7.6 ADDITIONAL QUESTIONS AND ANSWERS

7.6   ADDITIONAL QUESTIONS AND ANSWERS

1)  State S.A.S congruence rule

Solution:

Two triangles are congruent if two sides and included angle A

of first triangle is equal to two sides and included angle

of second triangle.

Explanation:

Two sides (AB and BC) and included ABC

of first triangle =  two sides (DE and EF)

and included DEF of the second triangle.

2)  State A.S.A congruence rule

Solution:  Two triangles are congruent if two angles and included side A                      D

of first triangle is equal to two angles and included side

of second triangle.

 

 

 

 

3)  Line segment PQ is parallel to another line segment RS.  O is the point where QR and PS meet.  O is also the mid point for the line PS.

Show that (i) POQ SOR

(ii)  O is the midpoint of QR also

Solution:

(i)  Consider POQ and SOR

We have PQO = SRO (angle)

(They are alternate interior angles formed as PQ  RS

and QR is the transversal)

PQO = SRO (vertically opposite angles)

And OP = OS given (side)

By A.A.S rule we get

POQ SOR

(ii)  Since POQ SOR

We get OQ = OR (c.p.c.t)

So O is the mid point of QR

4. In right triangle PQR, right angled at R, M is the mid-point of hypotenuse PQ. R is joined to M and produced to a point S.  M is the midpoint of line RS.   Point S is joined to point Q.  Show that:  (i)  PMR QMS  (ii)  SQR = 90⁰

Solution:

Proof:

Consider PMR and QMS we have SP

RM = SM (given)

PM = QM (given)

PMR = QMS(vertically opposite angles)

Using S.A.S congruence rule we get

PMR QMS

(ii)  We have PMR QMS

We get   MPR = MQS  (c.p.c.t)Q

R

MPR and MQS are a pair of alternate interior angles

So we get PR QS

QR is the transversal that intersects the two parallel lines PR and QS

QRP + SQR = 180⁰ (Angles on same side are supplementary)

90⁰   + SQR = 180⁰ ( QRP = 90⁰)

SQR =   90⁰

 

5. PQR is a triangle in which altitudes QS and RT are equal. Show that (i)  PQS PRT  (ii) PQ = PR

Solution:

Proof:  In PQS and PRT we have

PSQ = PTR  (each  = 90⁰) (angle)

P = P (common angle) (angle)

QS= RT (given) (side)

By A.A.S congruence rule

PQS PRT

(ii) Since PQS PRT

PQ = PR (c.p.c.t)

 

6. QM and RN are two equal altitudes of a triangle PQR. Using RHS congruence rule, prove that the triangle PQR is isosceles.

Solution:

Given: QM = RN

To prove: PQ = PR  (Hence PQR is an isosceles triangle)

Proof:

We know that QM  PR

QMR = 90⁰

Similarly RN  PQ

RNQ= 90⁰

Consider the two right angled QMR and RNQ

PQ = PR  (given)

QR = QR (common side)

By R.H.S congruence rule

QMR RNQ

(Since the triangles are congruent the corresponding parts are equal)

QRM =  RQN

or  QRP =  RQP (line extended only)

PQ = PR (sides opposite to equal angles of a triangle are equal)

The triangle is isosceles since two sides of a triangle are equal.

 

7.In Fig.  Q < P and R < S.  Show that PS < QR.

Solution:

Given: Q < P and R< S

To prove:  PS<QR

Proof: Q< P (given)

It means P> Q

Side opposite P is OQ

Side opposite Q is OP

OQ>OP  ————(1)

Similarly OR>OS  ———–(2)

From (1) and (2) we get

[OQ + OR] > [OP + OS]

QR>PS

It means PS <QR    (proved)