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# RD Chapter 11- Constructions Ex-11.3 Interview Questions Answers

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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.

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.

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]

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.

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]

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.

Steps of Construction :
Draw a line segment BC = 8 cm
From B draw an angle of 90°
Draw an arc 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]

Given, two concentric circles of radii 3 cm and 5 cm with centreO. We have to draw pair of tangents from point P on outer circle to the other. Steps of construction :
(i) Draw two concentric circleswith centre O and radii 3 cm and 5 cm.
(ii) Taking any point P onouter circle. Join OP.
(iii) Bisect OP, let M’ be themid-point of OP.
Taking M’ as centre and OM’ as radius draw a circle dotted which cuts the innercircle as M and P’.
(iv) Join PM and PP’. Thus, PMand PP’ are the required tangents.
(v) On measuring PM and PP’, wefind that PM = PP’ = 4 cm.
Actual calculation:
In right angle ΔOMP,
PMO =90°
PM2 =OP2 – OM2
[by Pythagoras theorem i.e. (hypotenuse)2 = (base)2 +(perpendicular)2]
PM2 =(5)2 – (3)2 = 25 – 9 = 16
PM = 4cm
Hence, the length of both tangents is 4 cm.

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