In the adjoining figure, X and Y are points on the sides PQ and
PR respectively of triangle PQR such that PX = 2.5 cm, XQ = 5 cm,
PY = 2 cm, YR = 4 cm and XY = 3.5 cm. The measure of QR
is?
Answers
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Page No 55:
Question 1:
In the adjoining figure the radius of a circle with centre C is 6 cm, line AB is a tangent at A. Answer the following questions.
(1) What is the measure of ∠CAB ? Why ?
(2) What is the distance of point C from line AB? Why ?
(3) d(A,B) = 6 cm, find d(B,C).
(4) What is the measure of ∠ABC ? Why ?

ANSWER:
(1) It is given that line AB is tangent to the circle at A.
∴ ∠CAB = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)
Thus, the measure of ∠CAB is 90º.

(2) Distance of point C from AB = 6 cm (Radius of the circle)
(3) ∆ABC is a right triangle.
CA = 6 cm and AB = 6 cm
Using Pythagoras theorem, we have
BC2=AB2+CA2⇒BC=62+62−−−−−−√ ⇒BC=62–√ cmBC2=AB2+CA2⇒BC=62+62 ⇒BC=62 cm
Thus, d(B, C) = 62–√62 cm
(4) In right ∆ABC, AB = CA = 6 cm
∴ ∠ACB = ∠ABC (Equal sides have equal angles opposite to them)
Also, ∠ACB + ∠ABC = 90º (Using angle sum property of triangle)
∴ 2∠ABC = 90º
⇒ ∠ABC = 90°290°2 = 45º
Thus, the measure of ∠ABC is 45º.
Page No 55:
Question 2:
In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, then
(1) What is the length of each tangent segment ?
(2) What is the measure of ∠MRO ?
(3) What is the measure of ∠ MRN ?

ANSWER:
(1) It is given that seg RM and seg RN are tangent segments touching the circle at M and N, respectively.
∴ ∠OMR = ∠ONR = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)

OM = 5 cm and OR = 10 cm
In right ∆OMR,
OR2=OM2+MR2⇒MR=OR2−OM2−−−−−−−−−−√ ⇒MR=102−52−−−−−−−√⇒MR=100−25−−−−−−−√=75−−√=53–√ cmOR2=OM2+MR2⇒MR=OR2-OM2 ⇒MR=102-52⇒MR=100-25=75=53 cm
Tangent segments drawn from an external point to a circle are congruent.
∴ MR = NR = 53–√53 cm
(2) In right ∆OMR,
tan∠MRO=OMMR⇒tan∠MRO=5 cm