RD SHARMA Solutions for Class 10 Maths Chapter 9 - Constructions

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Chapter 9 - Constructions Exercise Ex. 9.1

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions



Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 3
Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions



Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 4

Draw a line segment of length 8 cm and divide it internally in the radio 4:5.

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions 

1. Draw segment AB of length 8 cm.

2. Draw any ray AX making acute angle BAX with AB.

3. Draw BY parallel to AX by making ABY = BAX.

4. Mark four points A1, A2,…..,A4 on AX and five points B1, B2,…..,B5 on BY such that AA1=A1A2=…..= A3A4=BB1=B1B2=….=B4B5

5. Join A4B5; it intersects AB at P. P divides AB in the ratio 4:5. 

Chapter 9 - Constructions Exercise Ex. 9.2

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions


Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 5

Construct a triangle with sides 5 cm, 6cm and 7 cm and then another triangle whose sides are Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions of the corresponding sides of the first triangle. 

Solution 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Steps of construction

(1) Draw a line segment BC with 6 cm.

(2) Taking centres B and C and radii 5 cm and 7 cm respectively, draw two arcs (one from each centre) which intersect each other at A.

(3) Join AB and AC.

(4) At B, draw an angle CBX of any acute measure.

(5) Starting from B, cut 3 equal parts on BX such that BX1 = X1X2 = X2X3 = X3X4 = X4X5 = X5X6 = X6X7.

(6) Join X7C.

(7) Through X5, draw X5Q X7C.

(8) Through Q, draw QP CA.

PBQ ∼ ∆ABC.

Question 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions


Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 9

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 9

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 10

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 10

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 11

Construct a traingle similar to a given Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - ConstructionsXYZ with its sides equal to (3/4)th of the corresponding sides of Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - ConstructionsXYZ. Write the steps of construction.

Solution 11

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 12

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 12
Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions
Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions
Question 13

Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are 3/5 times the corresponding sides of the given triangle.

Solution 13

Steps of construction:

  1. Draw a line segment AB = 6.5 cm
  2. With A as the centre and radius AC = 5.5 cm, draw an arc.
  3. With B as the centre and radius BC = 5 cm, draw an arc intersecting the arc drawn in step 2 at C
  4. Join AC and BC to obtain ∆ABC.
  5. Below AB make an acute BAX.
  6. Along AX, mark off five points as the sides of triangle to be 3/5th of original triangle, the sides to be divided into five equal parts. Mark A1, A2, A3, A4, A5 along AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5
  7. Join A5B.
  8. Consider three parts out of five equal parts on AX. From point A3, draw A3B' A5B meeting AB at B' such that AA5B = AA3B'.
  9. From B', draw B'C' BC meeting AC at C' such that  ABC = AB'C'
  10. AB'C' is the required triangle, each of whose sides is 3/5th of the corresponding sides of ∆ABC

 

 

 

 

 

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 14

Construct a ∆PQR with side QR = 7 cm, PQ = 6 cm and mPQR = 60°. Then construct another triangle whose sides are 3/5 of the corresponding sides of ∆PQR.

Solution 14

Steps of construction:

  1. Draw a line segment PQ = 6 cm
  2. To construct PQR = 60°, taking Q as the centre and with an arbitrary radius draw an arc cutting PQ on S.
  3. Taking S as the centre, and the same radius, draw an arc cutting the previous arc at T. Join Q and T, extend QT further.
  4. With Q as the centre and a radius of 7 cm, draw an arc on the extended line segment QT at R.
  5. Join PR and QR to obtain ∆PQR.
  6. Below PQ make an acute angle QPX.
  7. Along PX, mark off five points as the sides of triangle to be 3/5th of original triangle, the sides to be divided into five equal parts. Mark P1, P2, P3, P4, P5 along PX, such that PP1 = P1P2 = P2P3 =P3P4 = P4P5.
  8. Join P5Q.
  9. Consider three parts out of the five equal parts on PX. From point P3, draw P3Q' P5Q meeting PQ at Q' by making QP5P = Q'P3P
  10. From Q', draw Q'R' QR by making PQR = PQ'R'
  11. PQ'R' is the required triangle, each of whose sides is 3/5th of the corresponding sides of ∆PQR

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 15

Draw a ∆ ABC in which base BC = 6cm, AB = 5 cm and ABC = 60o. Then construct another triangle whose sides are Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructionsof the corresponding sides of ∆ABC.

Solution 15

1. Draw ∆ABC with base BC = 6cm, AB = 5 cm andABC = 60˚.

2. Draw any ray BX making acute angle ABX with BC.

3. Mark four points B1, B2,B3,B4 on BY such that BB1=B1B2=B2B3=B3B4.

4. Join CB4 and draw a line parallel to CB4, through B3, to intersect AC at C'.

5. Join AC, and draw a line parallel to AC, through C', to intersect AB at A'.

6. ∆A'BC' is the required triangle.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions 

Question 16

Draw a right triangle in which the sides (other than the hypotenuse) are of lengths 4 cm and 3 cm. Now, construct another triangle whose sides areRd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructionstimes the corresponding sides of the given triangle.

Solution 16

1. Draw AB = 4cm.

2. Draw RA perpendicular to AB, and mark C on it such that AC = 3 cm.

3. Join BC. DABC is the required triangle.

4. Draw any ray AX making acute angle BAX with AB.

5. Mark five points A1, A2,…..,A5 on AX.

6. Join BA5, and draw a line parallel to it passing from A3, intersecting AB at B'.

7. Draw line parallel to CB through B' to intersect AC at C'.

8. ∆AB'C' is the required triangle.

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions 

Question 17

Construct an equilateral triangle with each side 5 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of ABC.

Solution 17

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Steps of construction

(1) Draw a line segment BC with 5 cm.

(2) Taking centres B and C with equal radii 5 cm, draw two arcs (one from each centre) which intersect each other at A.

(3) Join AB and AC.

(4) At B, draw an angle CBX of any acute measure.

(5) Starting from B, cut 3 equal parts on BX such that BX1 = X1X2 = X2X3.

(6) Join X3C.

(7) Through X2, draw X2Q X3C.

(8) Through Q, draw QP CA.

PBQ ∼ ∆ABC.

Chapter 9 - Constructions Exercise Ex. 9.3

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 2
Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions



Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Solution 3
Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions



Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Question 4

Draw two tangents to a circle of radius 3.5 cm from a point P at a distance of 6.2cm from its centre.

Solution 4

Steps of construction:

 

  1. Construct a line segment OP of length = 6.2 cm
  2. Taking O as the centre, construct a circle of radius3.5 cm.
  3. Taking O and P as the centres, draw arcs of circles above and below OP intersecting each other at points R and S respectively.
  4. Draw the perpendicular bisector of OP by joining A and  B. Mark the midpoint of OP as Q.
  5. Taking Q as the centre draw a circle of radius OQ or PQ and mark the points of intersection of the two circles as A and B.
  6. Join PA and extend it on both the sides.

Join PB and extend it on both the sides

PA and PB are the required tangents.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions 

 

 

 

Question 5

Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°.

Solution 5
  1. Draw a circle having a centre O and a radius of 4.5 cm.
  2. Take point P on the circle and join OP.
  3. Angle between the tangents = 45°

    Hence, the angle at the centre

    = 180° - 45° = 135° (supplement of the angle between the tangents)

    ∴Construct m∠POQ = 135°

  4. Keeping a radius of 4.5 cm, draw arcs of circle taking the points P, and Q as the centres.
  5. Name the points of intersection of arcs and circle as A and C respectively.
  6. Taking A as the centre, and with the same radius mark B such that OA = AB.
  7. Similarly, taking C as the centre and with the same radius mark D such that OC = CD.
  8. Taking A and B as the centres and the same radius draw two arcs intersecting each other at U.
  9. Join P, S and U and extend it on both the sides to draw a tangent at point P.
  10. Taking C and D as the centres and the same radius draw two arcs intersecting each other at V.
  11. Join Q, T and V and extend it on both the sides to draw a tangent at point Q.
  12. Extended tangents at P and Q intersect at R.
  13. Hence, the required tangents are UR and VR such that the angle between them is 45°.

 

 

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions 

Question 6

Draw a right ∆ABC in which AB = 6 cm, BC = 8 cm and mB = 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.

Solution 6

Steps of construction:

  1. Draw a line segment BC = 8 cm.
  2. Measure mB = 90˚ at point B.
  3. Taking B as the centre and a radius of 6 cm, draw an arc and hence, construct AB = 6 cm
  4. Join AC. Hence, ∆ABC right angled at B is constructed.
  5. Taking B as the centre and a radius more than BD, draw two arcs cutting AC at points E and F.
  6. Taking E and F as centres draw arcs to intersect each-other at P and Q respectively. Join and extend PQ on both the sides. This is the perpendicular BD from B on AC.
  7. BDCD, hence, ∆BCD is a right angled triangle. Therefore the circle passing through points B, C and D has the hypotenuse BC as the diameter.
  8. Construct a circle by taking O, the midpoint of BC as the centre (BC is the diameter of the circle)
  9. OB AB, extend AB on both the sides. OB is one of the tangents passing through point A and tangent at point B.
  10. Join OA. Take the centre of OA by drawing intersecting arcs from point O and A and which intersect at points R and S respectively.
  11. Join RS. The point at which RS intersects OA is G.
  12. Taking G as the centre and radius OG, construct a circle.
  13. The point of intersection of the two circles is H.
  14. Join H and A. Now extend AH on both the sides. AH is the second tangent drawn from point A.

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions 

Question 7

Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to the smaller circle from a point on the larger circle. Also, measure its length.

Solution 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

Chapter 9 - Constructions Exercise Ex. MCQs

Question 1

To draw a pair of tangents to a circle which are inclined to each other at an angle of 60o, it is required to draw tangents at end points of those two radii of the circle, the angle between then should be

(a) 135o

(b) 90o

(c) 60o

(d) 120o

Solution 1

The figure formed by two tangents and the two radii at points of contact, is a quadrilateral.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

ABCD is a quadrilateral where ABC = 90o and ADC = 90o

Therefore, BAD + BCD = 180o

BCD = 120o

Hence, the angle between them should be 120o.

Question 2

To divide a line segment AB in the ratio 5 : 7, first ray AX is drawn so that BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is

(a) 8

(b) 10

(c) 11

(d) 12

Solution 2

The minimum number of points required to divide a line segment in the ration 5 : 7, is (5 + 7 =) 12.

Hence, option (d) is correct.

Question 3

To divide a line segment AB in the ratio m : n (m, n are positive integers), draw a ray AX such that BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is

(a) greater than m and n

(b) m + n

(c) m + n - 1

(d) mn 

Solution 3

The minimum number of points required to divide a line segment in the ratio m : n is m + n.

Hence, option (b) is correct.

Question 4

To draw a pair of tangents to a circle which are inclined to each other at an angle of 35o, it is required to draw tangent at the end points of those two radii of the circle, the angle between which is

(a) 105o

(b) 70o

(c) 140o

(d) 145o

Solution 4

The figure formed by two tangents and the two radii at points of contact, is a quadrilateral.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Constructions

ABCD is a quadrilateral where ABC = 90o and ADC = 90o

Therefore, BAD + BCD = 180o

BCD = 145o

Hence, the angle between them should be 145o.

Question 5

To construct a triangle similar to a given ABC with its side 8/5 of the corresponding side of ABC draw a ray BX such that CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is

(a) 5

(b) 8

(c) 13

(d) 3

Solution 5

The minimum number of points to be located at equal distances on ray BX is (Max{8, 5} =) 8.

Hence, option (b) is correct.

Question 6

To divide a line segment AB in the ratio 4 : 7, a ray AX is drawn first such that BAX is an acute angle and then points A1, A2, A3, … are located at equal distances on the ray AX and the point B is joined to

(a) A12

(b) A11

(c) A10

(d) A9

Solution 6

As the line segment AB is divided in the ratio 4 : 7.

The minimum number of points required to is (4 + 7 =) 11.

So, B should be joined to A11.

Hence, option (b) is correct.

Question 7

To construct a triangle similar to a given ABC with its sides 3/7 of the corresponding sides of ABC, first draw a ray BX such that CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B1, B2, B3, … on BX at equal distances and next step is to join

(a) B10 to C

(b) B3 to C

(c) B7 to C

(d) B4 to C

Solution 7

The similar triangle has sides 3/7 of the original ABC.

After locating points B1, B2, B3, … B7 on BX at equal distances.

The next step is join B7 to C.

Hence, option (c) is correct.

Question 8

To divide a line segment AB in the ratio 5 : 6, draw a ray AX such that BAX is an acute angle, then draw a ray BY parallel to AX and the points A1, A2, … and B1, B2, … are located at equal distances on rays AX and BY respectively. Then the points joined are

(a) A5 and B6

(b) A6 and B5

(c) A4 and B5

(d) A5 and B4

Solution 8

The line segment AB is divided in the ratio 5 : 6.

Locate points A1, A2, … A5 on AX and B1, B2, … B6 on BY as A:B = 5:6.

So, the points joined to be are A5 and B6.

Hence, option (a) is correct.