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NCERT Solutions for Class 10 Maths Chapter 11 Constructions

NCERT Solutions Exercise 11.1 Class 10 Constructions

In each of the following, give the justification of the construction also:
Question 1.
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 a line segment AB = 7.6 cm.
  • 2. Draw an acute angle BAX on base AB. Mark the ray as AX.
  • 3. Locate 13 points A1, A2, A3, …… , A13 on the ray AX so that AA1 = A1A2 = ……… = A12A13
  • 4. Join A13 with B and at A5 draw a line ∥ to BA13, i.e. A5C. The line intersects AB at C.
  • 5. On measure AC = 2.9 cm and BC = 4.7 cm.

Justification:
In ∆AA5C and ∆AA13B,
 

∴ AC : BC = 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 2/3 of the corresponding sides of the first triangle.
 

Solution:
Steps of Construction:

  • 1. Construct a ∆ABC with AB = 4 cm, BC = 6 cm and AC = 5 cm.
  • 2. Draw an acute angle CBX on the base BC at point B. Mark the ray as BX.
  • 3. Mark the ray BX with B1, B2, B3 such that
  • BB1 = B1B2 = B2B3
  • 4. Join B3 to C.
  • 5. Draw B2C’ ∥ B3C, where C’ is a point on BC.
  • 6. Draw C’A’ ∥ AC, where A’ is a point on BA.
  • 7. ∆A’BC’ is the required triangle.

Justification: In ∆A’BC and ∆ABC,

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
 

Solution:
Steps of Construction:

  • 1. Draw a ∆ABC with AB = 5 cm, BC = 7 cm and AC = 6 cm.
  • 2. Draw an acute angle CBX below BC at point B.
  • 3. Mark the ray BX as B1, B2, B3, B4, B5, B6 and B7 such that BB1= B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
  • 4. Join B5 to C.
  • 5. Draw B7C’ parallel to B5C, where C’ is a point on extended line BC.
  • 6. Draw A’C’ ∥ AC, where A’ is a point on extended line BA.
  • A’BC’ is the required triangle.

Justification: In ∆ABC and ∆A’BC’,

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1 1/2 times the corresponding sides of the isosceles triangle.
 

Solution:
Steps of Construction:

  • 1. Draw base AB = 8 cm.
  • 2. Draw perpendicular bisector of AB. Mark CD = 4 cm, on ⊥ bisector where D is mid-point on AB.
  • 3. Draw an acute angle BAX, below AB at point A.
  • 4. Mark the ray AX with A1, A2, A3 such that AA1 =A1A2 = A2A3
  • 5. Join A2 to B. Draw A3B’ ∥ A2 B, where B’ is a point on extended line AB.
  • 6. At B’, draw B’C’ 11 BC, where C’ is a point on extended line AC.
  • 7. ∆AB’C’ is the required triangle.

Justification: In ∆ABC and ∆A’BC’,
 

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
 

Solution:
Steps of Construction:

  • 1. Draw a line segment BC = 6 cm and at point B draw an ∠ABC = 60°.
  • 2. Cut AB = 5 cm. Join AC. We obtain a ∆ABC.
  • 3. Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.
  • 4. Locate 4 points A1, A2, A3 and A4 on the ray BX so that BA1 = A1A2 = A2A3 = A3A4.
  • 5. Join A4 to C.
  • 6. At A3, draw A3C’ ∥ A4C, where C’ is a point on the line segment BC.
  • 7. At C’, draw C’A’ ∥ CA, where A’ is a point on the line segment BA.

∴ ∆A’BC’ is the required triangle.

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 43 times the corresponding sides of ∆ABC.
 

Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180°
⇒ 105° + 45° + ∠C = 180°
⇒ 150° + ∠C = 180°
⇒ ∠C = 30°
Steps of Construction:
1. Draw a line segment BC = 7 cm. At point B, draw an ∠B = 45° and at point C, draw an ∠C = 30° and get ΔABC.
2. Draw an acute ∠CBX on the base BC at point B. Mark the ray BX with B1, B2, B3, B4, such that BB1 = B1B2 = B2B3 = B3B4
3. Join B3 to C.
4. Draw B4C’ ∥ B3C, where C’ is point on extended line segment BC.
5. At C’, draw C’A’ ∥ AC, where A’ is a point on extended line segment BA.
6. ∆A’BC’ is the required triangle.

Justification: In ∆ABC and ∆A’BC’,

Question 7.
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 5/3F times the corresponding sides of the given triangle.
 

Solution:
Steps of Construction:

  • 1. Draw a right angled ∆ABC with AB = 4 cm, AC = 3 cm and ∠A = 90°.
  • 2. Make an acute angle BAX on the base AB at point A.
  • 3. Mark the ray AX with A1, A2, A3, A4, A5 such that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
  • 4. Join A3B. At A5, draw A5B’ ∥ A3B, where B’ is a point on extended line segment AB.
  • 5. At B’, draw B’C’ ∥ BC, where C’ is a point on extended line segment AC.
  • 6. ∆AB’C’ is the required triangle.

Justification:
In ∆ABC and ∆AB’C’,

NCERT Solutions Exercise 11.2 Class 10 Constructions

In each of the following, give the justification of the construction also:
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.
  • 2. Mark a point P at a distance 10 cm from the centre O.
  • 3. Here OP = 10 cm, draw perpendicular bisector of OP, which intersects OP at O’.
  • 4. Take O’ as centre, draw a circle of radius O’O, which passes through O and P and intersects the previous circle at points T and Q.
  • 5. Join PT and PQ, measure them, these are the required tangents.

Justification:
Join OT and OQ.

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ PT = PQ.

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 concentric circles of radius OA = 4 cm and OP = 6 cm having same centre O.
  • 2. Mark these circles as C and C’.
  • 3. Points O, A and P lie on the same line.
  • 4. Draw perpendicular bisector of OP, which intersects OP at O’.
  • 5. Take O’ as centre, draw a circle of radius OO’ which intersects the circle C at points T and Q.
  • 6. Join PT and PQ, these are the required tangents.
  • 7. Length of these tangents are approx. 4.5 cm.

Justification:
Join OT and OQ.

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ PT = PQ.

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, having centre O. Mark this circle as C.
  • 2. Mark points P and Q along its extended diameter such that OP = 7 cm and OQ = 7 cm.
  • 3. Draw perpendicular bisector of OQ, intersecting OQ at O’.
  • 4. Draw perpendicular bisector of OP intersecting OP at O”.
  • 5. Take O’ as centre and draw a circle of radius OO’ which passes through points O and Q, intersecting circle C at points R and T.
  • 6. Take O” as centre and draw a circle of radius O”P which passes through points O and P, intersecting the, circle C at points S and U.
  • 7. Join QR and QT; PS and PU, these are the required tangents.
     

Justification:
Join OR.
In ∆OQR,
OR ⊥ QR [Radius ± to tangent]
OQ2 = OR2 + QR2
⇒ (7)2 = (3)2 + QR2 ⇒ 49 = 9 + OR2
⇒ 40 = QR2 ⇒ QR = 2 √10 cm
Similary,
QT = 2 √10 cm
PS = 2 √10 cm
PU = 2 √10 cm
A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ QR = QT and PS = PU

Question 4.
Draw a pair of tangents to a circle of 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.
  • 2. As tangents are inclined to each other at an angle of 60°.
  • ∴ Angle between the radii of circle is 120°. (Use quadrilateral property)
  • 3. Draw radii OA and OB inclined to each other at an angle 120°.
  • 4. At points A and B, draw 90° angles. The arms of these angles intersect at point P.
  • 5. PA and PB are the required tangents.
     

Justification:
In quadrilateral AOBP,
AP and BP are the tangents to the circle.
Join OP.
In right angled AOAP, OA ⊥ PA [Radius is ⊥ to tangent]

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ PA = PB

Question 5.
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:

  • 1. Draw a line segment AB = 8 cm.
  • 2. Taking A as centre draw a circle C of radius 4 cm and taking B as centre draw a circle C’ of radius 3 cm.
  • 3. Draw perpendicular bisector of AB, which intersects AB at point O.
  • 4. Taking point O as centre draw a circle of radius 4 cm passing through points A and B which intersect circle C at P and S and circle C’ at points R and Q.
  • 5. Join AQ, AR, BP and BS. These are the required tangents.
     

Justification:
Join AP.

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ AQ = AR and BP = BS.

Question 6.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ZB = 90°. BD is the perpendicular Burn 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 right triangle ABC with AB = 6 cm, BC = 8 cm and ZB = 90°.
  • 2. From B, draw BD perpendicular to AC.
  • 3. Draw perpendicular bisector of BC which intersect BC at point O’.
  • 4. Take O’ as centre and O’B as radius, draw a circle C’ passes through points B, C and D.

5. Join O’A and draw perpendicular bisector of O’A which intersect O’A at point K.
6. Take K as centre, draw an arc of radius KO’ intersect the previous circle C’ at T.
7. Join AT, AT is required tangent.
Justification:
∠BDC = 90°
∴ BC acts as diameter.
AB is tangent to circle having centre O’

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ AB = AT.

Question 7.
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:
Steps of Construction:
1. Draw a circle C’ with the help of a bangle, for finding the centre, take three non-collinear points A, B and C, lying on the circle. Join AB and BC and draw perpendicular bisector of AB and BC, both intersect at a point O,
‘O’ is centre of the circle.

2. Take a point P outside the circle. Join OP.
3. Draw perpendicular bisector of OP, which intersects OP at point O’.
4. Take O’ as the centre with OO’ as radius draw a circle which passes through O and P, intersecting previous circle C’ at points R and Q.
5. Join PQ and PR.
6. PQ and PR are the required pair of tangents.
Justification:
Join OQ and OR.
In AOQP and AOPR, OQ = OR [Radii of the circle]
OP = OP [Common]
∠Q = ∠R = 90° [Radius is ⊥ to tangent]

∆OQP ≅ ∆ORP [by RHS]
∴ PQ = PR [By C.P.C.T]
A pair of tangents can be drawn to a circle from an external point lying outside the circle. These two tangents are equal in lengths.
∴ PQ = PR

Important Question

NCERT CBSE for Class 10 Maths Chapter 11 Constructions Important Questions

Constructions Class 10 Important Questions Short Answer-I (2 Marks)

Question 1.
Draw a line segment of length 6 cm. Using compasses and ruler, find a point P on it which divides it in the ratio 3 : 4.
Year of Question: (2011D)

Solution:
Hence, PA : PB = 3 : 4

Question 2.
Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that AP/AB=3/5.
Year of Question: (2011OD)

Solution:
AB = 7 cm, AB = AP/AB=3/5 . [Given
∴ AP : PB = 3 : 2
Hence, AP : AB = 3 : 5 or AP/AB=3/5

Constructions Class 10 Important Questions Short Answer-II (3 Marks)

Question 3.
Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Then construct a triangle whose sides are 5/7 times the corresponding sides of ∆ABC.
Year of Question: (2011D)
Solution:
In ∆ABC
AB = 5 cm
BC = 6 cm
∠ABC = 60°
Hence, ∆A’BC’ is the required ∆.
Question 4.
Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are 35 times the corresponding sides of the given triangle.
Year of Question: (20140D)

Solution:
∴ ∆A’BC’ is the required ∆.

Question 5.
Construct a right triangle in which the sides, (other than the hypotenuse) are of length 6 cm and 8 cm. Then construct another triangle, whose sides are 3/5 times the corresponding sides of the given triangle.
Year of Question: (2012D)
Solution:
Here AB = 8 cm, BC = 6 cm and
Ratio = 3/5 of corresponding sides
∴ ∆A’BC’ is the required triangle.
Question 6.
Draw a triangle PQR such that PQ = 5 cm, ∠P = 120° and PR = 6 cm. Construct another triangle whose sides are 3/4 times the corresponding sides of ∆PQR.
Year of Question: (2011D)

Solution:
In ∆PQR,
PQ = 5 cm, PR = 6 cm, ∠P = 120°
∴ ∆P’OR’ is the required ∆.

Question 7.
Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠C = 60°. Then construct another triangle, whose sides are 3/5 times the corresponding sides of ∆ABC.
Year of Question: (2012OD)
Solution:
Here, BC = 7 cm, ∠B = 45°, ∠C = 60° and ratio is 35 times of corresponding sides
∴ ∆A’BC’ is the required triangle.
Question 8.
Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of first triangle.
Year of Question: (2013D)

Solution:

Steps of Construction:
Draw ∆ABC with AC = 6 cm, AB = 5 cm, BC = 4 cm.
Draw ray AX making an acute angle with AÇ.
Locate 3 equal points A1, A2, A3 on AX.
Join CA3.
Join A2C’ ‖ CA3.
From point C’ draw B’C’ ‖ BC
. ∴ ∆A’BC’ is the required triangle.

Question 9.
Draw a pair of tangents to a circle of radius 3 cm, which are inclined to each other at an angle of 60°.
Year of Question:(2011OD)

Solution:
∴ PA & PB are the required tangents.

Question 10.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm.
Year of Question:(2013D)

Solution:
Steps of Construction:
Draw two circles with radius OA = 4 cm and OP = 6 cm with O as centre. Draw ⊥ bisector of OP at M. Taking M as centre and OM as radius draw another circle intersecting the smaller circle at A and B and touching the bigger circle at P. Join PA and PB. PA and PB are the required tangents.
Verification:
In rt. ∆OAP,
OA2 + AP2= OP2 . [Pythagoras’ theorem
(4)2 + (AP)2 = (6)2
AP2 = 36 - 16 = 20
AP = + √20=√4×5 = 2√5 = 2(2.236) = 4.472 = 4.5 cm By measurement, ∴ PA = PB = 4.5 cm

Question 11.
Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°.
Year of Question:(2013OD)

Solution:
Draw ∠AOB = 135°, ∠OAP = 90°, ∠OBP = 90°
∴ PA and PB are the required tangents.

Question 12.
Draw two tangents to a circle of radius 3.5 cm, from a point P at a distance of 6.2 cm from its cehtre.
Year of Question:(2013OD)

Solution:
OP = OC + CP = 3.5 + 2.7 = 6.2 cm
Hence AP & PB are the required tangents.

Question 13.
Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.
Year of Question:(2014D, 2015OD)

Solution:
Steps of Construction:
Draw BC = 8 cm.
From B draw an angle of 90°.
Draw an arc BA = 6 cm cutting the angle at A.
Join AC. ∴ ∆ABC is the required ∆.
Draw ⊥ bisector of BC cutting BC at M.
Take Mas centre and BM as radius, draw a circle.
Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.
AB and AE are the required tangents.
Justification: ∠ABC = 90° .[Given
Since, OB is a radius of the circle.
∴ AB is a tangent to the circle.
Also, AE is a tangent to the circle.

Question 14.
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.
Year of Question: (2014OD)

Solution:
Draw two circles on A and B as asked.
Z is the mid-point of AB.
From Z, draw a circle taking ZA = ZB as radius,
so that the circle intersects the bigger circle at M and N and smaller circle at X and Y.
Join AX and AY, BM and BN.
BM, BN are the reqd. tangents from external point B.
AX, AY are the reqd. tangents from external point A.
Justification:
∠AMB = 90° .[Angle in a semi-circle
Since, AM is a radius of the given circle.
∴ BM is a tangent to the circle
Similarly, BN, AX and AY are also tangents.

Constructions Class 10 Important Questions Long Answer (4 Marks).

Question 15.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45° and ∆A = 105°. Then construct a triangle whose sides are 35times the corresponding sides of ∆ABC.
Year of Question:(2011D)

Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180° . [angle sum property of a ∆
105° + 45° + C = 180°
∠C = 180° - 105° - 45o = 30°
BC = 7 cm
∴ ∆A’BC’ is the required ∆.

Question 16.
Draw a triangle ABC with side BC = 6 cm, ∠C = 30° and ∠A = 105°. Then construct another triangle whose sides are 2/3 times the corresponding sides of ∆ABC.
Year of Question:(2012D)

Solution:
Here, BC = 6 cm, ∠A = 105° and ∠C = 30°
In ∆ABC,
∠A + ∠B + ∠C = 180° .[Angle-sum-property of a ∆
105° + ∠B + 30° = 180°
∠B = 180° - 105° - 30° = 45°
∴ ∆A’BC’ is the required ∆.

Question 17.
Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of the first triangle.
Year of Question:(2012OD)

Solution:
Here, AB = 5 cm, BC = 7 cm, AC = 6 cm and ratio is 2/3 times of corresponding sides.
∴ ∆A’BC’ is the required triangle.

Question 18.
Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle whose sides are 3/4 times the corresponding sides of the isosceles triangle.
Year of Question:(2015D)
Solution:
∴ ∆A’BC’ is the required triangle.
Question 19.
Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle.
Year of Question:(2015D)

Solution:
Draw two circles on A and B as asked.
Z is the mid-point of AB.
From Z, draw a circle taking ZA = ZB as radius,
so that the circle intersects the bigger circle at M and N and smaller circle at X and Y.
Join AX and AY, BM and BN.
BM, BN are the required tangents from external point B.
AX, AY are the required tangents from external point A.
Justification:
∠AMB = 90° .[Angle in a semi-circle
Since, AM is a radius of the given circle.
∴BM is a tangent to the circle
Similarly, BN, AX and AY are also tangents.

Question 20.
Construct a ∆ABC in which AB = 6 cm, ∠A = 30° and ∆B = 60°. Construct another ∆AB’C’ similar to ∆ABC with base AB’ = 8 cm.
Year of Question:(2015OD)

Solution:
Steps of construction:
Draw a ∆ABC with side AB = 6 cm, ∠A = 30° and ∠B = 60°.
Draw a ray AX making an acute angle with AB on the opposite side of point C.
Locate points A1, A2, A3 and A4 on AX.
Join A3B. Draw a line through A4 parallel to A3B intersecting extended AB at B’.
Draw a line parallel to BC intersecting ray AY at C’.
Hence, ∆AB’C’ is the required triangle.

Question 21.
Construct a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Now construct another triangle whose sides are 5/7 times the corresponding sides of ∆ABC.
Year of Question:(2015OD)

Solution:
In ∆ABC, AB = 5 cm; BC = 6 cm; ∠ABC = 60°
∴ ∆A’BC’ is the required ∆.

Question 22.
Construct a triangle ABC in which BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are 3/4 times the corresponding sides of ∆ABC.
Year of Question:(2016D)

Solution:
Steps of Construction:
Draw ∆ABC with the given data.
Draw a ray BX downwards making an acute angle with BC.
Locate 4 points B1, B2, B3, B4, on BX, such that BB1 = B1B2 = B2B3 = B3B4.
Join CB4.
From B3 draw a line C’B3 ‖ CB4 intersecting BC at C’.
From C’ draw A’C’ ‖ AC intersecting AB at B’.
Then ∆A’BC’ is the required triangle.
Justification:

Question 23.
Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct a triangle whose sides are 4/5 times the corresponding sides of ∆ABC.
Year of Question:(2016D)

Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180° ..[Angle-sum-property of a ∆
105° + 45° + ∠C = 180°
∠C = 180° - 105° - 45° = 30°
∴ ∆A’BC’ is the required triangle.

Question 24.
Draw an isosceles ∆ABC in which BC = 5.5 cm and altitude AL = 3 cm. Then construct another triangle whose sides are 3/4 of the corresponding sides of ∆ABC.
Year of Question:(2016OD)

Solution:
∴ ∆A’BC’ is the required ∆.

Question 25.
Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then draw another triangle whose sides are 4/5 of the corresponding sides of first triangle.
Year of Question:(2016OD)

Solution:
∴ ∆A’BC’ is the required ∆.

Question 26.
Draw two concentric cirlces of radii 3 cm and 5 cm. Construct a tangent to smaller circle from a point on the larger circle. Also measure its length.
Year of Question:(2016D)

Solution:
We have, OD = 3 cm and OP = 5 cm
PA and PB are the required tangents
By measurement PA = PB = 4 cm.

Question 27.
Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other.
Year of Question:(2013, 2016OD)

Solution:
Draw a circle with O as centre and radius 4 cm.
Draw any ∠AOB = 120°. From A and B draw ∠PAO = ∠PBO = 90° which meet at P.
∴ PA and PB are the required tangents.

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