• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • RBSE Model Papers
    • RBSE Class 12th Board Model Papers 2022
    • RBSE Class 10th Board Model Papers 2022
    • RBSE Class 8th Board Model Papers 2022
    • RBSE Class 5th Board Model Papers 2022
  • RBSE Books
  • RBSE Solutions for Class 10
    • RBSE Solutions for Class 10 Maths
    • RBSE Solutions for Class 10 Science
    • RBSE Solutions for Class 10 Social Science
    • RBSE Solutions for Class 10 English First Flight & Footprints without Feet
    • RBSE Solutions for Class 10 Hindi
    • RBSE Solutions for Class 10 Sanskrit
    • RBSE Solutions for Class 10 Rajasthan Adhyayan
    • RBSE Solutions for Class 10 Physical Education
  • RBSE Solutions for Class 9
    • RBSE Solutions for Class 9 Maths
    • RBSE Solutions for Class 9 Science
    • RBSE Solutions for Class 9 Social Science
    • RBSE Solutions for Class 9 English
    • RBSE Solutions for Class 9 Hindi
    • RBSE Solutions for Class 9 Sanskrit
    • RBSE Solutions for Class 9 Rajasthan Adhyayan
    • RBSE Solutions for Class 9 Physical Education
    • RBSE Solutions for Class 9 Information Technology
  • RBSE Solutions for Class 8
    • RBSE Solutions for Class 8 Maths
    • RBSE Solutions for Class 8 Science
    • RBSE Solutions for Class 8 Social Science
    • RBSE Solutions for Class 8 English
    • RBSE Solutions for Class 8 Hindi
    • RBSE Solutions for Class 8 Sanskrit
    • RBSE Solutions

RBSE Solutions

Rajasthan Board Textbook Solutions for Class 5, 6, 7, 8, 9, 10, 11 and 12

  • RBSE Solutions for Class 7
    • RBSE Solutions for Class 7 Maths
    • RBSE Solutions for Class 7 Science
    • RBSE Solutions for Class 7 Social Science
    • RBSE Solutions for Class 7 English
    • RBSE Solutions for Class 7 Hindi
    • RBSE Solutions for Class 7 Sanskrit
  • RBSE Solutions for Class 6
    • RBSE Solutions for Class 6 Maths
    • RBSE Solutions for Class 6 Science
    • RBSE Solutions for Class 6 Social Science
    • RBSE Solutions for Class 6 English
    • RBSE Solutions for Class 6 Hindi
    • RBSE Solutions for Class 6 Sanskrit
  • RBSE Solutions for Class 5
    • RBSE Solutions for Class 5 Maths
    • RBSE Solutions for Class 5 Environmental Studies
    • RBSE Solutions for Class 5 English
    • RBSE Solutions for Class 5 Hindi
  • RBSE Solutions Class 12
    • RBSE Solutions for Class 12 Maths
    • RBSE Solutions for Class 12 Physics
    • RBSE Solutions for Class 12 Chemistry
    • RBSE Solutions for Class 12 Biology
    • RBSE Solutions for Class 12 English
    • RBSE Solutions for Class 12 Hindi
    • RBSE Solutions for Class 12 Sanskrit
  • RBSE Class 11

RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions

April 22, 2019 by Fazal Leave a Comment

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

Rajasthan Board RBSE Class 10 Maths Chapter 10 Locus Additional Questions

Multiple Choice Questions
Question 1.
Intersection point of three (RBSESolutions.com) altitudes of a triangles is called:
(A) Circumcentre
(B) incentre
(C) Orthocentre
(D) Centroid
Solution :
(C) is correct.

Question 2.
The centroid of triangle divides each median in the ratio :
(A) 2 : 1
(B) 1 : 2
(C) 2 : 3
(D) 3 : 2
Solution :
(A) is correct.

RBSE Solutions

Question 3.
In the following criterian, in which (RBSESolutions.com) criterion two triangles will not be congruent?
(A) All corresponding sides are equal
(B) All corresponding angles are equal
(C) Two corresponding sides and angle between them are equal.
(D) All corresponding angles and one corresponding side are equal.
Solution :
(B) is correct.

Question 4.
What will be the locus of mid-points of equal chords of a circle?
(A) Circle
(B) Chord
(C) Distance of center from chord
(D) Center point
Solution :
(A) is correct.

Question 5.
The locus of point.s located at (RBSESolutions.com) equal distance from two fixed points is called :
(A) line segment
(B) circle
(C) ⊥ bisector of line segment
(D) two points
Solution :
(C) is correct.

Question 6.
The locus of point in space can be imagined:
(A) sphere
(B) cuboid
(C) cylinder
(D) cone
Solution :
(A) is correct.

Question 7.
Locus of swinging in (RBSESolutions.com) swing wheel will be :
(A) ⊥ bisector
(B) line segment
(C) semicircle
(D) circle
Solution :
(D) is correct.

Question 8.
In figure, AB = AC and ∠ABD = ∠ACD, then ΔBDC will be :
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 1
(A) Equilateral triangle
(B) Isosceles triangle
(C) Equiangular triangle
(D) Scalene triangle
Solution :
From figure,
AB = AC (Given)
∠ABC = ∠ACB
and ∠ABD = ∠ACD (Given) …(ii)
Adding equations (i) and (ii),
∠ABC + ∠ABD = ∠ACB + ∠ACD
∠DBC = ∠DCB
BD = DC (Opposite sides of equal angles)
∴ ΔBDC will be an isosceles triangle.
Hence (B) is correct.

RBSE Solutions

Question 9.
The intersecting point of (RBSESolutions.com) medians of triangle is called :
(A) Centroid
(B) Orthocentre
(C) Circumcentre
(D) Incentre
Solution :
(A) is correct.

Question 10.
Distance of circumcentre from vertices of triangle will be :
(A) equal
(B) different
(C) zero
(D) different, equal
Solution :
(A) is correct.

Long Answer Type Questions

Question 1.
Find the locus of a point which (RBSESolutions.com) re-mains at a distance of 5 cm from line AB.
Solution :
Locus of a point which remains at a distance of 5 cm from line AB are two parallel lines on both sides of AB at a distance of 5 cm.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 2

Question 2.
Find the locus of the center of circle, passing through three non-collinear points A, R and C.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 3
Solution :
Required locus will be point of intersection of perpendicular bisectors of sides AB and BC.

Question 3.
The straight roads intersects each (RBSESolutions.com) other at O. A testing center have to formed such that its distance from O be 1 km and equidistant from two roads, Show the possible cases of testing center by figure.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 4
Solution :
Possible cases are four points P, Q, R and S. In which two points P and Q lie on bisector of ∠AOC and two points R and s on bisector of ∠BOC.
Thus intersecting point O lie at equidistant (RBSESolutions.com) from the roads and on the bisector of ∠AOC or ∠BOC.

Question 4.
In figure, BF = FC, ∠BAE – ∠CAE and ∠ADE – ∠GEC = 90°.
Then name one altitude, the median one angle bisector and one perpendicular bisector of side of triangle.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 5
Solution :
Altitude = AD
Median = AF
Angle bisector = AE
perpendicular bisector = GE.

RBSE Solutions

Question 5.
What will be the locus at equal distance (RBSESolutions.com) from two parallel lines in a plane.
Solution :
Two parallel lines are m and n. P is any point equidistant from there lines.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 6
From figure, required locus will be two parallel

Question 6.
Find the locus of the point in the interior of triangle (RBSESolutions.com) which is equidistant from three sides of triangle.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 7
Solution :
We know that point equidistant from three sides of of triangle is incentre triangle.

Question 7.
Find locus of that point which lie at equidistant from two (RBSESolutions.com) intersecting lines and at a distance of 3 cm from their point of intersecting.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 8
Solution :
Let AB and CD are two intersecting lines and their point of intersection is equidistant from these lines and P is any point at a distance 3 cm from O.
Thus locus of P will be line at a distance of 3 cm from point of intersection.

Question 8.
In figure, ∆ABC and ∆DBC are two isosceles triangles. (RBSESolutions.com) Prove that line segment AD lie on perpendicular bisector of base BC.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 9
Solution :
Given : Two isosceles triangles ABC and DBC having same base BC.
AB = AC and DB = DC
To Prove : AD lies on perpendicular (RBSESolutions.com) bisector of base BC.
Construction : Produce AD upto point M on BC.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 10
Proof : ∵ ∆DBC is isosceles triangle.
∴ DB = DC
Now ∠DBC = ∠DCB
In ∆DBM and ∆DCM,
DB = DC (Given)
∠DBM = ∠DCM, (In an isosceles triangle angles (RBSESolutions.com) opposite to equal sides are equal).
DM = DM (Common)
∵ ∆DBM ≅ ∆DCM
∴ ∠BMD = ∠CMD and BM = CM
Now ∠BMD + ∠CMD = 180°
∠BMD = ∠CMD = 90°
∴ DM ⊥ BC
∴ M is mid-point of BC.
∴ DM is perpendicular bisector of BC.
Now point D lies on AM
Similarly, ∴ ∆ABC is an isosceles triangle.
AM is perpendicular bisector of BC.
Thus, line segment AD, lie on ⊥ bisector of base BC.

Question 9.
Prove that sum of three altitudes of (RBSESolutions.com) triangle is less than its perimeter.
Solution :
Given : ∆ABC, in which from A, B, and C perpendicular AD, BE and CF are drawn on BC, AC and AB respectively.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 11
To Prove : AB + BC + CA > AD + BE + CF
Proof : ln ∆ADB,
∠ADB = 90°
Thus AB > AD …..(i)
Similarly, in ∆BEC,
BE ⊥ AC
Thus BC > BE …(ii)
In ∆CAF,
CF ⊥ AB
Thus AC > CF …(iii)
Adding equation (i), (ii) and (iii)
AB + BC + CA > AD + BE + CF

RBSE Solutions

Question 10.
Prove that median of a triangle divides (RBSESolutions.com) it Into two equal triangles of similar area.
Solution :
Given : ∆ABC, in which AD is median of side BC.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 12
To prove : ar ∆ABD = ar. ∆ADC
Construction : Draw ⊥ AM on side BC.
Proof : AD is median in ∆ABC.
BD = CD
Triangles ABD and ADC will be on the (RBSESolutions.com) same base BC (CD) and same vertex A. So having same height AM.
∴ Area of ∆ABD = \(\frac { 1 }{ 2 }\) BD × AM
Area of ∆ACD = \(\frac { 1 }{ 2 }\) CD × AM
But BD = CD [From equation (i)]
∴ ar. (∆ABD) = ar. (∆ACD)

Question 11.
In given figure D, E and F are midpoint of sides BC, CA and AB respectively. If AB = 4.3 cm, BC = 5.6 cm and AC = 3.9 cm then find perimeter of the
following:
(i) ∆DEF and
(ii) quadrilateral BDEF.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 13
Solution :
We know that line segment joining (RBSESolutions.com) the mid-points of two sides ¡s half the third side.
Thus in ∆DEF,
DE = \(\frac { 1 }{ 2 }\) AB
EF = \(\frac { 1 }{ 2 }\) BC
DF = \(\frac { 1 }{ 2 }\) AC
∴ Perimeter of ∆DEF = DE + EF + DF
= \(\frac { 1 }{ 2 }\) AB + \(\frac { 1 }{ 2 }\) BC + \(\frac { 1 }{ 2 }\) AC
= \(\frac { 1 }{ 2 }\) (AB + BC + AC)
= \(\frac { 1 }{ 2 }\) (4.3 + 5.6 + 3.9)
= \(\frac { 1 }{ 2 }\) (13 . 8)
= 6.9 cm
(ii) Perimeter of quadrilateral BDEF = BD + DE + EF + 8F
= \(\frac { 1 }{ 2 }\) BC + \(\frac { 1 }{ 2 }\)AB + \(\frac { 1 }{ 2 }\) BC + \(\frac { 1 }{ 2 }\)AB
= AB + BC
= 4.3 + 5.6
= 9.9 cm

Question 12.
Sides of equilateral (RBSESolutions.com) triangle ABC is 12 cm each If G is its centroid, then find AG.
Solution :
AB = AC = BC = 12 cm
∴ BD = \(\frac { BC }{ 2 }\) = \(\frac { 12 }{ 2 }\) = 6 cm
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 14
In right angled triangle ABD,
AB2 = BD2 + AD2
AD2 = AB2 – BD2
= 122 – 62
= 144 – 36
= 108
AD = \(\sqrt { 108 }\)
AD = 6\(\sqrt { 3 }\)
We know that
AG = \(\frac { 2 }{ 3 }\) AD
= \(\frac { 12 }{ 2 }\) × 6\(\sqrt { 3 }\) = 4\(\sqrt { 3 }\)

Question 13.
Show that, in an equilateral triangle, angle (RBSESolutions.com) bisectors of three angles, are three altitudes, three medians and perpendicular bisectors of three sides.
Solution :
AB = AC
In ΔABC AD bisector of bisector of ∠A ⊥ bisector of side, a median and altitude similarly, AB = BC and BC = AC.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 15
∴ In ∆ABC, ∠B and ∠C bisectors BE and CE are altitudes medians and perpendicular bisectors of sides of triangle.

Question 14.
In an Isosceles triangle show that bisector of (RBSESolutions.com) angle formed between two similar sides is also a altitude, a median and a perpendicular bisector of side of that triangle.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 16
Solution :
In ∆ABD and ACD,
AB = AC
∠BAD = ∠CAD [∵ AD, is bisector of ∠A]
and AD = AD
∴ ∆ABD ≅ ∆ACD
∴ BD = CD
AD is also a median
and ∠ADB = ∠ADC = 90°
Thus AD is perpendicular bisector of BC.

RBSE Solutions

Question 15.
Prove that medians of isosceles triangle, which (RBSESolutions.com) bisects the similar sides of triangle are also equal.
Solution :
∆ABC is an isosceles triangle in which AB = AC. CD is median of AB.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 17
∴ BD = AD
BE is median on AC
CE = AE
AB = AC
\(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) AC
BD = CE
In ∆DBC and ∆BCE,
ED = EC [From above)
∠B = ∠C [∵ AB = AC]
BC = BC (Common side)
By SAS rule,
∆BDC ≅ ∆BCE (By CPCT)
CD = BE
Similarly, other medians will be same.

Question 16.
BE and CF are medians of right (RBSESolutions.com) triangle ABC and ∠A is right angle. Prove that : 4(BE2 + CF2) = 5BC2. [M.S.B. Raj 2012, 14]
Solution :
Given : BE and CF are medians of right angled triangle ABC,
∠A = 90°.
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 18
To prove : 4(BE2 + CF2) = 5BC2.
Proof : In right angled ∆ABC
BC2 = AB2 + AC2 ……(i) (By Pythagoras theorem)
In right ∆ABE
BE2 = AE2 + AB2
RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions 19
Adding (RBSESolutions.com) eqn. (ii) and (iii)
4(8E2 + CF2) = AC2 + 4AB2 + 4AC2 + AB2
4(BE2 + CF2) = 5(AC2 + AB2)
4(BE2 + CF2) = 5BC2 [Using eqn. (i)]

RBSE Solutions

We hope the given RBSE Solutions for Class 10 Maths Chapter 10 Locus Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 10 Locus Additional Questions, drop a comment below and we will get back to you at the earliest.

Share this:

  • Click to share on WhatsApp (Opens in new window)
  • Click to share on Twitter (Opens in new window)
  • Click to share on Facebook (Opens in new window)

Related

Filed Under: Class 10

Reader Interactions

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

Recent Posts

  • RBSE Solutions for Class 7 Our Rajasthan in Hindi Medium & English Medium
  • RBSE Solutions for Class 6 Our Rajasthan in Hindi Medium & English Medium
  • RBSE Solutions for Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise
  • RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions
  • RBSE Solutions for Class 11 Psychology in Hindi Medium & English Medium
  • RBSE Solutions for Class 11 Geography in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 Hindi
  • RBSE Solutions for Class 3 English Let’s Learn English
  • RBSE Solutions for Class 3 EVS पर्यावरण अध्ययन अपना परिवेश in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 Maths in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 in Hindi Medium & English Medium

Footer

RBSE Solutions for Class 12
RBSE Solutions for Class 11
RBSE Solutions for Class 10
RBSE Solutions for Class 9
RBSE Solutions for Class 8
RBSE Solutions for Class 7
RBSE Solutions for Class 6
RBSE Solutions for Class 5
RBSE Solutions for Class 12 Maths
RBSE Solutions for Class 11 Maths
RBSE Solutions for Class 10 Maths
RBSE Solutions for Class 9 Maths
RBSE Solutions for Class 8 Maths
RBSE Solutions for Class 7 Maths
RBSE Solutions for Class 6 Maths
RBSE Solutions for Class 5 Maths
RBSE Class 11 Political Science Notes
RBSE Class 11 Geography Notes
RBSE Class 11 History Notes

Copyright © 2023 RBSE Solutions

 

Loading Comments...