CBSE 10th Maths - constructions - Exercise 11.2

Solution:

Construction Procedure

The construction to draw a pair of tangents to the given circle is as follows.

Step 1: Draw a circle with a radius = 6 cm with centre O.

Step 2: Locate a point P, which is 10 cm away from O.

Step 3: Join points O and P through the line.

Step 4: Draw the perpendicular bisector of the line OP.

Step 5: Let M be the mid-point of the line PO.

Step 6: Take M as the centre and measure the length of MO.

Step 7: The length MO is taken as the radius, and draw a circle.

Step 8: The circle drawn with the radius of MO intersect the previous circle at point Q and R.

Step 9: Join PQ and PR.

Step 10: Therefore, PQ and PR are the required tangents.

Justification

The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 6cm with centre O.

To prove this, join OQ and OR represented in dotted lines.

From the construction,

$\angle PQO$ is an angle in the semi-circle.

We know that angle in a semi-circle is a right angle, so it becomes

∴$\angle PQO$ = 90°

Such that

OQ $\perp$ PQ

Since OQ is the radius of the circle with a radius of 6 cm, PQ must be a tangent of the circle. Similarly, we can prove that PR is a tangent of the circle.

Hence, justified.

Solution:

Construction Procedure

For the given circle, the tangent can be drawn as follows.

Step 1: Draw a circle of 4 cm radius with centre “O”.

Step 2: Again, take O as the centre and draw a circle of radius 6 cm.

Step 3: Locate a point P on this circle.

Step 4: Join the points O and P through lines, such that it becomes OP.

Step 5: Draw the perpendicular bisector to the line OP

Step 6: Let M be the mid-point of PO.

Step 7: Draw a circle with M as its centre and MO as its radius,

Step 8: The circle drawn with the radius OM intersects the given circle at the points Q and R.

Step 9: Join PQ and PR.

Step 10: PQ and PR are the required tangents.

From the construction, it is observed that PQ and PR are of length 4.47 cm each.

It can be calculated manually as follows:

In ∆PQO,

Since PQ is a tangent,

$\angle PQO$ = 90°. PO = 6cm and QO = 4 cm

Applying Pythagoras’ theorem in ∆PQO, we obtain $PQ^2+QO^2 = PQ^2$

$PQ^2+(4)^2 = (6)^2$

$PQ^2$ +16 =36

$PQ^2$ = 36−16

$PQ^2$ = 20

PQ = $2\sqrt{5}$

PQ = 4.47 cm

Therefore, the tangent length PQ = 4.47

Justification

The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 4 cm with centre O.

To prove this, join OQ and OR represented in dotted lines.

From the construction,

$\angle PQO$ is an angle in the semi-circle.

We know that angle in a semi-circle is a right angle, so it becomes

∴ $\angle PQO$ = 90°

Such that

OQ $\perp$ PQ

Since OQ is the radius of the circle with a radius of 4 cm, PQ must be a tangent of the circle. Similarly, we can prove that PR is a tangent of the circle.

Hence, justified.

Solution:

Construction Procedure

The tangent for the given circle can be constructed as follows.

Step 1: Draw a circle with a radius of 3cm with a centre “O”.

Step 2: Draw a diameter of a circle, and it extends 7 cm from the centre, and mark it as P and Q.

Step 3: Draw the perpendicular bisector of the line PO and mark the midpoint as M.

Step 4: Draw a circle with M as the centre and MO as the radius.

Step 5: Now, join the points PA and PB in which the circle with radius MO intersects the circle of circle 3cm.

Step 6: Now, PA and PB are the required tangents.

Step 7: Similarly, from point Q, we can draw the tangents.

Step 8: From that, QC and QD are the required tangents.

Justification

The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 3 cm with centre O.

To prove this, join OA and OB.

From the construction,

$\angle PAO$ is an angle in the semi-circle.

We know that angle in a semi-circle is a right angle, so it becomes

∴ $\angle PAO$ = 90°

Such that

OA $\perp$ PA

Since OA is the radius of the circle with a radius of 3 cm, PA must be a tangent of the circle. Similarly, we can prove that PB, QC and QD are the tangents of the circle.

Hence, justified.

Solution:

Construction Procedure

The tangents can be constructed in the following manner:

Step 1: Draw a circle of radius 5 cm, with the centre as O.

Step 2: Take a point Q on the circumference of the circle and join OQ.

Step 3: Draw a perpendicular to QP at point Q.

Step 4: Draw a radius OR, making an angle of 120°, i.e., (180°−60°) with OQ.

Step 5: Draw a perpendicular to RP at point R.

Step 6: Now, both the perpendiculars intersect at point P.

Step 7: Therefore, PQ and PR are the required tangents at an angle of 60°.

Justification

The construction can be justified by proving that $\angle QPR$ = 60°.

By our construction,

$\angle OQP$ = 90°

$\angle ORP$ = 90°

And $\angle QOR$ = 120°

We know that the sum of all interior angles of a quadrilateral = 360°

$\angle OQP+\angle QOR + \angle ORP +\angle QPR$ = 360°

90°+120°+90°+$\angle QPR$ = 360°

Therefore, $\angle QPR$ = 60°

Hence, justified.

Solution:

Construction Procedure

The tangent for the given circle can be constructed as follows:

Step 1: Draw a line segment AB = 8 cm.

Step 2: Take A as the centre and draw a circle of radius 4 cm.

Step 3: Take B as the centre and draw a circle of radius 3 cm.

Step 4: Draw the perpendicular bisector of the line AB, and the midpoint is taken as M.

Step 5: Now, take M as the centre and draw a circle with the radius of MA or MB, which intersects the circle at the points P, Q, R and S.

Step 6: Now, join AR, AS, BP and BQ.

Step 7: Therefore, the required tangents are AR, AS, BP and BQ.

Justification

The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B with a radius of 3 cm), and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm).

From the construction, to prove this, join AP, AQ, BS, and BR.

$\angle ASB$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ $\angle ASB$ = 90°

BS $\perp$ AS

Since BS is the radius of the circle, AS must be a tangent of the circle.

Similarly, AR, BP, and BQ are the required tangents of the given circle.

Solution:

Construction Procedure

The tangent for the given circle can be constructed as follows:

Step 1: Draw the line segment with base BC = 8cm.

Step 2: Measure the angle 90° at the point B, such that $\angle B$ = 90°.

Step 3: Take B as the centre and draw an arc with a measure of 6cm.

Step 4: Let the point be A, where the arc intersects the ray.

Step 5: Join the line AC.

Step 6: Therefore, ABC is the required triangle.

Step 7: Now, draw the perpendicular bisector to the line BC, and the midpoint is marked as E.

Step 8: Take E as the centre, and BE or EC measure as the radius and draw a circle.

Step 9: Join A to the midpoint E of the circle.

Step 10: Now, again, draw the perpendicular bisector to the line AE, and the midpoint is taken as M.

Step 11. Take M as the centre, and AM or ME measure as the radius and draw a circle.

Step 12. This circle intersects the previous circle at points B and Q.

Step 13. Join points A and Q.

Step 14. Therefore, AB and AQ are the required tangents.

Justification

The construction can be justified by proving that AG and AB are the tangents to the circle.

From the construction, join EQ.

$\angle AQE$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ $\angle AQE$ = 90°

EQ $\perp$ AQ

Since EQ is the radius of the circle, AQ has to be a tangent of the circle. Similarly, $\angle B$ = 90°

AB $\perp$ BE

Since BE is the radius of the circle, AB has to be a tangent of the circle.

Hence, justified.

Solution:

Construction Procedure

The required tangents can be constructed on the given circle as follows:

Step 1: Draw a circle with the help of a bangle.

Step 2: Draw two non-parallel chords, such as AB and CD.

Step 3: Draw the perpendicular bisector of AB and CD.

Step 4: Take the centre as O, where the perpendicular bisector intersects.

Step 5: To draw the tangents, take a point P outside the circle.

Step 6: Join points O and P.

Step 7: Now, draw the perpendicular bisector of the line PO, and the midpoint is taken as M

Step 8: Take M as the centre and MO as the radius and draw a circle.

Step 9: Let the circle intersects intersect the circle at the points Q and R.

Step 10: Now, join PQ and PR.

Step 11: Therefore, PQ and PR are the required tangents.

Justification

The construction can be justified by proving that PQ and PR are tangents to the circle.

Since, O is the centre of a circle, we know that the perpendicular bisector of the chords passes through the centre.

Now, join the points OQ and OR.

We know that the perpendicular bisector of a chord passes through the centre.

It is clear that the intersection point of these perpendicular bisectors is the centre of the circle.

$\angle PQO$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ $\angle PQO$ = 90°

OQ $\perp$ PQ

Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly,

∴ $\angle PRO$ = 90°

OR $\perp$ PO

Since OR is the radius of the circle, PR has to be a tangent of the circle.

Therefore, PQ and PR are the required tangents of a circle.