# RBSE Solutions for Class 10 Maths Chapter 14 Constructions Additional Questions

RBSE Solutions for Class 10 Maths Chapter 14 Constructions Additional Questions is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 14 Constructions Additional Questions.

## Rajasthan Board RBSE Class 10 Maths Chapter 14 Constructions Additional Questions

Question 1.
Draw a line segment of length 7.6 cm and (RBSESolutions.com) divide It Into a ratio of 5 : 8. Measure both the parts.
Solution :
Given :
Line segment AB = 7.6 cm.
Construction : (1) Incircle of ∆ABC.
Steps of Construction:
(1) Draw a line segment AB = 7.6 cm
(2) Draw a ray AX making an acute angle with
AB at A.
(3) Locate 5 + 8 = 13 points A1, A2, A3,………A13 at equal distance on AX. Join A13B.

(4) From point A5 draw a line AA5P || AA13B to meet AB at P. i.e., ∠BA13A = ∠PA5A.
Then AP : PB = 5 : 8
Hence, AP and PB are the (RBSESolutions.com) requires parts of AB.
Proof : In ∆AA5P and ∆AA13B,
A13B || A5P
∴ by BPT, we get
$$\frac { { AA }_{ 5 } }{ { A }_{ 5 }{ A }_{ 13 } }$$ = $$\frac { AP }{ PB }$$
But $$\frac { { AA }_{ 5 } }{ { A }_{ 5 }{ A }_{ 13 } }$$ = $$\frac { 5 }{ 8 }$$ (By construction)
∴ $$\frac { AP }{ PB }$$ = $$\frac { 5 }{ 8 }$$
Hence, point P, divides AB into the ratio 5 : 8.
Measuring two parts, we get.
AP = 2.9 cm
and PB = 4.7 cm

Question 2.
Draw a circle center O and radius 6 cm. Construct a (RBSESolutions.com) pair of tangents from a point 10 cm a part of its center. Also measure the length of tangents drawn.
Solution :
Given : A circle center O and radius 6 cm. Also a point P at distance of 10 cm from the center of the circle.
Steps of construction :
(1) First of all we draw a circle with center O and radius 6 cm.
(2) Take a point P part of 10 cm from O. Join OP.

(3) Draw a perpendicular bisector of OP that intersects OP at M.
(4) Taking M as center and radius OM = OP. draw a (RBSESolutions.com) circle which intersects the given circle at T1 and T2.
(5) Join PT1 and PT2 which are the required tangents.
Proof : We know that a tangent is the perpendicular on the point of contact.
∴ ∠OT1P = ∠OT2P = 90°
Now, Join OT1 and OT2, In the circle OT1, PT2 OP is diameter.
∴ ∠OT1P is the angle of semicircle.
∴ ∠OT1P = 90°
Similarly ∠OT2P = 90°
Hence, PT1 and PT2 are the tangents to the given circle.
On measuring, we get
PT1 = PT2 = 8.0 cm

Question 3.
Draw a circle of radius 3 cm. Take two point P and Q on its (RBSESolutions.com) diameter produced at the equal distances of 7 cm from its center. Draw two tangents to the circle from these two points.
Solution :
Given :
A circle with radius 3 cm. Two points P and Q on its diameter produced at equal distance of 7 cm from its center. i.e., OP = OQ = 7 cm
Steps of construction :
(1) Draw a circle with radius 3 cm.

(2) Draw the diameter AB and produce it to point P and Q of the either side of circle such that OP = OQ = 7 cm.
(3) Draw the perpendicular bisectors of OP and QQ which intersect them of M1 and M2 respectively.
(4) Taking M1 as centre and radius M1O draw (RBSESolutions.com) a circle which intersects the given circle at T1 and T2.
(5) Join PT1 and PT2.
(6) Taking M2 as centre and radius M2O draw another circle which intersects the given circle at S1 and S2.
(7) Join QS1 and QS2.
Hence PT1, PT2, QS1 and QS2 are the required tangents.

Question 4.
Draw two tangents to a circle of radius 5 cm such that the angle between the tangents is 60°.
Solution :
Given :
A circle with (RBSESolutions.com) center O and radius 5 cm.
Steps of construction :
(1)Taking O as center and radius 5 cm draw a circle.

(2) Draw the diameter AB of the circle.
(3) Draw a ray of O making angle of 60° with OB to meet the circumference at P.
(4) At point A, draw a ray AX making and angle of 90° with OA.
(5) Similarly, draw another tangent PY from P. AX and PI intersect each other at T.
Hence PT and AT are the required tangents which make an angle of 60° between.

Question 5.
Draw a line segment AB of length 8 cm. Taking A as (RBSESolutions.com) center and radius 4 cm draw a circle and draw a pair of tangents to this circle from point B. (Rajasthan Board 2013)
Solution :
Steps of construction :
(1) Draw a line segment AB = 8 cm.
(2) Draw a circle of radius 4 cm from A as center.
(3) Draw the perpendicular bisector of AB at M.

(4) Taking M as center and radius MA draw (RBSESolutions.com) a circle which intersects the circle center A at T1 and T2.
(5) Join BT1 and BT2 and BT1 and BT2 are the required tangents.
On measuring, we get BT1 = BT2 = 6.93 cm.

Question 6.
Draw a circle of radius 2.5 cm. Take a point a part 6.5 cm from its center and from this point draw the tangents to the circle. Measure the length of tangent also find the length of tangents by calculation.
Solution :
(1) First of all, we draw a circle with center O and radius 2.5 cm
(2) Take a point P a part 6.5 cm from the center of circle and join OP.
(3) Draw a perpendicular bisector of PO and mark the point M.
(4) Taking M as center and radius and draw a (RBSESolutions.com) circle which intersects their circle center O at A and B.

(5) Join PA and PB.
Hence PA and PB are the required tangents.
On measuring we get PA = PB = 6.0 cm.
Calculation to find the length of tangents :
In ∆POA
∠PAO = 90°
So ∆POA is a right angled (RBSESolutions.com) triangle, So by the Pythagoras Theorem,
PO2 = AO2 + PA2
⇒ PA2 = PO2 – AO2
= (6.5)2 – (2.5)2
= 42.25 – 6.25
⇒ PA2 = 36.00
∴ PA = $$\sqrt { 36 }$$ = 6 cm
Hence, the length of each tangent is 6.0 cm.

Question 7.
Draw a circle of radius 4 cm. Draw a pair of tangents to this circle such that the angle between two tangents is 60°. Also give the verification. Find the distance between the center of the circle and intersecting point of two tangents. [NCERT Exemplar Problem]
Solution :
Steps of Construction :
(1) Taking O as center and (RBSESolutions.com) radius 4 cm draw a circle.

(2) Draw the diameter AB in the circle.
(3) Draw a ray OC making an angle of 60° at O of OA to meet the circle at C.
(4) Draw a ray making an angle of 90° at C of OC.
(5) Draw a ray making an angle of 90° at B of OB to meet the first ray at D.
Hence, CD and BD are the required tangents which make an angle of 60° between each others.
On Measuring the distance and intersecting point of two tangents we get 8 cm.

Question 8.
To a circle of radius 4.5, draw two (RBSESolutions.com) tangents such that the angle between them is 45°, Also verify it [CBSE 2013]
Solution :
Steps of construction:
(1) Draw a circle with center O and radius
(2) Draw a diameter A in the circle.
(3) At point A of OA, draw a ray OC making an angle of 45° to meet the circle.

(4) Draw line perpendicular (RBSESolutions.com) to OB at B also draw a perpendicular line at C of OC, which intersects each other at P.
PB and PC are the required tangents that make an angle of 45° between them,
Proof : O is the center of circle and PB and PC are the tangents.
The angle between them ∠BPC = 45°.
∴ ∠BOC = 180° – 45° = 135°
∴ ∠AOC = 180° – ∠BOC (By pair of linear equation)
= 180° – 135° = 45°

Question 9.
Draw a tangent to a circle of radius 4 cm from a point, which is 6 cm a part from its centre. [NCERT Exempbr Problem]
Solution :
Steps of construction :
(1) Draw a circle with (RBSESolutions.com) center O and radius 4 cm.
(2) Take a point P, 6 cm a part from its center and meet OP.
(3) Bisect OP at point M.

(4) With M as center and MP as radius, draw a circle that intersects the given circle at A and B.
(5) Join PA and PB.
Hence PA and PB are the required tangents.

Question 10.
Draw a line segment AB of length 85 cm. Taking A and B as (RBSESolutions.com) centers and radii respectively 5 cm and 3 cm draw two circle. From the center of each circle draw the tangents to the other circle.
Solution :
Given : Line segment AB = 8.5
cm. Two circles with radii 5 cm and 2 cm on the two ends of AB.
Steps of construction :
(1) First of all, we draw a line segment AB = 8.5 cm
(2) Taking A as center and radius 5 cm, draw circle draw another circle of radius 2 cm at B.
(3) Bisect line segment AB at M.

(4) With M as center and MA as radius, draw a circle (RBSESolutions.com) which intersects the circle center A at S and T and the circle center B at P and Q.
(5) Join SB, TB, PA and QA.
Hence SB and TB are the tangents to the circle center A from B also PA and QA are the tangents to the circle center B from A.

Question 11.
Draw a circumcircle of an equilateral triangle with side 5 cm. Also write the steps of construction.
Solution :
Given :
Equilateral ∆ABC whose each side is 5 cm in length.
Steps of construction :
(1) Draw a line segment BC = 5 cm.
(2) Taking B as center and (RBSESolutions.com) radius 5 cm draw an arc.

(3) Taking C as center and radius 5 cm draw another arc which intersects the first arc at A.
(4) Join AB and AC. ∠ABC is an equilateral triangle.
(5) Draw the perpendicular bisectors of AC and AB which intersect each other at O. O is the circumcentre of ∆ABC. Join O to B.
(6) Taking O as center and radius equal to OA draw a circle which passes through three vertices of ∆ABC.

Question 12.
Draw a right ∆ABC in (RBSESolutions.com) which AB = 4.5 cm, AC = 7.5 cm and ∠B = 90°. Draw the incircle of is A BC. Also write the steps of construction.
Solution :
Given :
∆ABC in which AB = 45 cm, AC = 7.5 cm and ∠P = 90°
Steps of construction :
(1) Draw a line segment AB = 45 cm
(2) Construct ∠ABX = 90°.
(3) Taking A as center and radius 7.5 draw an arc which intersects BX to C.
(4) Join AC ∆ABC is required triangle.
(5) Draw the bisectors of ∠A and ∠B which intersects each other at I.

(6) From I draw IP ⊥ AB.
(7) Taking I as center and IP as (RBSESolutions.com) radius draw the circle which touches the three sides of ∆ABC.
Hence this is the required in circle of ∆ABC.

Question 13.
Construct an incircle of an equilateral triangle whose one side is 4 cm in length.
Solution :
Given:
In ∆ABC, AB = BC = CA = 4 cm
Steps of construction:
(1) Construct ∆ABC with its one side 4 cm.
(2) Draw the bisectors of ∠ABC and ∠ACB which intersect each other O.

(3) Draw BC ⊥ OD from O.
(4) Taking O as center and OD as (RBSESolutions.com) radius draw a circle which touches three sides of ABC.
Hence this is the required m circle of ∆ABC.

Question 14.
The distance between the centers of two circles with radii 3.2 cm and 1.5 cm respectively Is 6.2 cm. Draw the direct common tangents these circles. Measure these tangents and check your answer by calculation.
Solution :
Steps of construction :
(1) First of all, we draw a line segment OO’ = 6.2 cm.
(2) Taking O as center and radius 3.2 cm, draw a circle.
(3) Taking O’ as center and radius 1.5 cm draw another circle.

(4) Taking OO’ as (RBSESolutions.com) diameter, draw a circle.
(5) Taking center, the center of the larger circle and radius equal to the difference of the radii of two circle. i.e., 3.2 – 15 = 17, draw another circle which intersects circle with diameter OO’ at A and B.
(6) Join OA and OB which intersects the larger circle when produced at A and B.
(7) From O’ the center of smaller circle draw O’Q || OP and O’Q || OP’.
(8) Join PQ, P’Q’, PQ and P’Q’are the required direct common tangents.
On measuring : we get.
PQ = P’Q’ = 6 cm
Calculation : PQ

Question 15.
The distance between the (RBSESolutions.com) centers of two circles with radii 2.0 cm and 4.0 cm is 7 cm. Draw their direct (transversal common) tangents of these circles. Measure these tangents and check your answer by calculation.
Solution :
Steps. of construction :
(1) Draw a line segment OO’ equal to the distance between two centers = 7 cm.
(2) Taking O as center and radius 2 cm draw a circle.
(3) Taking O’ as center and radius 4 cm, draw another circle.

(4) Taking OO’ as diameter draw another circle.
(5) Taking as center O the (RBSESolutions.com) center of smaller circle and the radius equal to the sum of two radii (i.e., 2 + 4 = 6) cm draw a circle which intrersects the circle diameter OO’ at A and B.
(6) Join OA and OB which intersects the smaller circle at P and P’.
(7) From the point O’ the centre of larger circle draw Q’Q || QA and O’Q’ || OB
(8) Join PQ and P’Q’. PQ and P’Q’ are the required tangents.
On measuring PQ = P’Q’ = 3.6 cm
Calculation : PQ

Question 16.
The radii of two circle are respectively 2.0 cm and 3.5 cm and the distance (RBSESolutions.com) between their centers is 7 cm. Draw direct common tangents of two circles.
Solution :
Steps of construction :
(1) Draw a line segment OO’ = 7 cm equal to distance between their centers.
(2) Taking O as center and radius 2.0 cm draw a circle.
(3) With O’ as center and radius of 3.5 cm draw another circle.
(4) With diameter OO’ draw another circle.
(5) With center ‘O’ the center of larger circle and with radius equal to difference of their radii i.e., r1 – r2 = 3.5 – 2.0 = 1.5 cm draw a circle, which intersects the circle OO’ at points A and B.

(6) Join O’A and O’B and (RBSESolutions.com) produce to touch the larger circle at point Q and Q’.
(7) From point O the center of smaller circle draw O’Q || OP and O’Q’ || OP’
(8) Join Q and P’Q’. which are the required tangents.

Question 17.
Construct a ∆ABC in which BC = 5 cm, ∠B = 90° and side CA = 7 cm. Draw a incircle of this triangle.
Solution:
Given :
In ∆ABC, BC =5 cm, ∠B = 90°, CA = 7 cm.
To contract: In circle of ∆ABC.
Steps of construction:
(1) First of all draw line segment BC = 5 cm.
(2) Draw a ray BX making an angle of 90° at B of BC.
(3) Taking C as (RBSESolutions.com) center and radius 7 cm draw an arc which intersects BX at A. Join A to C.
It is the ∆ABC given.
(4) Draw the perpendicular bisectors of AB and AC, which intersect each other at O. Join O to A.
(5) Taking O as center and OA as radius draw a circle which passes through three vertices. It is the required circle.

Question 18.
Construct a ∆ABC In which AB = 5 cm, BC = 6 cm and ∠B = 60°. Construct (RBSESolutions.com) an incircle of ∆ABC.
Solution :
Given :
In ∆ABC in which AB = 5 cm, BC = 6 cm and ∠B = 60°.
To construct: Incircle ∆ABC.
Steps of construction:
(1) Draw a line segment BC = 60 cm.
(2) Draw a ray BX making an angle of 60° at B of BC. Cut BA = 5 cm.

(3) Join A to C.
∆ABC is the given triangle.
(4) Draw the (RBSESolutions.com) bisectors of ∠B and ∠C which intersect each other at O.
(5) Draw OD ⊥ BC from O.
(6) Taking O as center and OD as radius draw a circle which touches three sides of ∆ABC.
This is the required incircle.

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