**Question 1.Select the correct alternative for each of the following questions.**

**i. The number of tangents that can be drawn to a circle at a point on the circle is ______(A) 3(B) 2(C) 1(D) 0Answer:**

(C)

**ii. The maximum number of tangents that can be drawn to a circle from a point outside it is ______(A) 2(B) 1(C) one and only one(D) 0Answer:**

(A)

(A) AABC is bigger.

(B) APQR is bigger.

(C) both triangles will be equal

(D) can not be decided

Answer:

(A)

**Question 2.Draw a circle with centre O and radius 3.5 cm. Take point P at a distance 5.7 cm from the centre. Draw tangents to the circle from point P.Solution:**

Analysis:

As shown in the figure, let P be a point in the exterior of circle at a distance of 5.7 cm.

Let PQ and PR be the tangents to the circle at points Q and R respectively.

∴ seg OQ ⊥ tangent PQ …[Tangent is perpendicular to radius]

∴ ∠OQP = 90°

∴ point Q is on the circle having OP as diameter. …[Angle inscribed in a semicircle is a right angle]

Similarly, point R also lies on the circle having OP as diameter.

∴ Points Q and R lie on the circle with OP as diameter.

On drawing a circle with OP as diameter, the points where it intersects the circle with centre O, will be the positions of points Q and R respectively.

**Question 3.Draw any circle. Take any point A on it and construct tangent at A without using the centre of the circle.Solution:**

Analysis:

As shown in the figure, line l is a tangent to the circle at point A.

seg BA is a chord of the circle and ∠BCA is an inscribed angle.

By tangent secant angle theorem,

∠BCA = ∠BAR

By converse of tangent secant angle theorem,

If we draw ∠BAR such that ∠BAR = ∠BCA, then ray AR (i.e. line l) is a tangent at point A.

**Question 4.Draw a circle of diameter 6.4 cm. Take a point R at a distance equal to its diameter from the centre. Draw tangents from point R.Solution:**

Diameter of circle = 6.4 cm 6.4

Radius of circle = 6.4/2 = 3.2 cm

Analysis:

As shown in the figure, let R be a point in the exterior of circle at a distance of 6.4 cm.

Let RQ and RS be the tangents to the circle at points Q and S respectively.

∴ seg PQ ⊥tangent RQ …[Tangent is perpendicular to radius]

∴ ∠PQR = 90°

∴ point Q is on the circle having PR as diameter. …[Angle inscribed in a semicircle is a right angle]

Similarly,

Point S also lies on the circle having PR as diameter.

∴ Points Q and S lie on the circle with PR as diameter.

On drawing a circle with PR as diameter, the points where it intersects the circle with centre P, will be the positions of points Q and S respectively.

**Question 5.Draw a circle with centre P. Draw an arc AB of 100° measure. Draw tangents to the circle at point A and point B.Solution:**m(arc AB) = ∠APB = 100°

Analysis:

seg PA ⊥ line l

seg PB ⊥line m … [Tangent is perpendicular to radius]

The perpendicular to seg PA and seg PB at points A and B respectively will give the required tangents at A and B.

Steps of construction:

i. With centre P, draw a circle of any radius and take any point A on it.

ii. Draw ray PA.

iii. Draw ray PB such that ∠APB = 100°.

iv. Draw line l ⊥ray PA at point A.

v. Draw line m ⊥ ray PB at point B.

Lines l and m are tangents at points A and B to the circle.

**Question 6.Draw a circle of radius 3.4 cm and centre E. Take a point F on the circle. Take another point A such that E – F – A and FA = 4.1 cm. Draw tangents to the circle from point A.Solution:**

Analysis:

Draw a circle of radius 3.4 cm

As shown in the figure, let A be a point in the exterior of circle at a distance of (3.4 + 4.1) = 7.5 cm.

Let AP and AQ be the tangents to the circle at points P and Q respectively.

∴ seg EP ⊥ tangent PA … [Tangent is perpendicular to radius]

∴ ∠EPA = 90°

∴ point P is on the circle having EA as diameter. …[Angle inscribed in a semicircle is a right angle]

Similarly, point Q also lies on the circle having EA as diameter.

∴ Points P and Q lie on the circle with EA as diameter.

On drawing a circle with EA as diameter, the points where it intersects the circle with centre E, will be the positions of points P and Q respectively.

**Question 7.Solution:**Analysis:

As shown in the figure,

Let B – C – N and B – A – L.

∆ABC ~ ∆LBN …[Given]

∴ ∠ABC ≅ ∠LBN …[Corresponding angles of similar triangles]

∴ sides of ∆LBN are longer than corresponding sides of ∆ABC.

∴ If seg BC is divided into 4 equal parts, then seg BN will be 7 times each part of seg BC.

So, if we construct ∆ABC, point N will be on side BC, at a distance equal to 7 parts from B.

Now, point L is the point of intersection of ray BA and a line through N, parallel to AC.

∆LBN is the required triangle similar to ∆ABC.

Steps of construction:

i. Draw ∆ABC of given measure. Draw ray BD making an acute angle with side BC.

ii. Taking convenient distance on compass, mark 7 points B

_{1}, B

_{2}, B

_{3}, B

_{4}, B

_{5}, B

_{6}and B

_{7}such that

BB

_{1}= B

_{1}B

_{2}= B

_{2}B

_{3 }B

_{3}= B4

_{4}= B

_{4}B

_{5}= B

_{5}B

_{6}= B

_{6}B

_{7}.

iii. Join B

_{4}C. Draw line parallel to B

_{4}C through B

_{7}to intersects ray BC at N.

iv. Draw a line parallel to side AC through N. Name the point of intersection of this line and ray BA as L.

∆LBN is the required triangle similar to ∆ABC.

**Question 8.Solution:**

Analysis:

As shown in the figure,

Let Y – Q – Z and Y – P – X.

∆XYZ ~ ∆PYQ …[Given]

∴ ∠XYZ ≅ ∠PYQ …[Corresponding angles of similar triangles]

∴ sides of ∆XYZ are longer than corresponding sides of ∆PYQ.

∴ If seg YQ is divided into 5 equal parts, then seg YZ will be 6 times each part of seg YQ.

So, if we construct ∆PYQ, point Z will be on side YQ, at a distance equal to 6 parts from Y.

Now, point X is the point of intersection of ray YP and a line through Z, parallel to PQ.

∆XYZ is the required triangle similar to ∆PYQ.

Steps of construction:

i. Draw ∆ PYQ of given measure. Draw ray YT making an acute angle with side YQ.

ii. Taking convenient distance on compass, mark 6 points Y

_{1}, Y

_{2}, Y

_{3}, Y

_{4}, Y

_{5}and Y

_{6}such that

YY

_{1}= Y

_{1}Y

_{2}= Y

_{2}Y

_{3}= Y

_{3}Y

_{4}= Y

_{4}Y

_{5}= Y

_{5}Y

_{6}.

iii. Join Y

_{5}Q. Draw line parallel to Y

_{5}Q through Y

_{6}to intersects ray YQ at Z.

iv. Draw a line parallel to side PQ through Z. Name the point of intersection of this line and ray YP as X.

∆XYZ is the required triangle similar to ∆PYQ.