Aldes
Then you could proceed as above
so non
piscctors, 72
ircle, ling
n the
in each of the followin
11. Draw a circle of ra
ar to
of a
this
EXERCISE 112
following, give also the justification of the construction
circle of radius 6 cm. From a point 10 cm away from its centre. construct the
ents to the circle and measure their lengths.
et a tangent to a circle of radius 4 cm from a point on the concentric cire
6 cm and measure its length. Also verify the measurement by actual calcul
a circle of radius 3 cm. Take two points P and ( on one of its extended dia
at a distance of 7 cm from its centre. Draw tangents to the circle from the
of tangents to the circle
2. Construct a tange
radius 6 cm and
3. Draw a circle of ra
NO
points P and Q.
aw a pair of tangents to a circle of radius 5 cm which are inclined to each othe
4. Draw a pair of to
angle of 60°
aw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radi
d taking B as centre, draw another circle of radius 3 cm. Construct tangents
circle from the centre of the other circle.
and
Answers
Answer:
Step-by-step explanation:
In each of the following, give also the justification of the construction: 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.Sol. Steps of construction: I. With O as centre and radius 6 cm, draw a circle. II. Take a point P at 10 cm away from the centre. III. Join O and P. IV. Bisect OP at M. V. Taking M as centre and MP or MO as radius, draw a circle. VI. Let the new circle intersects the given circle at A and B. VII.Join PA and PB. Thus, PA and PB are the required two tangents. By measurement, we have: PA = PB = 9.6 cm. Justification: Join OA and OB Since PO is a diameter. ∴ ∠OAF = 90° = ∠OBP [Angles in a semicircle] Also, OA and OR are radii of the same circle. ⇒ PA and PB are tangents to the circle.
Q.2. Construct a tangent to a circle of radius 4 cm from a point on the concentric and measure its length. Also verify the measurement by actual calculation.Sol. Steps of construction: I. Join PO and bisect it such that the mid point of PO is represented by M. II. Taking M as centre and OM or MP as radius, draw a circle such that this circle intersects the circle (of radius 4 cm) at A and B. III. Join A and P. Thus, PA is the required tangent. By measurement, we have: PA = 4.5 cm Justification: Join OA such that ∠PAO = 90° [Angle in a semi-circle] ⇒ PA ⊥ OA ∵ OA is a radius of the inner circle. ∴ PA has to be a tangent to the inner circle.
Q.3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance of 7 an from its centre. Draw tangents to the circle from these two points P and Q.Sol. Steps of construction: I. Join P and O. II. Bisect PO such that M be its mid-point. III. Taking M as centre and MO as radius, draw a circle. Let it at A and B. IV. Join PA and PB. Thus, PA and PB are the two required tangents from P. V. Now, join O and Q. VI. Bisect OQ such that N is its mid point. VII.Taking N as centre and NO as radius, draw a circle. Let it at C and D. VIII. Join QC and QD. Thus, QC and QD are the required tangents to the given Justification: Join OA such that ∠OAP = 90° [Angle in a semi-circle] ⇒ PA ⊥ OA PA is a tangent. Similarly, PB ⊥ OA ⇒ PB is a tangent Now, join OC such that ∠QCO = 90° [Angle in a semi-circle] ⇒ QC ⊥ OC ⇒ QC is a tangent. Similarly, QD ⊥ OC ⇒ QD is a tangent.
Q.4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.Sol. Steps of construction: I. With centre O and radius = 5 cm, draw a circle. II. Draw an angle ∠AOB = 120°. III. Draw a perpendicular on OA at A. IV. Draw another perpendicular on OB at B. V. Let the two perpendiculars. meet at C. CA and CB are the two required tangents to the given circle which are inclined to each other at 60°. Justification: In a quadrilateral OACB, using angle sum property, we have: 120° + 90° + 90° + ∠ACE = 360° ⇒ 300 + ∠ACE = 360° ⇒ ∠ACB = 360° – 300° = 60°.