## RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.3

### RD Sharma Class 10 Solutions Constructions Exercise 11.3

Question 1.

Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Solution:

Steps of construction :

(i) Draw a circle with O centre and 6 cm radius.

(ii) Take a point P, 10 cm away from the centre O.

(iii) Join PO and bisect it at M.

(iv) With centre M and diameter PO, draw a circle intersecting the given circle at T and S.

(v) Join PT and PS.

Then PT and PS are the required tangents.

Question 2.

Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Solution:

Steps of construction :

(i) Draw a circle with centre O and radius 3 cm.

(ii) Draw a diameter and produce it to both sides.

(iii) Take two points P and Q on this diameter with a distance of 7 cm each from the centre O.

(iv) Bisect PO at M and QO at N

(v) With centres M and N, draw circle on PO and QO as diameter which intersect the given circle at S, T and S’, T’ respectively.

(vi) Join PS, PT, QS’ and QT’.

Then PS, PT, QS’ and QT’ are the required tangents to the given circle.

Question 3.

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. [CBSE 2013]

Solution:

Steps of construction :

(i) Draw a line segment AB = 8 cm.

(ii) With centre A and radius 4 cm and with centre B and radius 3 cm, circles are drawn.

(iii) Bisect AB at M.

(iv) With centre M and diameter AB, draw a circle which intersects the two circles at S’, T’ and S, T respectively.

(v) Join AS, AT, BS’and BT’.

Then AS, AT, BS’ and BT’ are the required tangent.

Question 4.

Draw two tangents to a circle of raidus 3.5 cm from a point P at a distance of 6.2 cm from its centre.

Solution:

Steps of construction :

(i) Draw a circle with centre O and radius 3.5 cm

(ii) Take a point P which is 6.2 cm from O.

(iii) Bisect PO at M and draw a circle with centre M and diameter OP which intersects the given circle at T and S respectively.

(iv) Join PT and PS.

PT and PS are the required tangents to circle.

Question 5.

Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°. [CBSE 2013]

Solution:

Steps of construction :

Angle at the centre 180° – 45° = 135°

(i) Draw a circle with centre O and radius 4.5 cm.

(ii) At O, draw an angle ∠TOS = 135°

(iii) At T and S draw perpendicular which meet each other at P.

PT and PS are the tangents which inclined each other 45°.

Question 6.

Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.

Solution:

Steps of Construction :

Draw a line segment BC = 8 cm

From B draw an angle of 90°

Draw an arc \(\breve { BA }\) = 6cm cutting the angle at A.

Join AC.

ΔABC is the required A.

Draw ⊥ bisector of BC cutting BC at M.

Take M as centre and BM as radius, draw a circle.

Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.

AB and AE are the required tangents.

Justification :

∠ABC = 90° (Given)

Since, OB is a radius of the circle.

∴ AB is a tangent to the circle.

Also AE is a tangent to the circle.

Question 7.

Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to the smaller circle from a point on the larger circle. Also, measure its length. [CBSE 2016]

Solution:

Given, two concentric circles of radii 3 cm and 5 cm with centre O. We have to draw pair of tangents from point P on outer circle to the other.

Steps of construction :

(i) Draw two concentric circles with centre O and radii 3 cm and 5 cm.

(ii) Taking any point P on outer circle. Join OP.

(iii) Bisect OP, let M’ be the mid-point of OP.

Taking M’ as centre and OM’ as radius draw a circle dotted which cuts the inner circle as M and P’.

(iv) Join PM and PP’. Thus, PM and PP’ are the required tangents.

(v) On measuring PM and PP’, we find that PM = PP’ = 4 cm.

Actual calculation:

In right angle ΔOMP, ∠PMO = 90°

∴ PM^{2} = OP^{2} – OM^{2
}[by Pythagoras theorem i.e. (hypotenuse)^{2} = (base)^{2} + (perpendicular)^{2}]

⇒ PM^{2} = (5)^{2} – (3)^{2} = 25 – 9 = 16

⇒ PM = 4 cm

Hence, the length of both tangents is 4 cm.

### RD Sharma Class 10th Solutions Chapter 11 Constructions Exercise 11.3 Q1

RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.3 Q2

Q3

#### RD Sharma Class 10 Solutions