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.

**Soln. :** Steps of Construction :

I. Draw a line segment *AB* = 7.6 cm.

II. Draw a ray *AX* making an acute angle with *AB*.

III. Mark 13 (8 + 5) equal points on *AX*, and mark them as *X*_{1}, *X*_{2}, *X*_{3}, ........, *X*_{13}.

IV. Join points *X*_{13} and *B*.

V. From point *X*_{5}, draw *X*_{5}*C* || *X*_{13}*B*, which meets *AB* at *C*.

Thus, *C* divides *AB* in the ratio 5 : 8

On measuring the two parts, we get
* AC* = 2.9 cm and *CB* = 4.7 cm.

Justification:

In ∆*ABX*_{13} and ∆*ACX*_{5}, we have
*CX*_{5} || *BX*_{13}

[By Thales theorem]

⇒ *AC* : *CB* = 5 : 8.

In each of the following, give the justification of the construction also.

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.

**Soln. :** Steps of Construction :

I. Draw a ∆*ABC* such that *BC* = 6cm, *AC* = 5 cm and *AB* = 4 cm.

II. Draw a ray *BX* making an acute angle ∠*CBX*.

III. Mark three points *X*_{1}, *X*_{2}, *X*_{3} on

*BX* such that *BX*_{1} = *X*_{1}*X*_{2} = *X*_{2}*X*_{3}.

IV. Join *X*_{3}*C*.

V. Draw a line through *X*_{2} such that it is parallel to *X*_{3}*C* and meets *BC* at C'.

VI. Draw a line through *C*' parallel to *CA* to intersect *BA* at *A*'.

Thus, *A*'*BC*' is the required similar triangle.

Justification:

By construction, we have *X*_{3}*C* || *X*_{2}*C*'

[Using Thales theorem]

Adding 1 to both sides, we get

Now, in ∆*BC*′*A*′ and ∆*BCA*, we have *CA* ||*C*′*A*′

∴ Using *AA* similarity, we have

∆*BC*′*A*′ ~ ∆*BCA*

In each of the following, give the justification of the construction also.

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.

**Soln. :** Steps of Construction :

I. Construct a ∆*ABC* such that *AB* = 5 cm,

*BC* = 7 cm and *AC* = 6 cm.

II. Draw a ray *BX* such that ∠*CBX* is an acute angle.

III. Mark 7 points of *X*_{1}, *X*_{2}, *X*_{3}, *X*_{4}, *X*_{5}, *X*_{6} and *X*_{7} on *BX* such that *BX*_{1} = *X*_{1}*X*_{2} =*X*_{2}*X*_{3} = X_{3}*X*_{4} = *X*_{4}*X*_{5} = *X*_{5}*X*_{6} = *X*_{6}*X*_{7}

IV. Join *X*_{5} to *C*.

V. Draw a line through *X*_{7} intersecting *BC* (produced) at *C*′ such that *X*_{5}*C* || *X*_{7}*C*′

VI. Draw a line through *C*′ parallel to *CA* to intersect *BA* (produced) at *A*′.

Thus, ∆*A*'*BC*' is the required triangle.

Justification:

By construction, we have *C*′*A*′ || *CA*

∴ Using *AA* similarity, ∆*ABC* ~ ∆*A*'*BC*'

Also *X*_{7}*C*′ || *X*_{5}*C* [By construction]

∴ ∆*BX*_{7}*C*' ~ ∆*BX*_{5}*C*

In each of the following, give the justification of the construction also.

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

**Soln. :** Steps of Construction :

I. Draw *BC* = 8 cm

II. Draw the perpendicular bisector of *BC* which intersects *BC* at *D*.

III. Mark a point *A* on the above perpendicular such that *DA* = 4 cm.

IV. Join *AB* and *AC*.

Thus, ∆*ABC* is the required isosceles triangle.

V. Now, draw a ray *BX* such that ∠*CBX* is an acute angle.

VI. On *BX*, mark three points *X*_{1}, *X*_{2} and *X*_{3} such that:

*BX*_{1} = *X*_{1}*X*_{2} = *X*_{2}*X*_{3}

VII. Join *X*_{2}*C*.

VIII. Draw a line through *X*_{3} parallel to *X*_{2}*C* and intersecting *BC* (extended) to C′.

IX. Draw a line through *C*′ parallel to *CA* intersecting *BA* (extended) to *A*′, thus,

∆*A*'*BC*' is the required triangle.

Justification:

We have *C*′*A*′ || *CA* [By construction]

∴ Using *AA* similarity, ∆*A*′*BC*′ ~ ∆*ABC*

...(i)

Since, In ∆*BX*_{3}*C*′, *X*_{3}*C*′ || *X*_{2}*C* [By construction]

[By *BPT*]

Thus, By (i)

In each of the following, give the justification of the construction also.

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

**Soln. :** Steps of construction:

I. Construct a ∆*ABC* such that *BC* = 6 cm, *AB* = 5 cm and ∠*ABC* = 60°.

II. Draw a ray *BX* such that ∠*CBX* is an acute angle.

III. Mark four points in a line through *X*_{3}, parallel to *X*_{4}*C* to intersect *BC* at *C*′

*BX*_{1} = *X*_{1}*X*_{2} = *X*_{2}*X*_{3} = *X*_{3}*X*_{4}

IV. Join *X*_{4}*C* and draw a line through *X*_{3} parallel to *X*_{4}*C* to intersect *BC* at *C*'.

V. Also draw another line through *C*' and parallel to *CA* to intersect *BA* at *A*'.

Thus, ∆*A*'*BC*' is the required triangle.

Justification:

In ∆*BX*_{4}*C* we have
*X*_{4}*C* || *X*_{3}*C*' [By construction]

[By BPT]

[By construction]

...(i)

Now, we also have
*CA* || *C*'*A*' [By construction]

∴ ∆*BC*'*A*' ~ ∆*BCA* [using *AA* similarity]

[From (i)]

In each of the following, give the justification of the construction also.

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 ∆*ABC*.

**Soln. :** Steps of Construction:

I. Construct a ∆*ABC* such that *BC* = 7 cm, ∠*B* = 45°, ∠*A* = 105° and ∠*C* = 30°

II. Draw a ray *BX* making an acute angle ∠*CBX* with *BC*.

III. On *BX*, mark four points *X*_{1}, *X*_{2}, *X*_{3} and *X*_{4} such that

*BX*_{1} = *X*_{1}*X*_{2} = *X*_{2}*X*_{3} = *X*_{3}*X*_{4}.

IV. Join *X*_{3}*C*.

V. Draw a line through *X*_{4} parallel to *X*_{3}*C* intersecting *BC*(extended) at *C*′

VI. Draw a line through *C*′ parallel to *CA* intersecting the extended line segment *BA* at *A*′.

Thus, ∆*A*′*BC*′ is the required triangle.

Justification:

By construction, we have
*C*′*A*′ || *CA*

∴ ∆*ABC* ~ ∆*A*′*BC*′ [*AA* similarity]

...(i)

Also, In ∆*BX*_{4}*C*′,
*X*_{4}*C*′ || *X*_{3}*C* [By construction]

∴ ∆*BX*_{4}*C*′ ~ ∆*BX*_{3}*C* [*AA* similarity]

...(ii)

From (i) and (ii), we have

In each of the following, give the justification of the construction also.

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are times the corresponding sides of the given triangle.

**Soln. :** Steps of Construction:

I. Construct the right triangle *ABC* such that ∠*B* = 90°, *BC* = 4 cm and *BA* = 3 cm.

II. Draw a ray *BX* such that an acute angle ∠*CBX* is formed.

III. Mark 5 points *X*_{1}, *X*_{2}, *X*_{3}, *X*_{4} and *X*_{5} on *BX* such that *BX*_{1} = *X*_{1}*X*_{2} = *X*_{2}*X*_{3} = *X*_{4}*X*_{5}.

IV. Join *X*_{3}*C*.

V. Draw a line through *X*_{5} parallel to *X*_{3}*C*, intersecting the extended line segment *BC* at *C*'.

VI. Draw another line through *C*′ parallel to *CA* intersecting the extended line segment *BA* at *A*'.

Thus, ∆*A*′*BC*′ is the required triangle.

Justification:

By construction, we have:
*C*′*A*′ || *CA*

∴ ∆*ABC* ~ ∆*A*′*BC*′ [*AA* similarity]

...(i)

Also, in ∆*BX*_{5}*C*′, *X*_{5}*C*′ || *X*_{3}*C* [By construction]

∴ ∆*BX*_{5}*C*′ ~ ∆*BX*_{3}*C* [*AA* similarity]

...(ii)

From (i) and (ii), we get

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.

**Soln. :** Steps of Construction:

I. Draw a circle of radius 6 cm. Let its centre be *O*.

II. Take a point *P* such that *OP* = 10 cm. Join *OP*.

III. Bisect *OP* and let *M* be its midpoint.

IV. Taking *M* as centre and *MP* or *MO* as radius draw a circle.

Let the new circle intersects the given circle at *A* and *B*. Join *PA* and *PB*.

Thus, *PA* and *PB* are the required tangents.

By measurement, we have : *PA* = *PB* = 8 cm.

Justification:

Join *OA* and *OB*

Since *PO* is a diameter.

∴ ∠*OAP* = 90° = ∠*OBP*

[Angles in a semicircle]

Also, *OA* and *OB* are radii of the same circle.

⇒ *PA* and *PB* are tangents to the circle.

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

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.

**Soln. :** Steps of Construction :

I. Draw two concentric circles with centre *O* and radii 4 cm and 6 cm.

II. Take any point *P* on outer circle.

III. Join *PO* and bisect it and let the midpoint of *PO* is represented by *M*.

IV. 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*.

V. Join *AP*. Thus, *PA* is the required tangent.

By measurement, we have *PA* = 4.5 cm

Justification:

Join *OA*. As *PO* is diameter

∴ ∠*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. Verification of length of *PA*. In right ∆*PAO*, *PO* = 6 cm, *OA* = 4 cm.

Hence both lengths are approximately equal.

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

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 cm from its centre. Draw tangents to the circle from these two points *P* and *Q*.

**Soln. :** Steps of Construction :

I. Draw a circle of radius 3 cm with centre *O* and draw a diameter.

II. Extend its diameter on both sides and cut *OP* = *OQ* = 7 cm

III. Bisect *PO* such that *M* be its mid-point.

IV. Taking *M* as centre and *MO* as radius, draw a circle. Let it intersect the given circle at *A* and *B*.

V. Join *PA* and *PB*.

Thus, *PA* and *PB* are the two required tangents from *P*.

VI. Now bisect *OQ* such that *N* is its mid point.

VII. Taking *N* as centre and *NO* as radius, draw a circle. Let it intersect the given circle at *C* and *D*.

VIII.Join *QC *and *QD*.

Thus, *QC* and *QD* are the required tangents from *Q*.

Justification :

Join *OA* to get ∠*OAP* = 90°

[Angle in a semi-circle]

⇒ *PA* ┴ *OA* ⇒ *PA* is a tangent.

Similarly, *PB* ┴ *OA* ⇒ *PB* is a tangent.

Now, join *OC* to get ∠*QCO* = 90° [Angle in a semi-circle]

⇒ *QC* ┴ *OC* ⇒ *QC* is a tangent.

Similarly, *QD* ┴ *OC* ⇒ *QD* is a tangent.

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

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

**Soln. :** Steps of Construction :

I. With centre *O* and radius = 5 cm, draw a circle.

II. Taking a point *A* on the circle draw ∠*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*.

Thus *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° + ∠*ACB* = 360°

⇒ 300° + ∠*ACB* = 360°

⇒ ∠*ACB* = 360° - 300° = 60°.

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

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.

**Soln. :** Steps of Construction:

I. Draw a line segment *AB* = 8 cm

II. Draw a circle with centre *A* and radius 4 cm, draw another circle with centre *B* and radius 3 cm.

III. Bisect the line segment *AB*. Let its mid point be *M*.

IV. With centre as *M* and *MA* (or *MB*) as radius, draw a circle such that it intersects the two circles at points *P*, *Q*, *R* and *S*.

V. Join *BP* and *BQ*.

Thus, *BP* and *BQ* are the required two tangents from *B* to the circle with centre *A*.

VI. Join *RA* and *SA*.

Thus, *RA* and *SA* are the required two tangents from *A* to the circle with centre *B*.

Justification:

Let us join *A* and *P*.

∵ ∠*APB* = 90° [Angle in a semi-circle]

∴ *BP* ┴ *AP*

But *AP* is radius of the circle with centre *A*.

⇒ *BP* has to be a tangent to the circle with centre *A*.

Similarly,

*BQ* has to be tangent to the circle with centre *A*.

Also *AR* and *AS* are tangents to the circle with centre *B*.

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

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

**Soln. :** Steps of construction:

I. Draw ∆*ABC* such that *AB* = 6 cm, *BC* = 8 cm and ∠*B* = 90°.

II. Draw *BD* ┴ *AC*. Now bisect *BC* and let its midpoint be *O*.

So *O* is centre of the circle passing through *B*, *C* and *D*.

III. Join *AO*

IV. Bisect *AO*. Let *M* be the mid-point of *AO*.

V. Taking *M* as centre and *MA* as radius, draw a circle intersecting the given circle at *B* and *E*.

VI. Join *AB* and *AE*. Thus, *AB* and *AE* are the required two tangents to the given circle from *A*.

Justification:

Join *OE*, then ∠*AEO* = 90°

[Angle in a semi circle]

∴ *AE* ┴ *OE*.

But *OE* is a radius of the given circle.

⇒ *AE* has to be a tangent to the circle.

Similarly, *AB* is also a tangent to the given circle.

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

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.

**Soln. :** Steps of Construction:

I. Draw the given circle using a bangle.

II. Take two non parallel chords *PQ* and *RS* of this circle.

III. Draw the perpendicular bisectors of *PQ* and *RS* such that they intersect at *O*. Therefore, *O* is the centre of the given circle.

IV. Take a point *P* outside this circle.

V. Join *OP* and bisect it. Let *M* be the mid-point of *OP*.

VI. Taking *M* as centre and *OM* as radius, draw a circle. Let it intersect the given circle at *A* and *B*.

VII. Join *PA* and *PB*. Thus, *PA* and *PB* are the required two tangents.

Justification :

Join *OA* and *OB*.

Since ∠*OAP* = 90° [Angle in a semi circle]

∴ *PA* ┴ *OA*.

Also *OA* is a radius.

∴ *PA* has to be a tangent to the given circle.

Similarly, *PB* is also a tangent to the given circle.