Q: Draw a line segment PQ= 8 cm. Construct the perpendicular bisector of the line segment PQ. Let the perpendicular bisector drawn meets PQ at point R. Measure the length of PR and QR. Is PR=QR?

Steps of Construction : \$1.\$ With \$P\$ and \$Q\$ as centers, draw arcs on both sides of \$PQ\$ with equal radii. The radius should be more than half the length of \$PQ\$. \$2.\$ Let these arcs cut each other at points \$R\$ and \$RS\$ \$3.\$ Join \$RS\$ which cuts \$PQ\$ at \$D\$. Then \$RS = PQ\$. Also \$\angle{POR} = {90}^{\circ}\$.Hence, the line segment \$RS\$ is the perpendicular bisector of \$PQ\$ as it bisects \$PQ\$ at \$P\$ and is also perpendicular to \$PQ\$. On measuring the lengths of \$PR = 4\$cm, \$QR = 4\$ cm Since \$PR = QR\$, both are \$4\$cm each \$\therefore PR = QR\$.

Q: Draw a circle of radius \$3\$cm.At a point \$P\$ on it, draw a tangent to the circle without using centre.

\$1.\$Draw a circle with radius \$3\$cm\$2.\$Take a point \$P\$ on the circle.\$3.\$Through \$P,\$ draw any chord \$PQ\$\$4.\$Mark a point \$R\$ either in major arc or in minor arc.\$5.\$Join \$PR\$ and \$QR\$\$6.\$At \$P,\,\,\$construct \$\angle{QPT}\$ equal to \$\angle{PRQ}\$ on the opposite side of the chord \$PQ\$\$7.\$Produce \$TP\$ to \${T}^{\prime}\$ to get a straight line \${T}^{\prime}PT\$.

Q: With the same centre O, draw two circles of radii \$4\$cm and \$2.5\$cm.

The required circle can be drawn as follows.Step \$1\$First, open the compasses for the required radius \$4\$cm.Step \$2\$Make a point 'O' where we want the centre of the circle to be.Step \$3\$Place the pointer of compasses on O.Step \$4\$Turn the compasses slowly to draw the circle.Step \$5\$Now, open the compasses for \$2.5\$cm.Step \$6\$Again put the pointer of the compasses on point 'O' and turn the compasses slowly to draw the circle.

Q: Draw any line segment \$AB\$. Mark any point \$M\$ on it. Through \$M\$, draw a perpendicular using ruler and compass.

Steps of construction:\$(i)\$ With \$M\$ as centre and a convenient radius, draw an arc intersecting the line \$AB\$ at two points \$C\$ and \$B\$.\$(ii)\$ With \$C\$ and \$D\$ as centres and a radius greater than \$MC\$, draw two arcs, which cut each other at \$P\$.\$(iii)\$ Join \$PM\$. Then \$PM\$ is perpendicular to \$AB\$ through the point \$M\$.

Q: Construct \$\overline{AB}\$ of length \$7.8\ cm\$. Find this, cut off \$AC\$ of length \$4.7\ cm\$ . Measure \$\overline {BC}\$

Step\$1\$ Draw a line \$l\$.Mark a point \$A\$ on it.Step\$2\$ Since length of \$AB\$ is \$7.8\$cm, we measure \$7.8\$cm using ruler and compass.Step\$3\$Now keeping compass opened the same length.We keep pointed end at point \$A\$, and draw an arc on line \$l\$.So the required line segment is \$\bar{AB}=7.8\$cmStep\$4\$Now, we need to cut of \$AC\$ of length \$4.7\$cm.We measure \$4.7\$cm using ruler and compass.Step\$5\$Now keeping compass opened the same length.We keep pointed end at point \$A\$, and draw an arc on line \$l\$.Result:The required line segment \$\bar{AC}=4.7\$cm

Question 1

Draw a line segment of length 6 cm. Construct its perpendicular bisector. Measure the two parts of the line segment.

Accepted Answer

Steps of construction:Step I - draw a line segment AB= 6 cm with the help of ruler.Step II - With vertex A, by using compass take more than half of AB and draw arcs, one each side of side of AB. Step III- With vertex B, by using compass with same length, cut the previous arcs at X and Y respectively.Each part measures $=3\ cm$

Question 2

Construct an angle of ${90}^{o}$ at the initial point of a given ray and justify the construction.

Accepted Answer

Steps of construction Draw a line segment OA.Taking O as center and any radius, draw an arc cutting OA at B.Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.With C as center and the same radius, draw an arc cutting the arc at D.With C and D as center and radius more than $\dfrac{1}{2}$CD, draw two arc intersecting at P.Join OP.Thus, $\angle{AOP}=90^{o}$$\underline{Justification}$ Join OC and BCThus,$OB=BC=OC$             [Radius of equal arcs]$\therefore \triangle{OCB}$ is an equilateral triangle$\therefore \angle{BOC}=60^{o}$Join OD,OC and CDThus,$OD=OC=DC$             [Radius of equal arcs]$\therefore \triangle{DOC}$ is an equilateral triangle$\therefore \angle{DOC}=60^{o}$Join PD and PCNow,In $\triangle{ODP}$ and $\triangle{OCP}$$OD=OC$             [Radius of same arcs]$DP=CP$         [Arc of same radii]$OP=OP$        [Common]$\therefore\triangle{ODP}\cong\triangle{OCP}$       [SSS congruency]$\therefore \angle{DOP}=\angle{COP}$    [CPCT]So, we can say that$\angle{DOP}=\angle{COP}=\dfrac{1}{2}\angle{DOC}$$\angle{DOP}=\angle{COP}=\dfrac{1}{2}\times60=30^{o}$Now,$\angle{AOP}=\angle{BOC}+\angle{COP}$$\angle{AOP}=60+30$$\angle{AOP}=90^{o}$Hence justified.

Question 3

Which of the following angles cannot be constructed using a ruler and compass?

Accepted Answer

We know that using ruler and compass we can only draw angles which are multiples of 15 degree.15,45,75 degree all are multiples of 15 and only 85 is not the multiple so we cannot draw it using ruler and compass.Hence, option ( D ) is the correct answer.

Question 4

Draw a line segment PQ= 8 cm. Construct the perpendicular bisector of the line segment PQ. Let the perpendicular bisector drawn meets PQ at point R. Measure the length of PR and QR. Is PR=QR?

Accepted Answer

Steps of Construction : $1.$ With $P$ and $Q$ as centers, draw arcs on both sides of $PQ$ with equal radii. The radius should be more than half the length of $PQ$. $2.$ Let these arcs cut each other at points $R$ and $RS$ $3.$ Join $RS$ which cuts $PQ$ at $D$. Then $RS = PQ$. Also $\angle{POR} = {90}^{\circ}$.Hence, the line segment $RS$ is the perpendicular bisector of $PQ$ as it bisects $PQ$ at $P$ and is also perpendicular to $PQ$. On measuring the lengths of $PR = 4$cm, $QR = 4$ cm Since $PR = QR$, both are $4$cm each $\therefore PR = QR$.

Question 5

Given a line AB of 5 cm. Draw the perpendicular bisector of AB.

Accepted Answer

1. Draw a line segment AB = 5 cm.2. With any radius greater than 2.5 cm, and compass fixed at A, draw 2 arcs, one above and one below AB.3. With the same radius, compass at B, cut the 2 arcs drawn in step 2.4. The line joining these two arcs is the perpendicular bisector of AB.    i.e. CD is the perpendicular bisector of AB.

Question 6

Draw  a circle of radius $3$cm.At a point $P$ on it, draw a tangent to the circle without using centre.

Accepted Answer

$1.$Draw a circle with radius $3$cm$2.$Take a point $P$ on the circle.$3.$Through $P,$ draw any chord $PQ$$4.$Mark a point $R$ either in major arc or in minor arc.$5.$Join $PR$ and $QR$$6.$At $P,\,\,$construct $\angle{QPT}$ equal to $\angle{PRQ}$ on the opposite side of the chord $PQ$$7.$Produce $TP$ to ${T}^{\prime}$ to get a straight line ${T}^{\prime}PT$.

Question 7

With the same centre O, draw two circles of radii $4$cm and $2.5$cm.

Accepted Answer

The required circle can be drawn as follows.Step $1$First, open the compasses for the required radius $4$cm.Step $2$Make a point 'O' where we want the centre of the circle to be.Step $3$Place the pointer of compasses on O.Step $4$Turn the compasses slowly to draw the circle.Step $5$Now, open the compasses for $2.5$cm.Step $6$Again put the pointer of the compasses on point 'O' and turn the compasses slowly to draw the circle.

Question 8

Draw $\overline {AB}$ of length $7.3\ cm$ and find its axis of symmetry.

Accepted Answer

Let's first draw a line segment of length 7.3 cm .1) Draw a line l, mark point A on it.2) Since length is 7.3 cm, we measure 7.3 cm using ruler and compass.3) Now keeping compass opened the same length, we keep pointed end at point A and draw an arc on line l.So, the required line segment is AB = 7.3 cm .Now, we need to find its axis of symmetry.1) Given a line segment AB.2) Now with A as centre, and radius more than half AB, draw an arc on top and bottom of AB.3) Now with B as centre, and same radius, draw an arc on top and bottom of AB.4) Where the two arcs intersect above AB is point C and where the two arcs intersect below AB is point D.Join CD.CD is the line of symmetry of line segment AB.

Question 9

Draw any line segment $AB$. Mark any point $M$ on it. Through $M$, draw a perpendicular using ruler and compass.

Accepted Answer

Steps of construction:$(i)$ With $M$ as centre and a convenient radius, draw an arc intersecting the line $AB$ at two points $C$ and $B$.$(ii)$ With $C$ and $D$ as centres and a radius greater than $MC$, draw two arcs, which cut each other at $P$.$(iii)$ Join $PM$. Then $PM$ is perpendicular to $AB$ through the point $M$.

Question 10

Construct $\overline{AB}$ of length $7.8\ cm$. Find this, cut off $AC$ of length $4.7\ cm$ . Measure $\overline {BC}$

Accepted Answer

Step$1$ Draw a line $l$.Mark a point $A$ on it.Step$2$ Since length of $AB$ is $7.8$cm, we measure $7.8$cm using ruler and compass.Step$3$Now keeping compass opened the same length.We keep pointed end at point $A$, and draw an arc on line $l$.So the required line segment is $\bar{AB}=7.8$cmStep$4$Now, we need to cut of $AC$ of length $4.7$cm.We measure $4.7$cm using ruler and compass.Step$5$Now keeping compass opened the same length.We keep pointed end at point $A$, and draw an arc on line $l$.Result:The required line segment $\bar{AC}=4.7$cm