Gujarat Board GSEB Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2

Question 1.

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 circle of radius 6 cm and take O as centre.

2. Take a point P which is 10 cm away from O. Join OP.

3. Bisect OP. Let M be the mid-point of OP.

4. Take M as centre and MO as radius, draw another circle intersecting the previous circle at Q and R.

5. Join PQ and PR

PQ and PR are the required tangents

PQ = PR = 8cm.

Justification:

Join OQ and OR.

∠OQP = ∠ORP = 90° [Angles in semicircies]

Since OQ and OR are radii of the circle and PQ and PR will be the tangents to the circle at Q and R respectively. Circle with diameter OP intersects the given circle in only two points.

Hence, only two tangents can be drawn.

Question 2.

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 circle of radius 4 cm with centre O.

2. Taking O as centre draw another circle of radius 6 cm.

3. Take a point P an outer circle. Join OP.

4. Bisect OP. Let M be the mid-point of OP.

5. Take M as centre and MO as radius, draw a circle. Let it intersect the given circle at the point Q and R.

6. Join PQ.

PQ is the required tangent. PQ = 4.5 cm.

By actual Calculation

As ∠OQP = 90° (Angle in a semicircle)

By Pythagoras theorem:

OP^{2} = PQ^{2} + OQ^{2}

PQ = \(\sqrt{\mathrm{OP}^{2}-\mathrm{OQ}^{2}}\)

= \(\sqrt{6^{2}-4^{2}}\)

= \(\sqrt{36-16}\) = \(\sqrt{20}\)

= 4.47 cm (approx)

Justification:

Since ∠OQP = 90° (Angle in a semicircle)

⇒ PQ ⊥ OQ

Since OQ is the radius of the given circle, PQ has to be a tangent to the circle.

Question 3.

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:

Steps of construction:

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

2. Take two points P arid Q on one of its extended diameter each at a distance 7 cm from its centre O.

3. Bisect PO. Let M be the mid-point of PO.

4. Take M as centre and MO as radius, draw a circle intersecting the given circle at A and B

5. Join PA and PB.

6. Bisect QO. Let N be the mid-point of QO.

7. Take N as centre and NO as radius, draw a circle intersecting the given circle of C and D.

8. Join QC and QD.

PA, PB, QC and QD are the required tangents.

Justification:

Join OA and OB.

⇒ ∠PAO = 90° [Angle in Semicircle]

PA ⊥ OA

Since OA is the radius of the given circle PA has to be a tangent to the circle. Similarly, PB in also tangent to the circle. ,.

With the same above explanation, QC and QD are also tangent to the circle.

Question 4.

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

Solution:

Steps of construction

1. Draw a circle of radius 5 cm and take O as centre.

2. Take a point A on a circle and draw ∠AOB = 120°.

3. At A and B draw angles of 90° which other arms meet at C. Then, AC and BC are the required tangents inclining each other at an angle of 60°.

Justification:

∠OAC = 90° [By construction]

∠OBC = 90°

⇒ OA ⊥ AC

OB ⊥ BC

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

∴ AC and BC is a tangent to the circle. In quadrilateral ΔOBC,

∠AOB + ∠OBC + ∠BCA + ∠CAO = 360°

[Angle sum property of a quadrilaterall

⇒ 120° + 90° + ∠BCA + 90° = 360°

⇒ ∠BCA + 300° = 360°

⇒ ∠BCA = 360° – 300° = 60°

Question 5.

Draw a line segment AB of length 8 cm. Taking A as the 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:

1. Draw a line segment AB = 8 cm.

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

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

4. Bisect AB. Let M be the mid-point of AB.

5. Taking M as a centre and AM as radius, draw a circle, intersecting the circle with centre A at P and Q; and intersecting the circle with centre B at R and S.

6. Join BP and BQ

7. Join AR and AS BP, BQ, AR and AS are the required tangents.

Justification:

∠APB = 90° [Angles in the semicircle]

∠ARB = 90°

BP ⊥ AP

and AR ⊥BR

Since AP and BR are the radii of the circles with A and B as centre respectively. So, BP and AR are the tangents to the circle with centre A and B respectively.

Question 6.

Let ABC be a right triangle in which AB = 6cm, BC = 8cm and ∠B = 90°. BD is 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 ABC with AB 6 cm, BC = 8 cm and ∠B = 90°.

2. Draw BD ⊥ AC.

3. Through points B, C and D, draw a circle and mark the centre as O.

4. Join AO and bisect it. Let M be the midpoint of AO.

5. Taking M as centre and MO as radius, draw a circle intersecting the given circle at B and E.

6. Join AB and AE.

Thus, AB and AE are the required tangents.

Justification:

By joining OE.

∠AEO = 90° [Angle in the semicircie]

⇒ AE ⊥ OE

Now since OE is a radius of the given circle, AE will be a tangent to the circle.

Question 7.

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

Steps of Construction:

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

2. Take two chords AB and CD (non-parallel to each other)

3. Draw the perpendicular bisectors of AB and CD intersecting at O. Then O is the centre of the given circle.

4. Take a point P outside the circle. Join OP

5. Bisect OP. Let M be the mid-point of OP.

6. Taking M as centre and MO as radius draw a circle intersecting the given circle at Q and R.

7. Join PQ and PR

∴ PQ and PR are the required tangents.

Justification:

Join OQ and OR

∠PQO = 90° [Angle in the semicircle]

⇒ PQ ⊥ OQ

Since OQ in a radius of the given circle. PQ will be a tangent to the circle. Similarly, PR will also be a tangent to the circle.