Chapter 11 Constructions (Class 10 - Maths Old NCERT Textbook Solutions)

The scale factor is $\frac{3}{4}$. Since the scale factor is less than 1, the triangle to be constructed will be smaller than the given $\triangle$ABC.

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Construction Required

To construct a triangle A'BC' similar to $\triangle$ABC such that each side of $\triangle$A'BC' is $\frac{3}{4}$ of the corresponding side of $\triangle$ABC.

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Steps of Construction

1. Draw any triangle ABC.

2. Draw a ray BX from B on the side opposite to vertex A, making an acute angle with side BC.

3. Locate 4 points (the greater of 3 and 4) $B_1, B_2, B_3, B_4$ on ray BX such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

4. Join $B_4$ (the point corresponding to the denominator) to C.

5. Draw a line through $B_3$ (the point corresponding to the numerator) parallel to $B_4C$, to intersect BC at a point C'.

6. Draw a line through C' parallel to the side AC to intersect AB at a point A'.

7. Then, $\triangle$A'BC' is the required triangle.

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Justification

By construction, $A'C' \parallel AC$.

In $\triangle ABC$ and $\triangle A'BC'$,

$\angle ABC = \angle A'BC'$ (Common angle)

$\angle BCA = \angle BC'A'$ (Corresponding angles, since $A'C' \parallel AC$)

Therefore, by AA similarity criterion, $\triangle A'BC' \sim \triangle ABC$.

The ratio of corresponding sides is equal:

$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$

Also, by construction, $B_3C' \parallel B_4C$.

In $\triangle BB_4C$, by the Basic Proportionality Theorem (Thales' Theorem), we have:

$\frac{BC'}{C'C} = \frac{BB_3}{B_3B_4} = \frac{3}{1}$

This gives $\frac{BC'}{BC} = \frac{BB_3}{BB_4} = \frac{3}{4}$.

Therefore, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{3}{4}$.

This justifies the construction.

The scale factor is $\frac{5}{3}$. Since the scale factor is greater than 1, the triangle to be constructed will be larger than the given $\triangle$ABC.

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Construction Required

To construct a triangle A'BC' similar to $\triangle$ABC such that each side of $\triangle$A'BC' is $\frac{5}{3}$ of the corresponding side of $\triangle$ABC.

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Steps of Construction

1. Draw any triangle ABC.

2. Draw a ray BX from B on the side opposite to vertex A, making an acute angle with side BC.

3. Locate 5 points (the greater of 5 and 3) $B_1, B_2, B_3, B_4, B_5$ on ray BX such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.

4. Join $B_3$ (the point corresponding to the denominator) to C.

5. Draw a line through $B_5$ (the point corresponding to the numerator) parallel to $B_3C$. Extend the line segment BC to intersect this parallel line at C'.

6. Draw a line through C' parallel to the side AC. Extend the line segment BA to intersect this parallel line at A'.

7. Then, $\triangle$A'BC' is the required triangle.

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Justification

By construction, $A'C' \parallel AC$.

In $\triangle ABC$ and $\triangle A'BC'$,

$\angle ABC = \angle A'BC'$ (Common angle)

$\angle BCA = \angle BC'A'$ (Corresponding angles, since $A'C' \parallel AC$)

Therefore, by AA similarity criterion, $\triangle A'BC' \sim \triangle ABC$.

The ratio of corresponding sides is equal:

$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$

Also, by construction, $B_3C \parallel B_5C'$.

In $\triangle BB_5C'$, by the Basic Proportionality Theorem, we have:

$\frac{BC}{BC'} = \frac{BB_3}{BB_5} = \frac{3}{5}$

Taking the reciprocal, we get $\frac{BC'}{BC} = \frac{5}{3}$.

Therefore, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{5}{3}$.

This justifies the construction.

Given

A line segment of length 7.6 cm to be divided in the ratio 5 : 8.

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Construction Required

To divide a line segment AB of length 7.6 cm into two parts AP and PB such that AP : PB = 5 : 8.

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Steps of Construction

1. Draw a line segment AB of length 7.6 cm.

2. Draw a ray AX making an acute angle with AB.

3. Locate $5+8=13$ points $A_1, A_2, A_3, \dots, A_{13}$ on the ray AX such that $AA_1 = A_1A_2 = \dots = A_{12}A_{13}$.

4. Join the last point $A_{13}$ to B.

5. From the 5th point, $A_5$, draw a line parallel to $A_{13}B$ (by making an angle equal to $\angle AA_{13}B$ at $A_5$) which intersects AB at a point P.

6. The point P divides the line segment AB in the ratio 5 : 8.

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Justification

By construction, $A_5P \parallel A_{13}B$.

In $\triangle ABA_{13}$, by the Basic Proportionality Theorem (Thales' Theorem), we have:

By construction, the points on ray AX are equally spaced. So, $AA_5 = 5$ units and $A_5A_{13} = 13 - 5 = 8$ units.

From equations (i) and (ii), we get:

This justifies the construction.

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Measurement

On measuring the two parts with a ruler, we find:

AP $\approx 2.9$ cm

PB $\approx 4.7$ cm

Verification by Calculation:

AP = $\frac{5}{5+8} \times 7.6 = \frac{5}{13} \times 7.6 = \frac{38}{13} \approx 2.92$ cm.

PB = $\frac{8}{5+8} \times 7.6 = \frac{8}{13} \times 7.6 = \frac{60.8}{13} \approx 4.68$ cm.

The measured values are approximately correct.

Given

A triangle with sides 4 cm, 5 cm, and 6 cm.

Scale factor = $\frac{2}{3}$.

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Construction Required

To construct a triangle ABC with sides 4 cm, 5 cm, 6 cm, and then construct a similar triangle A'BC' whose sides are $\frac{2}{3}$ of the corresponding sides of $\triangle$ABC.

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Steps of Construction

1. Draw a line segment BC of length 6 cm.

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

3. With C as centre and radius 5 cm, draw another arc to intersect the first arc at A.

4. Join AB and AC. $\triangle$ABC is the given triangle.

5. Draw a ray BX from B, making an acute angle with BC on the opposite side of vertex A.

6. Locate 3 points (the greater of 2 and 3) $B_1, B_2, B_3$ on ray BX such that $BB_1 = B_1B_2 = B_2B_3$.

7. Join $B_3$ (the point corresponding to the denominator) to C.

8. Draw a line through $B_2$ (the point corresponding to the numerator) parallel to $B_3C$, to intersect BC at C'.

9. Draw a line through C' parallel to AC to intersect AB at A'.

10. $\triangle$A'BC' is the required triangle.

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Justification

By construction, $A'C' \parallel AC$. Therefore, $\triangle A'BC' \sim \triangle ABC$ by AA similarity.

So, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$.

Also by construction, $B_2C' \parallel B_3C$. In $\triangle BB_3C$, by the Basic Proportionality Theorem:

$\frac{BC'}{BC} = \frac{BB_2}{BB_3} = \frac{2}{3}$

Hence, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{2}{3}$.

Given

A triangle with sides 5 cm, 6 cm, and 7 cm.

Scale factor = $\frac{7}{5}$.

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Construction Required

To construct a triangle ABC with sides 5 cm, 6 cm, 7 cm, and then construct a similar triangle A'BC' whose sides are $\frac{7}{5}$ of the corresponding sides of $\triangle$ABC.

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Steps of Construction

1. Draw a line segment BC of length 7 cm.

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

3. With C as centre and radius 6 cm, draw another arc to intersect the first arc at A.

4. Join AB and AC. $\triangle$ABC is the given triangle.

5. Draw a ray BX from B, making an acute angle with BC on the opposite side of vertex A.

6. Locate 7 points (the greater of 7 and 5) $B_1, B_2, \dots, B_7$ on ray BX such that $BB_1 = B_1B_2 = \dots = B_6B_7$.

7. Join $B_5$ (the point corresponding to the denominator) to C.

8. Draw a line through $B_7$ (the point corresponding to the numerator) parallel to $B_5C$. Extend BC to intersect this line at C'.

9. Draw a line through C' parallel to AC. Extend BA to intersect this line at A'.

10. $\triangle$A'BC' is the required triangle.

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Justification

By construction, $A'C' \parallel AC$. Therefore, $\triangle A'BC' \sim \triangle ABC$ by AA similarity.

So, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$.

Also by construction, $B_5C \parallel B_7C'$. In $\triangle BB_7C'$, by the Basic Proportionality Theorem:

$\frac{BC}{BC'} = \frac{BB_5}{BB_7} = \frac{5}{7}$

Taking the reciprocal, we get $\frac{BC'}{BC} = \frac{7}{5}$.

Hence, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{7}{5}$.

Given

An isosceles triangle with base 8 cm and altitude 4 cm.

Scale factor = $1\frac{1}{2} = \frac{3}{2}$.

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To Construct

1. The specified isosceles triangle.

2. A similar triangle whose sides are $\frac{3}{2}$ times the corresponding sides of the first triangle.

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Steps of Construction

1. Draw a line segment BC of length 8 cm.

2. Draw the perpendicular bisector of BC, let it intersect BC at D.

3. From D, mark a point A on the perpendicular bisector such that AD = 4 cm.

4. Join AB and AC. $\triangle$ABC is the required isosceles triangle.

5. Draw a ray BX from B, making an acute angle with BC on the opposite side of vertex A.

6. Locate 3 points (the greater of 3 and 2) $B_1, B_2, B_3$ on ray BX such that $BB_1 = B_1B_2 = B_2B_3$.

7. Join $B_2$ (the point corresponding to the denominator) to C.

8. Draw a line through $B_3$ (the point corresponding to the numerator) parallel to $B_2C$. Extend BC to intersect this line at C'.

9. Draw a line through C' parallel to AC. Extend BA to intersect this line at A'.

10. $\triangle$A'BC' is the required triangle.

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Justification

By construction, $A'C' \parallel AC$. Therefore, $\triangle A'BC' \sim \triangle ABC$ by AA similarity.

So, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$.

Also by construction, $B_2C \parallel B_3C'$. In $\triangle BB_3C'$, by the Basic Proportionality Theorem:

$\frac{BC}{BC'} = \frac{BB_2}{BB_3} = \frac{2}{3}$

Taking the reciprocal, we get $\frac{BC'}{BC} = \frac{3}{2}$.

Hence, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{3}{2}$.

Given

A triangle ABC with BC = 6 cm, AB = 5 cm, and $\angle ABC = 60^\circ$.

Scale factor = $\frac{3}{4}$.

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To Construct

1. The specified triangle ABC.

2. A similar triangle whose sides are $\frac{3}{4}$ of the corresponding sides of $\triangle$ABC.

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Steps of Construction

1. Draw a line segment BC of length 6 cm.

2. At point B, construct an angle of $60^\circ$. Let the ray be BX.

3. With B as the centre, draw an arc of radius 5 cm that intersects the ray BX at point A.

4. Join AC. $\triangle$ABC is the given triangle.

5. Draw a ray BY from B, making an acute angle with BC on the opposite side of vertex A.

6. Locate 4 points (the greater of 3 and 4) $B_1, B_2, B_3, B_4$ on ray BY such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

7. Join $B_4$ (the point corresponding to the denominator) to C.

8. Draw a line through $B_3$ (the point corresponding to the numerator) parallel to $B_4C$ to intersect BC at C'.

9. Draw a line through C' parallel to AC to intersect AB at A'.

10. $\triangle$A'BC' is the required triangle.

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Justification

By construction, $A'C' \parallel AC$. Therefore, $\triangle A'BC' \sim \triangle ABC$ by AA similarity.

So, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$.

Also by construction, $B_3C' \parallel B_4C$. In $\triangle BB_4C$, by the Basic Proportionality Theorem:

$\frac{BC'}{BC} = \frac{BB_3}{BB_4} = \frac{3}{4}$

Hence, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{3}{4}$.

Given

A triangle ABC with BC = 7 cm, $\angle B = 45^\circ$, and $\angle A = 105^\circ$.

Scale factor = $\frac{4}{3}$.

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To Construct

1. The specified triangle ABC.

2. A similar triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $\triangle$ABC.

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Steps of Construction

1. First, find $\angle C$. In $\triangle$ABC, $\angle A + \angle B + \angle C = 180^\circ$.

$105^\circ + 45^\circ + \angle C = 180^\circ \implies 150^\circ + \angle C = 180^\circ \implies \angle C = 30^\circ$.

2. Draw a line segment BC of length 7 cm.

3. At point B, construct an angle of $45^\circ$. Let the ray be BX.

4. At point C, construct an angle of $30^\circ$. Let the ray be CY.

5. The rays BX and CY will intersect at point A. $\triangle$ABC is the given triangle.

6. Draw a ray BZ from B, making an acute angle with BC on the opposite side of vertex A.

7. Locate 4 points (the greater of 4 and 3) $B_1, B_2, B_3, B_4$ on ray BZ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

8. Join $B_3$ (the point corresponding to the denominator) to C.

9. Draw a line through $B_4$ (the point corresponding to the numerator) parallel to $B_3C$. Extend BC to intersect this line at C'.

10. Draw a line through C' parallel to AC. Extend BA to intersect this line at A'.

11. $\triangle$A'BC' is the required triangle.

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Justification

By construction, $A'C' \parallel AC$. Therefore, $\triangle A'BC' \sim \triangle ABC$ by AA similarity.

So, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$.

Also by construction, $B_3C \parallel B_4C'$. In $\triangle BB_4C'$, by the Basic Proportionality Theorem:

$\frac{BC}{BC'} = \frac{BB_3}{BB_4} = \frac{3}{4}$

Taking the reciprocal, we get $\frac{BC'}{BC} = \frac{4}{3}$.

Hence, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{4}{3}$.

Given

A right triangle with perpendicular sides of 4 cm and 3 cm.

Scale factor = $\frac{5}{3}$.

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To Construct

1. The specified right triangle.

2. A similar triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the first triangle.

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Steps of Construction

1. Draw a line segment BC of length 4 cm.

2. At point B, construct a right angle ($90^\circ$). Let the ray be BX.

3. With B as the centre, draw an arc of radius 3 cm that intersects the ray BX at point A.

4. Join AC. $\triangle$ABC is the given right triangle.

5. Draw a ray BY from B, making an acute angle with BC on the opposite side of vertex A.

6. Locate 5 points (the greater of 5 and 3) $B_1, B_2, B_3, B_4, B_5$ on ray BY such that $BB_1 = B_1B_2 = \dots = B_4B_5$.

7. Join $B_3$ (the point corresponding to the denominator) to C.

8. Draw a line through $B_5$ (the point corresponding to the numerator) parallel to $B_3C$. Extend BC to intersect this line at C'.

9. Draw a line through C' parallel to AC. Extend BA to intersect this line at A'.

10. $\triangle$A'BC' is the required triangle.

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Justification

By construction, $A'C' \parallel AC$. Therefore, $\triangle A'BC' \sim \triangle ABC$ by AA similarity.

So, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$.

Also by construction, $B_3C \parallel B_5C'$. In $\triangle BB_5C'$, by the Basic Proportionality Theorem:

$\frac{BC}{BC'} = \frac{BB_3}{BB_5} = \frac{3}{5}$

Taking the reciprocal, we get $\frac{BC'}{BC} = \frac{5}{3}$.

Hence, $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{5}{3}$.

Given

A circle with radius 6 cm.

An external point P at a distance of 10 cm from the centre.

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To Construct

A pair of tangents from the external point to the circle and measure their lengths.

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Steps of Construction

1. Draw a circle with centre O and radius 6 cm.

2. Mark a point P such that OP = 10 cm.

3. Draw the perpendicular bisector of the line segment OP. Let M be the midpoint of OP.

4. With M as the centre and radius MO (or MP), draw another circle.

5. This new circle will intersect the given circle at two points, say Q and R.

6. Join PQ and PR.

7. PQ and PR are the required pair of tangents to the circle.

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Measurement and Verification

On measuring the lengths of the tangents using a ruler, we find that:

PQ = 8 cm

PR = 8 cm

To verify this by calculation, consider the right-angled triangle OQP (since the radius OQ is perpendicular to the tangent PQ at the point of contact).

By the Pythagorean theorem:

$OP^2 = OQ^2 + PQ^2$

$10^2 = 6^2 + PQ^2$

$100 = 36 + PQ^2$

$PQ^2 = 100 - 36 = 64$

$PQ = \sqrt{64} = 8$ cm.

The calculated length matches the measured length, thus verifying the construction.

Given

Two concentric circles with radii 4 cm and 6 cm.

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To Construct

A tangent to the inner circle from a point on the outer circle, measure its length, and verify.

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Steps of Construction

1. Draw two concentric circles with a common centre O and radii 4 cm and 6 cm.

2. Take any point P on the outer circle (radius 6 cm).

3. Join OP.

4. Draw the perpendicular bisector of OP. Let M be the midpoint of OP.

5. With M as the centre and radius MO (or MP), draw another circle.

6. This circle will intersect the inner circle (radius 4 cm) at two points, say Q and R.

7. Join PQ. PQ is the required tangent.

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Measurement and Verification

On measuring the length of the tangent PQ using a ruler, we find that:

PQ $\approx 4.5$ cm

To verify by calculation, consider the right-angled triangle OQP (since the radius OQ is perpendicular to the tangent PQ).

Here, OP is the radius of the outer circle (6 cm) and OQ is the radius of the inner circle (4 cm).

By the Pythagorean theorem:

$OP^2 = OQ^2 + PQ^2$

$6^2 = 4^2 + PQ^2$

$36 = 16 + PQ^2$

$PQ^2 = 36 - 16 = 20$

$PQ = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$ cm.

Using $\sqrt{5} \approx 2.236$, we get $PQ \approx 2 \times 2.236 = 4.472$ cm.

The calculated length is approximately 4.5 cm, which verifies the measurement from the construction.

Given

A circle of radius 3 cm.

Two points P and Q on an extended diameter, each 7 cm away from the centre.

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To Construct

Tangents to the circle from points P and Q.

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Steps of Construction

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

2. Draw a diameter and extend it on both sides. Mark points P and Q on the extended diameter such that OP = OQ = 7 cm.

3. Draw the perpendicular bisector of OP. Let M be the midpoint of OP.

4. With M as the centre and radius MO, draw a circle. This circle will intersect the original circle at points R and S.

5. Join PR and PS. These are the tangents from point P.

6. Draw the perpendicular bisector of OQ. Let N be the midpoint of OQ.

7. With N as the centre and radius NO, draw a circle. This circle will intersect the original circle at points T and U.

8. Join QT and QU. These are the tangents from point Q.

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Justification

Join OR. In the circle with diameter OP, the angle $\angle ORP$ is an angle in a semicircle, so $\angle ORP = 90^\circ$. Since OR is the radius of the given circle, and PR is perpendicular to the radius at the point of contact R, PR is a tangent. The same justification applies to the other three tangents PS, QT, and QU.

Given

A circle of radius 5 cm.

The angle between the pair of tangents is 60°.

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To Construct

A pair of tangents to the circle that are inclined to each other at 60°.

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Logic for Construction

If the tangents from an external point P touch the circle at A and B, and the centre is O, then the quadrilateral OAPB is formed. The sum of the angle between the tangents ($\angle APB$) and the angle subtended by the radii at the centre ($\angle AOB$) is 180°.

$\angle APB + \angle AOB = 180^\circ$

$60^\circ + \angle AOB = 180^\circ$

$\angle AOB = 180^\circ - 60^\circ = 120^\circ$

So, we need to construct two radii that are inclined at 120° to each other and then draw tangents at the endpoints of these radii.

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Steps of Construction

1. Draw a circle with centre O and radius 5 cm.

2. Draw any radius OA.

3. At the centre O, construct an angle $\angle AOB = 120^\circ$.

4. At point A, construct a line perpendicular to the radius OA. This is the first tangent.

5. At point B, construct a line perpendicular to the radius OB. This is the second tangent.

6. Let the two tangents intersect at point P.

7. PA and PB are the required tangents, and the angle between them, $\angle APB$, will be 60°.

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Justification

In the quadrilateral OAPB, we have:

$\angle OAP = 90^\circ$ (Tangent is perpendicular to radius)

$\angle OBP = 90^\circ$ (Tangent is perpendicular to radius)

$\angle AOB = 120^\circ$ (By construction)

Sum of angles in a quadrilateral = 360°

$\angle APB = 360^\circ - (\angle OAP + \angle OBP + \angle AOB)$

$\angle APB = 360^\circ - (90^\circ + 90^\circ + 120^\circ) = 360^\circ - 300^\circ = 60^\circ$.

The construction is justified.

Given

A line segment AB of length 8 cm.

A circle with centre A and radius 4 cm.

A circle with centre B and radius 3 cm.

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To Construct

Tangents to the circle with centre A from point B.

Tangents to the circle with centre B from point A.

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Steps of Construction

1. Draw a line segment AB of length 8 cm.

2. With A as centre, draw a circle of radius 4 cm.

3. With B as centre, draw a circle of radius 3 cm.

4. Draw the perpendicular bisector of the line segment AB. Let M be the midpoint of AB.

5. With M as the centre and radius MA (or MB), draw a third circle.

6. This circle (with diameter AB) will intersect the circle with centre A at two points, say P and Q. Join BP and BQ. These are the tangents from B to the circle with centre A.

7. The same circle (with diameter AB) will intersect the circle with centre B at two points, say R and S. Join AR and AS. These are the tangents from A to the circle with centre B.

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Justification

Join AP. The angle $\angle APB$ is an angle in the semicircle with diameter AB, so $\angle APB = 90^\circ$. Since AP is the radius of the circle with centre A, and BP is perpendicular to it at P, BP is a tangent. The justification is similar for the other tangents BQ, AR, and AS.

Given

A right triangle ABC with AB = 6 cm, BC = 8 cm, $\angle B = 90^\circ$.

BD is perpendicular to AC.

A circle passes through B, C, and D.

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To Construct

The tangents from point A to the circle passing through B, C, and D.

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Steps of Construction

1. Draw a line segment BC of length 8 cm.

2. At B, construct a right angle ($\angle XBC = 90^\circ$).

3. With B as centre and radius 6 cm, cut an arc on the ray BX at point A. Join AC. This forms the right triangle ABC.

4. Draw the perpendicular BD from B to AC.

5. Now, we need to draw the circle through B, C, and D. Since $\angle BDC = 90^\circ$, the circle passing through B, C, D must have the hypotenuse BC of the right triangle BDC as its diameter.

6. Find the midpoint of BC. Let it be O. This is the centre of the circle.

7. With O as centre and radius OB (or OC), draw the circle. This circle will pass through B, C, and D.

8. Now, we need to construct tangents from the external point A to this circle.

9. Join AO.

10. Draw the perpendicular bisector of AO. Let M be the midpoint of AO.

11. With M as centre and radius MA, draw a circle. This circle will intersect the circle through B, C, D at two points. One point will be B, and let the other be E.

12. Join AB and AE. These are the required tangents.

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Justification

We need to justify why AB and AE are the tangents.

The centre of the circle passing through B, C, D is O, the midpoint of BC. The radius is OB.

In $\triangle ABC$, $\angle ABC = 90^\circ$. This means the line AB is perpendicular to the radius OB at the point B on the circle. Therefore, AB is a tangent to the circle at B.

For the other tangent AE, join OE. The angle $\angle AEO$ is an angle in the semicircle with diameter AO, so $\angle AEO = 90^\circ$. Since OE is the radius of the circle and AE is perpendicular to it at E, AE is also a tangent to the circle.

Given

A circle (drawn with a bangle, so the centre is unknown).

A point P outside the circle.

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To Construct

A pair of tangents from point P to the circle.

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Steps of Construction

1. Draw a circle using a bangle.

2. To find the centre of this circle, draw two non-parallel chords, say AB and CD.

3. Draw the perpendicular bisectors of both chords AB and CD.

4. The point where the two perpendicular bisectors intersect is the centre of the circle. Let this point be O.

5. Take a point P outside the circle.

6. Join OP.

7. Draw the perpendicular bisector of OP. Let M be the midpoint of OP.

8. With M as the centre and radius MO, draw a circle.

9. This circle will intersect the original circle at two points, say Q and R.

10. Join PQ and PR. These are the required tangents.

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Justification

The centre O is found correctly because the perpendicular bisector of any chord passes through the centre of the circle. The intersection of two such bisectors uniquely identifies the centre.

The construction of tangents is justified because the angle in a semicircle is a right angle. $\angle OQP = 90^\circ$, which means the radius OQ is perpendicular to the line PQ at the point of contact Q. Hence, PQ is a tangent. The same logic applies to PR.