RBSE Solutions for Class 10 Maths Chapter 10 Locus Ex 10.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 10 Locus Exercise 10.2.

## Rajasthan Board RBSE Class 10 Maths Chapter 10 Locus Ex 10.2

Question 1.

Find the locus of the point equidistant from (RBSESolutions.com) three vertices and three sides of triangle.

Solution :

(i) Locus of point equidistant from three vertices :

Let in ΔABC, O is a moving point which is equidistant from three vertices A, B, and C.

∴ O is equidistant from A and B.

It means O is perpendicular bisector of AB

Again, O is equidistant from A and C. It means O is perpendicular bisector of AC.

So O is perpendicular bisector of AC.

So, O is intersecting point of perpendicular bisectors of AB and AC. Thus O is center of circle, passes through three vertices. This circle passes through three vertices of triangle and we called it circumcenter of circle.

Hence required locus will be circumcenter of the circle.

(ii) Locus of a point equidistant from (RBSESolutions.com) three sides : Let O be equidistant from path P, Q, and R of sides BC, AC and AB respectively.

∴ O is equidistant from P and Q.

It means O is t bisector of PR.

Again O is equidistant from point R It means O is ⊥ bisector of PR.

So, O is intersecting point of PQ and PR.

Thus, O is centre of a circle touching three sides of triangle.

It means O is incentre.

Hence, required locus is incentre of circle.

Question 2.

In ∆ABC, medians AD, BE and CF intersects at (RBSESolutions.com) point G. If AG = 6 cm, BE = 9 cm and GF = 4.5 cm, then find GD, 8G and CF.

Solution :

We know that centroid of triangle divides its medians in the ratio 2 : 1.

⇒ [latex]\frac { CF }{ GF }[/latex] = [latex]\frac { 3 }{ 1 }[/latex]

⇒ [latex]\frac { CF }{ 4.5 }[/latex] = [latex]\frac { 3 }{ 1 }[/latex]

⇒ CF = 3 × 4.5 = 13.5 cm

Question 3.

In ∆ABC, medians AD, BE, CF intersect at (RBSESolutions.com) point G. Prove that AD + BE > [latex]\frac { 3 }{ 2 }[/latex]AB. [Hint : AG + BG > AB]

Solution :

We know that sum of two sides of a triangle is greater than third side.

So In ∆ABG,

AG + BG > AB

We know that centroid of triangle (RBSESolutions.com) divides median in the ratio 2 : 1.

Question 4.

Prove that sum of two medians of a (RBSESolutions.com) triangle is more than third median.

Solution :

Given : In ∆ABC, medians AD, BE and CF intersect at point G.

To Prove :

AD + BE > CF

BE + CF > AD

and AD + CF > BE

Construction : Produce AD upto K when AG = GK

Join KB and KC.

Proof : In ∆ABK.

F, in mid-point of AB (Given)

G is mid point of AK (By construction)

∴ FG || BK [∵ Line joining the mid-point of (RBSESolutions.com) two lines of triangle is parallel to third side]

⇒ GC || BK …..(i)

E is mid-point of AC (Given)

G is mid-point of AK (By construction)

∴ GE || KC

⇒ BG || KC …(ii)

Thus, by quadrilateral BKCG

GC || BK (From eqn. (i))

and BG || KC (From eqn. (ii))

Now quadrilateral BKCG is a parallelogram

⇒ BK = CG …(iii)

Since sum of any two sides of triangle is (RBSESolutions.com) greater than third side.

Now, in ∆BKG, BG + GK > BK

⇒ BG + AG > CG [AG = GK by construction]

Thus, BK = CG from eqn. (iii)]

⇒ [latex]\frac { 2 }{ 3 }[/latex]BE + [latex]\frac { 2 }{ 3 }[/latex]AD > [latex]\frac { 2 }{ 3 }[/latex]CF

[∵ AG = [latex]\frac { 2 }{ 3 }[/latex] AD, BG = [latex]\frac { 2 }{ 3 }[/latex]BE and CG = [latex]\frac { 2 }{ 3 }[/latex]CF]

BE + AD > CF

BE + AD > CF

Similarly BE + CF >AD

and AD + CF > BE Hence proved.

Question 5.

In a triangle ABC, median AD, BE and CF intersect (RBSESolutions.com) at point G. Prove that :

4(AD + BE + CF) > 3(AB + BC + CA)

Solution :

Given : In ∆ABC, AD, BE, CF medians intersect at point G. We know that sum of any two sides of triangle is more than third side.

Now, in ∆BGC + BG + GC > BC

⇒ [latex]\frac { 2 }{ 3 }[/latex] BE + [latex]\frac { 2 }{ 3 }[/latex] CF > BC

Question 6.

P is orthocentre of ∆ABC. Prove (RBSESolutions.com) that A is the orthocentre of ∆ABC.

Solution :

Given :

Point P is orthocentre of ∆ABC.

To Prove : A is othocentre of triangle PBC.

Proof : Let AP, BP and CP are extended such (RBSESolutions.com) that they intersect BC, AC and AB at D, E and F respectively,

Then AD ⊥ BC, BE ⊥ AC, CF ⊥ AB

⇒ AD ⊥ BC, AB ⊥ CP, AC ⊥ BP

Hence, A is the orthocentre of ∆ABC

Question 7.

In ∆ABC, medians AD, BE and CF passes through G.

(a) If GF = 4 cm, then find GC.

(b) If AD = 7.5 cm, then find GD.

Solution :

We know that centroid of triangle divides its median into 2 : 1.

Question 8.

∆ABC is isosceles triangle in which AB = AC, D is mid-point (RBSESolutions.com) of BC. Prove that circumcentre orthocentre, incentre and centroid all lie on line AD.

Solution :

For cirumcentre fo a triangle It lies on the perpendicular bisector, so we have to show that AD is perpendicular bisector of the side BC.

Since ∆ABD ≅ ∆ACD (by SSS)

⇒ ∠ADB = ∠ADC

= 90° [∠ADB + ∠ADC = 180°]

⇒ AD ⊥ BC

Now, AD ⊥ BC and BD = CD

⇒ AD is perpendicular bisector of BC

Hence, Cirumcentre lies on AD

For incentre of a triangle

It lies on the angle bisector, (RBSESolutions.com) so we have to shown that AD is bisector of ∠BAC.

Since ∆ABD ≅ ∆ACD [by SSS]

⇒ ∠BAD = ∠CAD (CPCT)

⇒ AD is bisector of angle BAC

Hence, incentre lies on AD.

For orthocentre of a triangle

It lies on the altitude, so we have to show that AD is altitude corresponding to side BC’.

Since ∆ABD ≅ ∆ACD [by SSS]

⇒ ∠ADB = ∠ADC

= 90° [∠ADD + ∠ADC = 180°]

AD is perpendicular to BC’. i.e.. AD is altitude corresponding to side BC.

Hence, orthocentre lies on AD.

For centroid of a triangle

It lies on the medians so we have to (RBSESolutions.com) prove that AD is median corresponding to side BC.

Since, it is given that D is the mid-point of BC. AD is the median of triangle

Hence, centroid lies on AD Hence proved.

Question 9.

H is orthocentre of ∆ABC. X, Y and Z are mid-points in AH, BH and CH. Prove that orthocentre of ∆XYZ is also H.

Solution :

In ∆ABC, perpendiculars AM, BN and CO are drawn from A, B and C respectively which intersects at point H.

Join point H to vertices of (RBSESolutions.com) triangle A, B, and C. X, Y, Z are mid points of AH, BH and CH.

In ∆HBC

BC || YZ

From vertex X, XP is altitude on YZ.

in ∆AHC,

AC || XZ

From vertex Y, YQ is altitude on ZX.

Similarly in ∆ABH,

AB || XY

From vertex Z, ZR is altitude on XY.

These three altitudes intersecting at H.

Thus, orthocentre of XYZ will be point H.

Question 10.

How we will find the point in side BC of ∆ABC which is (RBSESolutions.com) equidistant from sides AB and AC?

Solution :

In ∆ABC, draw AX, bisector of ∠A which cuts BC at D.

Take any point P on AX. Draw ⊥ PN and PM (RBSESolutions.com) from P to AB and P to AC respectively.

PN ⊥ AB and PM ⊥ AC

From ∆APN and ∆APM

∠PNA = ∠PMA = 90° (By construction)

∠PAN = ∠PAM (AX is bisector of ∠A)

AP = AP (Common)

by A.A.S. rule

∆APN ≅ ∆APM

⇒ PN = PM (By CPCT)

Thus P is equidistant from AB and AC.

∴ Any point on AX will be equidistant from AB and AC.

Therefore point D on line BC, will (RBSESolutions.com) be equidistant from AB and AC.

We hope the given RBSE Solutions for Class 10 Maths Chapter 10 Locus Ex 10.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 10 Locus Exercise 10.2, drop a comment below and we will get back to you at the earliest.

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