RD SHARMA Solutions for Class 10 Maths Chapter 9  Constructions
Chapter 9  Constructions Exercise Ex. 9.1
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 A_{1}, A_{2},…..,A_{4} on AX and five points B_{1}, B_{2},…..,B_{5} on BY such that AA_{1}=A_{1}A_{2}=…..= A_{3}A_{4}=BB_{1}=B_{1}B_{2}=….=B_{4}B_{5}
5. Join A_{4}B_{5}; it intersects AB at P. P divides AB in the ratio 4:5.
Chapter 9  Constructions Exercise Ex. 9.2
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 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}.
(6) Join X_{7}C.
(7) Through X_{5}, draw X_{5}Q ‖ X_{7}C.
(8) Through Q, draw QP ‖ CA.
∴ ∆PBQ ∼ ∆ABC.
Steps of construction:
 Draw a line segment AB = 6.5 cm
 With A as the centre and radius AC = 5.5 cm, draw an arc.
 With B as the centre and radius BC = 5 cm, draw an arc intersecting the arc drawn in step 2 at C
 Join AC and BC to obtain ∆ABC.
 Below AB make an acute ∠BAX.
 Along AX, mark off five points as the sides of triangle to be 3/5^{th} of original triangle, the sides to be divided into five equal parts. Mark A_{1}, A_{2, }A_{3, }A_{4, }A_{5} along AX such that AA_{1 }= A_{1}A_{2 }= A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}
 Join A_{5}B.
 Consider three parts out of five equal parts on AX. From point A_{3}, draw A_{3}B' ∥ A_{5}B meeting AB at B' such that ∠AA_{5}B = ∠AA_{3}B'.
 From B', draw B'C' ∥ BC meeting AC at C' such that ∠ABC = ∠AB'C'
 ∆AB'C' is the required triangle, each of whose sides is 3/5^{th} of the corresponding sides of ∆ABC
Steps of construction:
 Draw a line segment PQ = 6 cm
 To construct ∠ PQR = 60°, taking Q as the centre and with an arbitrary radius draw an arc cutting PQ on S.
 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.
 With Q as the centre and a radius of 7 cm, draw an arc on the extended line segment QT at R.
 Join PR and QR to obtain ∆PQR.
 Below PQ make an acute angle QPX.
 Along PX, mark off five points as the sides of triangle to be 3/5^{th} of original triangle, the sides to be divided into five equal parts. Mark P_{1}, P_{2, }P_{3, }P_{4, }P_{5} along PX, such that PP_{1 }= P_{1}P_{2} = P_{2}P_{3} =P_{3}P_{4} = P_{4}P_{5}.
 Join P_{5}Q.
 Consider three parts out of the five equal parts on PX. From point P_{3}, draw P_{3}Q' ∥ P_{5}Q meeting PQ at Q' by making ∠QP_{5}P = ∠Q'P_{3}P
 From Q', draw Q'R' ∥ QR by making ∠PQR = ∠ PQ'R'
 ∆PQ'R' is the required triangle, each of whose sides is 3/5^{th} of the corresponding sides of ∆PQR
1. Draw ∆ABC with base BC = 6cm, AB = 5 cm and∠ABC = 60^{˚}.
2. Draw any ray BX making acute angle ∠ABX with BC.
3. Mark four points B_{1}, B_{2},B_{3},B_{4} on BY such that BB_{1}=B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4}.
4. Join CB_{4} and draw a line parallel to CB_{4}, through B_{3}, 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.
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 A_{1}, A_{2},…..,A_{5} on AX.
6. Join BA_{5}, and draw a line parallel to it passing from A_{3}, intersecting AB at B'.
7. Draw line parallel to CB through B' to intersect AC at C'.
8. ∆AB'C' is the required triangle.
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 BX_{1} = X_{1}X_{2} = X_{2}X_{3}.
(6) Join X_{3}C.
(7) Through X_{2}, draw X_{2}Q ‖ X_{3}C.
(8) Through Q, draw QP ‖ CA.
∴ ∆PBQ ∼ ∆ABC.
Chapter 9  Constructions Exercise Ex. 9.3
Steps of construction:
 Construct a line segment OP of length = 6.2 cm
 Taking O as the centre, construct a circle of radius3.5 cm.
 Taking O and P as the centres, draw arcs of circles above and below OP intersecting each other at points R and S respectively.
 Draw the perpendicular bisector of OP by joining A and B. Mark the midpoint of OP as Q.
 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.
 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.
 Draw a circle having a centre O and a radius of 4.5 cm.
 Take point P on the circle and join OP.

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°
 Keeping a radius of 4.5 cm, draw arcs of circle taking the points P, and Q as the centres.
 Name the points of intersection of arcs and circle as A and C respectively.
 Taking A as the centre, and with the same radius mark B such that OA = AB.
 Similarly, taking C as the centre and with the same radius mark D such that OC = CD.
 Taking A and B as the centres and the same radius draw two arcs intersecting each other at U.
 Join P, S and U and extend it on both the sides to draw a tangent at point P.
 Taking C and D as the centres and the same radius draw two arcs intersecting each other at V.
 Join Q, T and V and extend it on both the sides to draw a tangent at point Q.
 Extended tangents at P and Q intersect at R.

Hence, the required tangents are UR and VR such that the angle between them is 45°.
Steps of construction:
 Draw a line segment BC = 8 cm.
 Measure m∠B = 90˚ at point B.
 Taking B as the centre and a radius of 6 cm, draw an arc and hence, construct AB = 6 cm
 Join AC. Hence, ∆ABC right angled at B is constructed.
 Taking B as the centre and a radius more than BD, draw two arcs cutting AC at points E and F.
 Taking E and F as centres draw arcs to intersect eachother at P and Q respectively. Join and extend PQ on both the sides. This is the perpendicular BD from B on AC.
 BD⊥CD, hence, ∆BCD is a right angled triangle. Therefore the circle passing through points B, C and D has the hypotenuse BC as the diameter.
 Construct a circle by taking O, the midpoint of BC as the centre (BC is the diameter of the circle)
 OB ⊥ AB, extend AB on both the sides. OB is one of the tangents passing through point A and tangent at point B.
 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.
 Join RS. The point at which RS intersects OA is G.
 Taking G as the centre and radius OG, construct a circle.
 The point of intersection of the two circles is H.
 Join H and A. Now extend AH on both the sides. AH is the second tangent drawn from point A.
Chapter 9  Constructions Exercise Ex. MCQs
The figure formed by two tangents and the two radii at points of contact, is a quadrilateral.
ABCD is a quadrilateral where ∠ABC = 90^{o} and ∠ADC = 90^{o}
Therefore, ∠BAD + ∠BCD = 180^{o}
∴ ∠BCD = 120^{o}
Hence, the angle between them should be 120^{o}.
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.
The minimum number of points required to divide a line segment in the ratio m : n is m + n.
Hence, option (b) is correct.
The figure formed by two tangents and the two radii at points of contact, is a quadrilateral.
ABCD is a quadrilateral where ∠ABC = 90^{o} and ∠ADC = 90^{o}
Therefore, ∠BAD + ∠BCD = 180^{o}
∴ ∠BCD = 145^{o}
Hence, the angle between them should be 145^{o}.
The minimum number of points to be located at equal distances on ray BX is (Max{8, 5} =) 8.
Hence, option (b) is correct.
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 A_{11}.
Hence, option (b) is correct.
The similar triangle has sides 3/7 of the original ∆ABC.
After locating points B_{1}, B_{2}, B_{3}, … B_{7} on BX at equal distances.
The next step is join B_{7} to C.
Hence, option (c) is correct.
The line segment AB is divided in the ratio 5 : 6.
Locate points A_{1}, A_{2}, … A_{5} on AX and B_{1}, B_{2}, … B_{6} on BY as A:B = 5:6.
So, the points joined to be are A_{5} and B_{6}.
Hence, option (a) is correct.
Other Chapters for CBSE Class 10 Mathematics
Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pairs of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Coordinate Geometry Chapter 7 Triangles Chapter 8 Circles Chapter 10 Trigonometric Ratios Chapter 11 Trigonometric Identities Chapter 12 Heights and Distances Chapter 13 Areas Related to Circles Chapter 14 Surface Areas and Volumes Chapter 15 Statistics Chapter 16 ProbabilityRD SHARMA Solutions for CBSE Class 10 Subjects
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