Triangle

NCERT Exercise 6.6

Part 2

Question 6: Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Solution: ABCD is a parallelogram in which AB = CD and AD = BC

Perpendicular AN is drawn on DC and perpendicular DM is drawn on AB extended up to M.

In ∆ AMD; `AD^2 = DM^2 + AM^2` ……… (1)

In ∆ BMD; `BD^2 = DM^2 + (AM + AB)^2`

Or, `BD^2 = DM^2 + AM^2 + AB^2 + 2AM.AB` ………. (2)

Substituting the value of AM² from equation (1) in equation (2), we get;

`BD^2 = AD^2 + AB^2 + 2AM.AB` ………. (3)

In triangle AND;

`AD^2 = AN^2 + DN^2` ………. (4)

In triangle ANC;

`AC^2 = AN^2 + (DC – DN)^2`

Or, `AC^2 = AN^2 + DN^2 + DC^2 – 2DC.DN` ……… (5)

Substituting the value of AD² from equation (4) in equation (5), we get;

`AC^2 = AD^2 + DC^2 – 2DC.DN` ……….. (6)

We also have; AM = DN and AB = CD.

Substituting these values in equation (6), we get;

`AC^2 = AD^2 + DC^2 – 2AM.AB` ……… (7)

Adding equations (3) and (7), we get;

`AC^2 + BD^2 = AD^2 + AB^2 + 2AM.AB + AD^2 + DC^2 – 2AM.AB`

Or, `AC^2 + BD^2 = AB^2 + BC^2 + DC^2 + AD^2` proved

Question 7: In the given figure, two chords AB and CD intersect each other at the point P. Prove that:

(a) Δ APC ∼ Δ DPB

Solution: In Δ APC and Δ DPB;

∠ CAP = ∠ BDP (Angles on the same side of a chord are equal)

∠ APC = ∠ DPB (Opposite angles)

Hence; Δ APC ∼ Δ DPB (AAA Criterion)

(b) AP.PB = CP. DP

Solution: Since the two triangles are similar, hence;

`(AP)/(CP)=(DP)/(PB)`

Or, `AP.PB=CP.DP` proved

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

(a) Δ PAC ∼ Δ PDB

Solution: In triangle Δ PAC ∼ Δ PDB;

∠ PAC + ∠ CAB = 180° (Linear pair of angles)

∠ CAB + ∠ BDC = 180° (Opposite angles of cyclic quadrilateral are supplementary)

Hence; ∠ PAC = ∠ PDB

Similarly; ∠ PCA = ∠ PBD can be proven

Hence; Δ PAC ∼ Δ PDB

(b) PA.PB = PC.PD

Solution: Since the two triangles are similar, so;

`(PA)/(PC)=(PD)/(PB)`

Or, `PA.PB=PC.PD` proved

Question 9: In the given figure, D is a point on side BC of Δ ABC such that `(BD)/(CD)=(AB)/(AC)` Prove that AD is the bisector of ∠ BAC.

Solution: Draw a line CM which meets BA extended up to AM so that AM = AC

This means; ∠ AMC = ∠ ACM (Angles opposite to equal sides)

`(BD)/(CD)=(AB)/(AC)` given

Or, `(BD)/(CD)=(AB)/(AM)` because AM = AC

Hence; Δ ABD ∼ Δ MBC

Hence; AD || MC

This means;

∠ DAC = ∠ ACM (Alternate angles)

∠ BAD = ∠ AMC (Corresponding angles)

Since, ∠ AMC = ∠ ACM

Hence; ∠ DAC = ∠ BDA

Hence; AD is the bisector of ∠ BAC

Question 10: Nazima is fly fishing in 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 taut, how much string does she have out? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Solution: Here; AD = 1.8 m, BD = 2.4 m and CD = 1.2 m

Retracting speed of string = 5 cm per second

In triangle ABD; length of string, i.e. AB can be calculated as follows:

`AB^2 = AD^2 + BD^2`

`= (1.8)^2 + (2.4)^2`

`= 3.24 + 5.76 = 9`

Or, `AB = 3`

Let us assume that the string reaches at point M after 12 seconds

Length of retracted string in 12 seconds `= 5 xx 12 = 60` cm

Remaining length of string `= 3 m – 0.6 m = 2.4` m

In triangle AMD, we can find MD by using Pythagoras Theorem.

`MD^2 = AM^2 – AD^2`

`= 2.4^2 – 1.8^2`

`= 5.76 – 3.24= 2.52`

Or, `MD = 1.58`

Hence; horizontal distance between the girl and the fly `= CD + MD = 1.2 + 1.58 = 2.78` m