# Question-1

**In each of the following, give the justification of the construction also:**

Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

**Solution:**

**Steps of construction**

1. Draw any ray AX, making an acute angle with AB.

2. Locate 13(= 5 + 8) points A

_{1}, A

_{2}, A

_{3}……… A

_{13}on AX so that AA

_{1}= A

_{1}A

_{2}…….A

_{12 }A

_{13}.

3. Join BA

_{13}

4. Through the point A

_{5}(m = 5), draw a line parallel to BA

_{13 }(by making an angle equal to ∠ AA

_{13}B at A

_{5}intersecting AB at C. Then AC : CB = 5 : 8)

Let us see how this method gives us the required division.

Since A

_{5}C is parallel to A

_{13}B therefore = . (By the Basic proportionality theorem)

By construction, = . Therefore =

This shows C divides AB in the ratio 5 : 8.

# Question-2

**Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.****Solution:**

**Steps of construction**

1. Draw a line segment BC = 5 cm

2. With B as centre and radius equal to 4 cm draw an arc

3. With C as centre and radius equal to 6 cm draw an arc

4. Join AB and AC. Then, Δ ABC is the required triangle.

5. Below BC, make an acute angle ∠ CBX

6. Along BX, mark off three points B_{1}, B_{2}, B_{3} such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}

7. Join B_{3}C

8. From B_{2}, draw B_{2}C’||B_{3}C, meeting BC at C’

9. From C’ draw C’ A|| CA, meeting BA at A’

10. Then Δ A’BC’ is the required triangle, each of whose sides is two-third of the corresponding sides of Δ ABC.

Justification

Since A’C’||AC, so Δ ABC ∼ Δ A’BC’

# Question-3

**Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.****Solution:**

**Steps of construction**

1. Draw a line segment BC = 6 cm

2. With B as centre and with radius 5 cm, draw an arc.

3. With C as centre and with radius 7 cm, draw another arc, intersecting the previously drawn arc at A.

4. Join AB and AC. Then, Δ ABC is the required triangle.

5. Below BC, make an acute angle ∠ CBX.

6. Along BX, mark off seven points B

_{1}, B

_{2}, B

_{3}…..B

_{7}such that BB

_{1}= B

_{1}B

_{2}…… B

_{6}B

_{7}

7. Join B

_{5}to C (5 being smaller of 5 and 7 in and draw a line through B

_{7}parallel to B

_{5}C, intersecting the extended line segment BC at C’.

8. Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’. Then A’BC’ is the required triangle.

For justification of construction

Δ ABC ∼ Δ A’BC’

Therefore = =

But = =

So = and thus = = =

# Question-4

**Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ ABC = 60°. Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.****Solution:**

**Steps of 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 points B

_{1}, B

_{2}, B

_{3}, B

_{4}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 B

_{3}(the 3

^{rd}point, 3 being smaller of 3 and 4 in 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 Δ A’BC’ is the required triangle.

Justification of the construction

Δ ABC ∼ Δ A’BC’. Therefore = =

But = =

So = = =

# Question-5

**Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are****times the corresponding sides of Δ ABC.****Solution:**

**Steps of 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 points B

_{1}, B

_{2}, B

_{3}, B

_{4}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 B

_{3}(the 3

^{rd}point, 3 being smaller of 3 and 4 in 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 Δ A’BC’ is the required triangle.

Justification of the construction

Δ ABC ∼ Δ A’BC’. Therefore = =

But = =

So = = =

# Question-6

**Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1 times the corresponding sides of the isosceles triangle.****Solution:**

Given

Given

An isosceles triangle whose base is 8 cm and altitude 4 cm. Scale factor:

**1 =**

**Required**

To construct a similar triangle to above whose sides are 1 times the above triangle.

**Steps of 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.

# Question-7

**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.**

**Solution:**

**Steps of construction**

1. 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.

2. With M as centre and AM as radius, draw a circle. Let it intersect the given circle at the points P and Q.

3. Join PB and QB. Thus PB and QB are the required two tangents.

__Justification of construction__:

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 Rt Δ 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.

# Question-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:**

**Steps of construction**

1. Draw a line segment of length OA = 4 cm. With O as centre and OA as radius, draw a circle.

2. With O as centre draw a concentric circle of radius 6 cm(OB).

3. Let C be any point on the circle of radius 6 cm, join OC.

4. Bisect OC such that M is the mid point of OC.

5. 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.

6. Join CP and CQ. Thus CP and CQ are the required two tangents.

Justification of construction:

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 Δ COP, Ð P = 90°

⇒ CO^{2} = CP^{2 }+ OP^{2}

∴ CP^{2 }= CO^{2 }- OP^{2 }= 6^{2} - 4^{2}

∴ CP = 2√5 cm

# Question-9

**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.****Solution:**

**Given**

Two points P and Q on the diameter of a circle with radius 3 cm

OP = OQ = 7 cm.

**Required**

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

**Steps of construction**

1. Draw a circle of radius 3 cm with centre O.

2. Extend its diameter both sides and cut OP = OQ = 7 cm.

3. Bisect OP and OQ. Let M and N be the mid- points of OP and OQ respectively.

4. 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 two points C and D.

5. Join PA, PB, QC and QD. These are the required tangents from P and Q to circle (0, 3).

# Question-10

**Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60**°**.****Solution:**

We have to draw tangents at the ends of two radius which are inclined to each other at 120°

**Steps of Construction**

(i) Draw a circle of radius 5 cm with centre O.

(ii) Take a point Q on the circle and join it to O.

(iii) From OQ, Draw ∠ QOR = 120°.

(iv) Take an external point P.

(v) Join PR and PQ perpendicular to OR and OQ respectively intersecting at P.

Therefore the required tangents are RP and QP.

# Question-11

**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.****Solution:**

Steps of construction

Steps of construction

1. Draw a line segment AB of length 8 cm. Also bisects AB such that, M is the mid point of AB.

2. With M as centre draw a circle such that it touches A and B.

3. With A as centre draw a circle of radius 4 cm, and with B as centre draw a circle of radius 3 cm.

4. These two circles touches the bigger circle at P, Q, R and S.

5. Now join RB and SB they are the tangents from the point B.

6. Similarly join PA and QA they are the tangents from the point A.

# Question-12

**Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ Β = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.**

**Solution:**

**Steps of construction**

1. Draw a triangle with AB = 6 cm , BC = 8 cm and ∠ B = 90°.

2. Construct BD perpendicular to AC.

3. Draw a circle through B, C, D

4. Let M be the mid-point of BC obtained by bisecting BC.

5. Join AM and bisect it. Let N be the mid point of AM.

6. With N as centre and AN as radius draw a circle such that it touches the circle through B, C, D at the points E and F.

7. Join AE and AF.

8. Thus AE and AF are the required tangents.

# Question-13

**Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.**

**Solution:**

**Given**

Bangle, Point P outside the circle.

**Required**

To construct the pair of tangents from P to the circle.

**Steps of construction**

1. Draw a circle with the help of a bangle.

2. Draw two chords AB and AC. Perpendicular bisectors of AB and AC intersect each other at O, which is the centre of the circle.

3. Taking a point P, outside the circle, join OP.

4. Let M be the mid point of OP. Taking M as centre and OM as radius draw a circle which intersect the given circle at Q and R.

5. Join PQ and PR. Thus PQ and PR are the required tangents.