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.

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 :

(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.

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.

Question 2.

Find the locus of the center of circle, passing through three non-collinear points A, R and C.

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.

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.

Solution :

Altitude = AD

Median = AF

Angle bisector = AE

perpendicular bisector = GE.

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.

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.

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.

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.

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.

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.

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

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.

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.

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

In right angled triangle ABD,

AB^{2} = BD^{2} + AD^{2}

AD^{2} = AB^{2} – BD^{2}

= 12^{2} – 6^{2}

= 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.

∴ 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.

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.

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.

∴ 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(BE^{2} + CF^{2}) = 5BC^{2}. [M.S.B. Raj 2012, 14]

Solution :

Given : BE and CF are medians of right angled triangle ABC,

∠A = 90°.

To prove : 4(BE^{2} + CF^{2}) = 5BC^{2}.

Proof : In right angled ∆ABC

BC^{2} = AB^{2} + AC^{2} ……(i) (By Pythagoras theorem)

In right ∆ABE

BE^{2} = AE^{2} + AB^{2}

Adding (RBSESolutions.com) eqn. (ii) and (iii)

4(8E^{2} + CF^{2}) = AC^{2} + 4AB^{2} + 4AC^{2} + AB^{2}

4(BE^{2} + CF^{2}) = 5(AC^{2} + AB^{2})

4(BE^{2} + CF^{2}) = 5BC^{2} [Using eqn. (i)]

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