Class 10 RD SHARMA Solutions Maths Chapter 9 - Constructions

Ex. 9.1

Ex. 9.2

Ex. 9.3

Ex. MCQs

Constructions Exercise Ex. 9.1

Solution 1

Solution 2

Solution 3

Solution 4

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.

Constructions Exercise Ex. 9.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

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.

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

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 A₁, A_2,A_3,A_4,A₅ along AX such that AA₁= A₁A₂= A₂A₃ = A₃A₄ = A₄A₅
Join A₅B.
Consider three parts out of five equal parts on AX. From point A₃, draw A₃B' ∥ A₅B meeting AB at B' such that ∠AA₅B = ∠AA₃B'.
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

Solution 14

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 P₁, P_2,P_3,P_4,P₅ along PX, such that PP₁= P₁P₂ = P₂P₃ =P₃P₄ = P₄P₅.
Join P₅Q.
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
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

Solution 15

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 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.

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.

Solution 17

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.

Constructions Exercise Ex. 9.3

Solution 1

Solution 2

Solution 3

Solution 4

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.

Solution 5

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°.

Solution 6

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 each-other 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.

Solution 7

Constructions Exercise Ex. MCQs

Solution 1

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.

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.

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.

Solution 4

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.

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.

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.

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.

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.

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Class 10 RD SHARMA Solutions Maths Chapter 9 - Constructions

Constructions Exercise Ex. 9.1

Solution 1

Solution 2

Solution 3

Solution 4

Constructions Exercise Ex. 9.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Constructions Exercise Ex. 9.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Constructions Exercise Ex. MCQs

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

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