Class 10th, construction Exe-11.2 (Solution2)

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                             Exercise-11.2 Solution based on NCERT
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5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Construction Procedure: The tangent for the given circle can be constructed as follows.

1. Draw a line segment AB = 8 cm.
2. Take A as centre and draw a circle of radius 4 cm
3. Take B as centre, draw a circle of radius 3 cm
4. Draw the perpendicular bisector of the line AB and the midpoint is taken as M.
5. Now, take M as centre draw a circle with the radius of MA or MB which the intersects the circle at the points P, Q, R and S.
6. Now join AR, AS, BP and BQ
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 radius is 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.

∠ASB is an angle in the semi-circle.

We know that an angle in a semi-circle is a right angle.
∴ ∠ASB = 90° ⇒ BS ⊥ 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.

6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Construction Procedure: The tangent for the given circle can be constructed as follows

1. Draw the line segment with base BC =8cm
2. Measure the angle 90° at the point B, such that ∠ B = 90°.
3. Take B as centre and draw an arc with a measure of 6cm
4. Let the point be A where the arc intersects the ray
5. Join the line AC.
6. Therefore, ABC be the required triangle.
7. Now, draw the perpendicular bisector to the line BC and the midpoint is marked as E.
8. Take E as centre and BE or EC measure as radius draw a circle.
9. Join A to the midpoint E of the circle
10. Now, again draw the perpendicular bisector to the line AE and the midpoint is taken as M
11. Take M as Centre and AM or ME measure as radius, draw a circle.
12. This circle intersects the previous circle at the points B and Q
13. Join the points A and Q
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.

∠AQE is an angle in the semi-circle.

We know that an angle in a semi-circle is a right angle.

∴ ∠AQE = 90° ⇒ EQ ⊥ AQ

Since EQ is the radius of the circle, AQ has to be a tangent of the circle.

Similarly, ∠B = 90° ⇒ AB ⊥ BE Since BE is the radius of the circle, AB has to be a tangent of the circle. Hence, justified.

7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Construction Procedure: The required tangents can be constructed on the given circle as follows.

1. Draw a circle with the help of a bangle.
2. Draw two non-parallel chords such as AB and CD
3. Draw the perpendicular bisector of AB and CD
4. Take the centre as O where the perpendicular bisector intersects.
5. To draw the tangents, take a point P outside the circle.
6. Join the points O and P.
7. Now draw the perpendicular bisector of the line PO and midpoint is taken as M
8. Take M as centre and MO as radius draw a circle.
9. Let the circle intersects intersect the circle at the points Q and R
10. Now join PQ and PR
11. Therefore, PQ and PR are the required tangents.

Justification:

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

Since, ∠PQO is an angle in the semi-circle.

We know that an angle in a semi-circle is a right angle.

∴ ∠PQO = 90° ⇒ OQ ⊥ PQ

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

Similarly, ∴ ∠PRO = 90° ⇒ OR ⊥ 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.

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