# Triangles : Exercise - 6.6 Optional (Mathematics NCERT Class 10th)

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**(*These exercises are not for examination point of view)**

**Q.1Â Â Â Â In Figure, PS is the bisector of of . Prove that **

** ****Sol.**

Given PQR is a triangle and PS is the internal bisector of meeting QR at S.

Therefore

To prove :

Construction : Draw RT || SP to cut QP produced at T.

Proof : Since PS || TR and PR cuts them, hence, we have

............ (1)Â [Alternate ]

and, ............... (2)Â [Corresponding ]

But, [Given]

Therefore [From (1) and (2)]

PT = PR .......... (3) [Because sides opp. to equal are equal]

Now, in , we have

RT || SP [By construction]

Therefore [By Basic Proportionality Theorem]

[From (3)]

**Q.2Â Â Â In figure, D is a point on hypotenuse AC of , , and . Prove that**

Â Â Â Â Â Â Â Â Â Â Â Â Â **(i) **

Â Â Â Â Â Â Â Â Â Â Â Â Â **(ii) **

** **

**Sol.**

We have, and

AB || DM

Similarly, we have

and

CB || DN

Hence, quadrilateral BMDN is a rectangle.

Therefore BM = ND

(i)Â Â In , we have

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Similarly, in , we have

Â Â Â Â Â Â

Â Â Â Â Â Â Since . Therefore,

Â Â Â Â Â Â

Â Â Â Â Â Â Now, and

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Also, and

Â Â Â Â Â Â

Â Â Â Â Â Â Thus, in BMD and DMC, we have

Â Â Â Â Â Â and

Â Â Â Â Â Â Therefore By AA-criterion of similarity, we have

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â [Because BM = ND]

Â Â Â Â Â Â Â

(ii)Â Â Proceeding as in (i), we can prove that

Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â [Because BN = DM]

Â Â Â Â Â Â Â Â

**Q.3Â Â Â Â Â In figure, ABC is a triangle in which and produced. Prove that **

** **

**Sol.**

Given : ABC is a triangle in which and produced.

To Prove :

Proof : Since is a right triangle, right-angled at D, therefore, by Pythagoras theorem, we have

........... (1)

Again, is a right triangle, right-angled at D. Therefore, by Pythagoras theorem, we have

[Using (1)]

which proves the required result.

**Q.4Â Â Â Â Â In figures, ABC is a triangle in which and . Prove that **

** **

**Sol.**

Given : ABC is a triangle in which and .

To prove :

Proof : Since is a right triangle, right-angled at D, therefore, by Pythagoras theorem, we have,

.............. (1)

Again, is a right triangle, right-angled at D, therefore, by Pythagoras theorem, we have,

[Using (1)]

which proves the required result.

**Q.5 Â Â Â Â In figure, AD is a median of a triangle ABC and . Prove that**

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(i) **

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(ii) **

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(iii) **

** **

**Sol.**

Since , therefore, and .

Thus, is acute and is obtuse.

(i)Â Â Â In , is an obtuse angle.

Â Â Â Â Â Â Â Â Â Therefore

Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â .............. (1)

(ii)Â Â Â In , is an acute angle.

Â Â Â Â Â Â Â Â Â Therefore

Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â ............ (2)

(iii)Â Â Â From (1) and (2), we get

Â Â Â Â Â Â Â Â Â Â

**Q.6Â Â Â Â Â Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.**

**Sol.**

We know that if AD is a median of , then

[See above question part (iii)]

Sine the diagonals of a parallelogram bisect each other, therefore, BO and DO are medians of

ABC and ADC respectively.

Therefore ........... (1)

and, .......... (2)

Adding (1) and (2), we get

[Because ]

**Q.7Â Â Â Â Â In figure, two chords AB and CD intersect each other at the point P. Prove that**

Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(i) **

Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(ii) AP.PB = CP.DP**

**Sol.**

(i)Â Â In APC and DPB, we have

Â Â Â Â Â Â [Vert. opp. ]

Â Â Â Â Â [Angles in the same segment of a circle are equal]

Â Â Â Â Â Â Therefore by AA-criterion of similarity, we have

Â Â Â Â Â Â

(ii)Â Since

Â Â Â Â Â Â Â Therefore

Â Â Â Â Â Â Â

**Q.8Â Â Â Â Â In figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle, Prove that**

Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(i) **

Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(ii) PA.PB = PC.PD**

** **

**Sol.**

(i)Â Â Â In PAC and PDB, we have

Â Â Â Â Â Â Â [Common]

Â Â Â Â Â Â Â [Because and ]

Â Â Â Â Â Â Â Therefore by AA-criterion of similarity, we have

Â Â Â Â Â Â Â

(ii)Â Â Since

Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â PA.PB = PC.PD

**Q.9Â Â Â Â In figure, D is a point on side BC of such that . Prove that AD is the bisector of .**

**Sol.**

Given : ABC is a triangle and D is a point on BC such that

To prove : AD is the internal bisector of .

Construction : Produce BA to E such that AE = AC. Join CE.

Proof : In , since AE = AC, hence

........... (1)

[Because Angles opp. to equal sides of a are equal]

Now, [Given]

[Because AE = AC, construction]

Therefore by converse of Basic Proportionality Theorem, we have

DA || CE

Now, since CA is a transeversal, we have

......... (2) [Corresponding ]

and, ...... (3) [Alternate angles]

Also, [From (1)]

Hence, [From (2) and (3)]

Thus, AD bisects, internally.

**Q.10 Nazima is fly fishing i a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taur, how much string does she have out (see figure) ? If she pulls in the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds ?**

** **

* Sol.*

In fact, we want to find AC.

By Pythagoras theorem, we have

AC = 3 m

Therefore length of string she have out = 3 m.

Length of the string pulled at the rate of 5 cm/sec in 12 seconds.

= (5 Ã— 12) cm

= 60 cm = 0.60 m

Therefore remaining string left out = (3 â€“ 0.6) m = 2.4 m

In 2nd case let us find PB

Â Â Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

(nearly)

Hence, the horizontal distance of the fly from Nazima after 12 seconds

= (1.59 + 1.2) m

= 2.79 m (nearly)