60^=4/12=8\,cm.

Hence $\angle APB=60^{\circ}$APB=60^ and $OP=8\,\text{cm}$OP=8\,cm.

(i) The incentre is located by drawing the internal bisectors of the angles of $\triangle ABC$ABC (any two suffice); their intersection is the centre. The radius is the perpendicular distance from that centre to a side.

(ii) Since $\angle B=90^{\circ}$B=90^, by Pythagoras
$AC=\sqrt{AB^2+BC^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\,\text{cm}.$AC=√AB^2+BC^2=√6^2+8^2=√36+64=√100=10\,cm.

(iii) Area $=\tfrac12\times AB\times BC=\tfrac12\times6\times8=24\,\text{cm}^2$=12× AB× BC=12×6×8=24\,cm^2. Semi-perimeter $s=\dfrac{6+8+10}{2}=12\,\text{cm}$s=6+8+10/2=12\,cm.
$r=\frac{\text{Area}}{s}=\frac{24}{12}=2\,\text{cm}.$r=Area/s=24/12=2\,cm.

(iv) The designer must set the compasses to the inradius, $r=2\,\text{cm}$r=2\,cm, with the point at the incentre.