**Exercise 11.1**

**1) Draw a line segment of length 7.6 cm and divide it in 5: 8 ration. In addition, find the measure of two parts.**

**Solution:**

Procedure for construction:

- Draw any ray AX, making an acute angle with AB.
- Locate 13(= 5 + 8) points A
_{1}, A_{2}, A_{3}……… A13 on AX so that AA_{1}= A_{1}A_{2}A_{l2}A_{13}. - Join BA
_{13}. - Through the point A5(m = 5), draw a line parallel to BA13 (by making an angle equal to L AA13 B at A5 intersecting AB at C. Then AC: CB = 5 : 8)

**2) Construct a triangle with sides 4 cm, 5 cm and 6 cm and then a similar triangle to it whose sides are 2/3 of the corresponding sides of the first one.**

**Solution:**

Procedure for construction:

- Draw a line segment BC with length 5 cm.
- With B as centre and radius of 4 cm draw an arc.
- With C as centre and radius of 6 cm draw an arc.
- Join AB and AC. Then, ∆ABC is the required triangle.
- Below BC, make an acute angle ∠CBX
- Along BX, mark up three points B1, B2, B3 such that BB1 = B
_{1}B_{2}= B_{2}B_{3} - Join B
_{3}C - From B2, draw B
_{2}C’llB_{3}c, meeting BC at C’ - From C’ draw C’ All CA, meeting BA at A’
- Then ∆A’BC’ is the required triangle, each of whose sides is two-third of the corresponding sides of ∆ABC.

**3) Construct a triangle with side lengths 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.**

**Solution:**

Procedure for construction :

- Draw a line segment BC with length 6 cm.
- With B as centre and keeping radius as 5 cm, draw an arc.
- With C as centre and keeping radius as 7 cm, draw another arc, intersecting the previously drawn arc at Point A.
- Join AB and AC. Then, ∆ABC is the required triangle.
- Below BC, make an acute angle∠CBX.
- Along BX, mark up seven points B
_{1}, B_{2}, B_{3}….. B_{7}such that BB_{1}= B_{1},B_{2}, B_{6}B_{7}. - Join B
_{5}to C (5 being smaller of 5 and 7 in7/5) and draw a line through B_{7}parallel to B_{5}C, intersecting the extended line segment BC at C’. - Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’. Then A’BC’ is the required triangle.

**4) Draw a triangle ABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of ∆ABC.**

**Solution:**

**Procedure for construction:**

(i) Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

(ii) Draw any ray BX making an acute angle with BC on the side opposite to the vertex X.

(iii) Locate 4(the greater of 3 and 4 in ¾) points B1, B2, B3, B4 on BX so that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

(iv) Join B_{4}C and draw a line through B3(the 3rd point, 3 being smaller of 3 and 4 in ¾) parallel to B_{4}C to intersect BC at C’.

(v) Draw a line through C’ parallel to the line CA to intersect BA at A’. Then ∆A’BC’ is the required triangle.

Justification of construction

∆ABC ~ ∆A’BC’ , Therefore,

\(\frac{AB}{A’B} = \frac{AC}{A’C’} = \frac{BC}{BC’}\)But, \(\frac{BC}{BC’} = \frac{BB_{3}}{BB_{4}} = \frac{3}{4}\)

So, \(\frac{AB}{A’B} = \frac{AC}{A’C’} = \frac{BC}{BC’} = \frac{3}{4}\)

**5) Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are4/3 times the corresponding sides of ∆ABC.**

**Solution:**

**Procedure for construction : **

(i) Draw a triangle ABC with BC = 7cm, ∠B = 45° and ∠A = 105°.

(ii) Draw any ray BX making an acute angle with BC on the side opposite to the vertex X.

(iii) Locate 4(the greater of 3 and 4 in 4/3) points B_{1}, B_{2}, B_{3}, B_{4} on BX so that 3 BB_{1} = B_{1} B_{2} = B_{2}B_{3} = B_{3}B_{4}.

(iv) Join NC’ and draw a line through B3(the 3rd point, 3 being smaller of 3 and 4 in 1) parallel to NC’ to intersect BC’ at C. 3

(v) Draw a line through C’ parallel to the line CA to intersect BA at A’. Then A NBC’ is the required triangle.

**6) Construct a triangle of isosceles type, whose base is 8 cm and height 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.**

**Solution:**

**Given**:An isosceles triangle whose base is 8 cm and height 4 cm. Scale factor: 1 =

**Required:** To construct a similar triangle to above whose sides are 1.5 times the above triangle.

**Procedure for construction:**

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

(ii) Draw a perpendicular bisector AD of BC.

(iii) Join AB and AC we get a isosceles ∆ABC.

(iv) Construct an acute angle∠CBX downwards.

(v) On BX make 3 equal parts.

(vi) Join C to B_{2} and draw a line through B_{3} parallel to B_{2}C intersecting the extended line segment BC at C’.

(vii) Again draw a parallel line C’A’ to AC cutting BP at A’.

(viii) ∆A’BC’ is the required triangle.

**Exercise 11.2**

**7) Draw a circle with 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:**

**Procedure for construction:**

- Draw a line segment of length AB = 10 cm. Bisect AB by constructing a perpendicular bisector of AB. Let M be the mid-point of AB.
- With M as centre and AM as radius, draw a circle. Let it intersect the given circle at the points P and Q.
- Join PB and QB. Thus, PB and QB are the required two tangents.

**Justification:** Join AP. Here ∠APB is an angle in the semi-circle. Therefore, ∠APB = 90°. Since AP is a radius of a circle, PB has to be a tangent to a circle. Similarly, QB is also a tangent to a circle.

In a Right ∆APB, AB^{2} = AP^{2} + PB^{2} (By using Pythagoras Theorem)

PB^{2} = AB^{2} – AP^{2} = 10^{2} — 6^{2} = 100 – 36 = 64

PB = 8 cm.

**8) Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.**

**Solution:**

**Procedure for construction:**

- Draw a line segment of length OA = 4 cm. With O as centre and OA as radius, draw a circle.
- With O as centre draw a concentric circle of radius 6 cm(0B).
- Let C be any point on the circle of radius 6 cm, join OC.
- Bisect OC such that M is the mid point of OC.
- With M as centre and OM as radius, draw a circle. Let it intersect the given circle of radius 4 cm at the points P and Q.
- Join CP and CQ. Thus, CP and CQ are the required two tangents.

**Justification:**

Join OP. Here ∠OPC is an angle in the semi-circle. Therefore, ∠OPC = 90°. Since OP is a radius of a circle, CP has to be a tangent to a circle. Similarly, CQ is also a tangent to a circle.

In ∆COP, ∠P = 90°

\(CO^{2} = CP^{2} + OP^{2}\) \(CP^{2} = CO^{2} – OP^{2}\)=\(6^{2} – 4^{2}\)

\(CP = 2\sqrt{5}cm\)

**9) Draw a circle with radius 3 cm. On one of its extended diameter, take two points P and Q each at a distance of 7 cm from its centre. From two points P and Q, draw tangents to the circle.**

**Solution:**

**Given:**

Two points P and Q on the diameter of a circle with radius 3 cm OP = OQ = 7 cm.

**Aim**:

To construct the tangents to the circle from the given points P and Q.

**Procedure for construction:**

- Draw a circle with radius 3 cm with centreO.
- Extend its diameter both the sides and cut OP = OQ = 7 cm.
- Bisect OP and OQ.Let mid-points of OP and OQ be M and N.
- With M as centre and OM as radius, draw a circle. Let it intersect (0, 3) at two points A and B. Again taking N as centre ON as radius draw a circle to intersect circle(0, 3) at points C and D.
- Join PA, PB, QC and QD. These are the required tangents from P and Q to circle (0, 3).

**10) Draw a pair of tangents to a circle which is of radius 5 cm, such that they are inclined to each other at an angle of 60°.**

**Solution:**

**To determine**: To draw tangents at the ends of two radius which are inclined to each other at 120°

**Procedure for construction : **

- Keeping O as centre, draw a circle of radius 5 cm.
- Take a point Q on the circle and join it to O.
- From OQ, Draw∠QOR = 120°.
- Take an external point P.
- Join PR and PQ perpendicular to OR and OQ respectively intersecting at P.

The required tangents are RP and QP.