10th Maths 6.2

Exercise 6.2

Q1. In the figure i) and ii)

$DE∥BC.$ Find the

$EC$ in (i) and

$AD$ in (ii).

Sol.

i) Given

$DE∥BC$

$(AD)/(DE)=(AE)/(EC)$

$(1.5)/3=1/(EC)$

$EC=3/(1.5)=2$ cm.

Answer :

$EC=2$ cm.

ii) Given

$DE∥BC$

$(AD)/(DE)=(AE)/(EC)$

$(AD)/(7.2)=(1.8)/(5.4)$

$AD=(1.8)/(5.4)times7.2$

$AD=2.4$ cm.

Answer :

$AD=2.4$ cm.

Q2.

$E$ and

$F$ are points on the sides

$PO$ and

$PR$ respectively of a

$triangle PQR.$ For each of the following cases, state whether

$EF∥QR$ :

$PE=3.9$ cm,

$EQ=3$ cm,

$PF=3.6$ cm and

$FR=2.4$ cm

ii)

$PE=4$ cm,

$QE=4.5$ cm,

$PF=8$ cm, and

$RF=9$ cm

iii)

$PQ=1.28$ cm,

$PR=2.56$ cm,

$PE=0.18$ cm, and

$PF=0.36$ cm.

Sol.

$trianglePQR$

Points

$E$ and

$F$ are on sides

$PQ$ and

$PR$

$(PE)/(EQ)=(3.9)/3$

$(PE)/(EQ)=1.3$

and

$(PF)/(FR)=(3.6)/(2.4)$

$(PF)/(FR)=1.5$

$(PE)/(EQ)ne(PF)/(FR)$

$EF∦QR$

Answer :

$EF∦QR$ .

ii)

$(PE)/(EQ)=(4)/(4.5)$

$(PE)/(EQ)=(8)/(9)$

and

$(PF)/(FR)=(8)/(9)$

$(PE)/(EQ)=(PF)/(FR)$

$EF∥QR$

Answer :

$EF∥QR$ .

iii)

$(PE)/(PQ)=(0.18)/(1.28)$

$(PE)/(PQ)=(18)/(128)=9/(64)$

and

$(PF)/(PR)=(0.36)/(2.56)$

$(PF)/(PR)=(36)/(256)=(9)/(64)$

$(PE)/(PQ)=(PF)/(PR)$

$EF∥QR$

Answer :

$EF∥QR$

Q3. In Figure, if

$LM∥CB$ and

$LN∥CD$ , prove that

$(AM)/(AB)=(AN)/(AD)$

Sol. :

$triangleABC$

Given

$LM∥CB$

$(AM)/(AB)=(AL)/(AC)$ ______________(1)

and in

$triangleACD,$

Given

$LN∥CD$

$(AN)/(AD)=(AL)/(AC)$ ______________(2)

from equation (1) & (2)

$(AM)/(AB)=(AN)/(AD)$

Proved.

Q4. In Figure,

$DE∥AC$ and

$DF∥AE.$ Prove that

$(BF)/(FE)=(BE)/(EC)$

Sol. :

$triangleBAC$

Given

$DE∥AC$

$(BE)/(EC)=(BD)/(DA)$ ____________(1)

and in

$triangleBAE$

Given

$DF∥AE$

$(BE)/(FE)=(BD)/(DA)$ _____________(2)

from equation (1) & (2)

$(BF)/(FE)=(BE)/(EC)$

Proved.

Q5. In Figure,

$DE∥OQ$ and

$DF∥OR.$ Show that

$EF∥QR.$

Sol. :

$trianglePQO$

Given

$DE∥OQ$

$(PE)/(EQ)=(PD)/(DO)$ ____________(1)

and in

$trianglePOR$

Given

$DF∥OR$

$(PF)/(FR)=(PD)/(DO)$ ____________(2)

from equation (1) and (2)

$(PE)/(EQ)=(PF)/(FR)$

So that in

$trianglePQR$

$EF$ divides

$PQ$ and

$PR$ the two sides in same ratio.

then

$EF∥QR$

Proved.

Q6. In Figure,

$A, B,$ and

$C$ are points on

$OP, OQ,$ and

$OR$ respectively such that

$AB∥PQ$ and

$AC∥PR.$ Show that

$BC∥QR.$

Sol. :

$triangleOPQ$

Given

$AB∥PQ$

$(OB)/(BQ)=(OA)/(AP)$ ____________(1)

and in

$triangleOPR$

Given

$AC∥PR$

$(OC)/(CR)=(OA)/(AP)$ ____________(2)

from equation (1) and (2)

$(OB)/(BQ)=(OC)/(CR)$

So that in

$triangleOQR$

$BC$ divides

$OQ$ and

$OR$ the two sides in same ratio.

then

$BC∥QR$

Proved.

Q7. Using Theorem

$6.1$ , prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. ( Recall that you have proved it in Class IX ).

Sol. :

Let

$trianglePQR$ which have a line

$ST∥QR$ is drawn from the point

$S$ , a mid-point of side

$PQ$ .

$ST$ intersect the side

$PR$ at point

$T$ .

Given

$S$ is the mid-point of

$PQ$ .

$PS=SQ$

$(PS)/(SQ)=1$ ____________(1)

and in the

$trianglePQR, ST∥QR$

$(PT)/(TR)=(PS)/(SQ)$ _____(2) (By theroem

$6.1$ )

[Theorem $6.1$ : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.]

from equation (1) & (2)

$(PT)/(TR)=(PS)/(SQ)$

$(PT)/(TR)=1$

$PT=TR$

$ST$ is the bisector of side

$PR$

So that, in a triangle a line is drawn through the mid-point of one side of the triangle parallel to another side bisect the third side.

Prove.

Q8. Using Theorem

$6.2$ , prove that the line joining the mid-pints of any two sides of a triangle is parallel to the third side. ( Recall that you have done it in Class IX ).

Sol. :

Let

$trianglePQR$ which have

$S$ and

$T$ are mid-points of its sides

$PQ$ and

$PR$ respectively. Line

$ST$ is drawn.

Given

$PS=SQ$ and

$PT=TR$

$(PS)/(SQ)=(PT)/(TR)=1$

Line

$ST$ , divides

$PQ$ and

$PR$ the two sides of

$trianglePQR$ at points

$S$ and

$T$ in the same ratio,

$1:1$ .

$ST∥QR$ (By Theorem

$6.2$ )

[ Theorem $6.2$ : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.]

So that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Proved.

Q9.

$ABCD$ is a trapezium in which

$AB∥DC$ and diagonals intersect each other at the point

$O$ . Show that

$(AO)/(BO)=(CO)/(DO)$ .

Sol. :

In trapezium

$ABCD$ which have

$AB∥DC$ and its diagonals

$AC$ and

$BD$ intersect each other at point

$O$ .

Construction: Draw a line

$EF∥AB∥DC$ passing through

$O$ .

$triangleADC$ ,

$EF∥DC$ (By construction)

$(AE)/(ED)=(AO)/(CO)$ _____________(1) (By Theroem

$6.1$ )

[Theorem $6.1$ : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.]

$triangleBAD$

$EF∥AB$ (By construction)

$(AE)/(ED)=(BO)/(DO)$ _____________(2) (By Theroem

$6.1$ )

from equation (1) and (2)

$(AO)/(CO)=(BO)/(DO)$

$(AO)/(BO)=(CO)/(DO)$

Proved.

Alternate method:

In Trapezium

$ABCD$ ,

$AB∥DC$ and diagonals

$AC$ and

$BD$ intersect each other at a point

$O$ .

Given

$AB∥DC$ and

$AC$ is transversal

$angleOAB=angleOCD$ ___________(1) (Alternate angle)

Given

$AB∥DC$ and

$BD$ is transversal

$angleOBA=angleODC$ ___________(2) (Alternate angle)

$angleAOB=angleCOD$ ___________(3) (Vertiacally opposite angles)

By equation (1), (2) and (3)

$triangleOAB∼triangleOCD$ (By AAA Similarity criterion)

$(BO)/(DO)=(AO)/(CO)$

$(AO)/(BO)=(CO)/(DO)$

Proved.

Q10. The diagonals of a quadrilateral

$ABCD$ intersect each other at the point

$O$ such that

$(AO)/(BO)=(CO)/(DO).$ Show that

$ABCD$ is a trapezium.

Sol. :

In quadrilateral

$ABCD$ diagonal

$AC$ and

$BD$ intersect each other at point

$O$ and

Given

$(AO)/(BO)=(CO)/(DO)$

$triangleOAB$ and

$triangleOCD$ ,

$angleAOB=angleCOD$ (Vertically opposite angles)

Given

$(AO)/(BO)=(CO)/(DO)$

here sides are proportional and including the equal angles.

$triangleOAB∼triangleOCD$ (By

$SAS$ similarity criterion)

$angleOAB=angleOCD$ (Corresponding angles are equal)

By alternate angles

$AB∥DC$

So that, the quadrilateral

$ABCD$ is a trapezium.

Proved.

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10th Maths 6.2

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Exercise 6.2

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