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Question 1:

In the given figure, two chords AB and CD intersect each other at the point P.

Prove that:

$(i) \triangle APC\sim \triangle DPB$

$(ii) AP\times PB=CP\times DP$

Solution 1:

Let us join CB.

(i) In ΔAPC and ΔDPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB)

ΔAPC ∼ ΔDPB (By AA similarity criterion)

(ii) We have already proved that

ΔAPC ∼ ΔDPB

We know that the corresponding sides of similar triangles are proportional.

$\therefore \frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}$

$\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}$

∴ AP. PB = PC. DP

Comments

Tasneem

Question 2:

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:

$(i)\triangle PAC\sim \triangle PDB$

(ii) $PA\times PB=PC\times PD$

Solution 2:

(i) In ΔPAC and ΔPDB,

∠P = ∠P (Common)

∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)

∴ ΔPAC ∼ ΔPDB

(ii)We know that the corresponding sides of similar triangles are proportional.

$\therefore \frac{PA}{PD}=\frac{AC}{DB}=\frac{PC}{PB}$

$\Rightarrow \frac{PA}{PD}=\frac{PC}{PB}$

$\therefore PA\times PB=PC\times PD$

Comments

Tasneem

Question 3:

In the given figure, D is a point on side BC of â?? ABC such that $\frac{BA}{CD}=\frac{AB}{AC}$ .

Prove that AD is the bisector of $\angle BAC$

Solution 3:

Let us extend BA to P such that AP = AC. Join PC.

It is given that,

$\frac{BD}{CD}=\frac{AB}{AC}$

$\Rightarrow \frac{BD}{CD}=\frac{AP}{AC}$

By using the converse of basic proportionality theorem, we obtain

AD || PC

⇒ ∠BAD = ∠APC (Corresponding angles) … (1)

And, ∠DAC = ∠ACP (Alternate interior angles) … (2)

By construction, we have

AP = AC

⇒ ∠APC = ∠ACP … (3)

On comparing equations (1), (2), and (3), we obtain

∠BAD = ∠APC

⇒ AD is the bisector of the angle BAC.

Comments

Tasneem

Question 4:

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 (see the figure)? 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 4:

Let AB be the height of the tip of the fishing rod from the water surface. Let BC be the horizontal distance of the fly from the tip of the fishing rod.

Then, AC is the length of the string.

AC can be found by applying Pythagoras theorem in ΔABC.

AC² = AB² + BC²

AB² = (1.8 m)² + (2.4 m)²

AB² = (3.24 + 5.76) m²

AB² = 9.00 m²

Thus, the length of the string out is 3 m.

She pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m

Let the fly be at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC − String pulled by Nazima in 12 seconds

= (3.00 − 0.6) m

= 2.4 m

In ΔADB,

AB² + BD² = AD²

(1.8 m)² + BD² = (2.4 m)²

BD² = (5.76 − 3.24) m² = 2.52 m²

BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m

= 2.787 m

= 2.79 m

Comments

Tasneem

Question 5:

In the given fig, AD is a median of a triangle ABC and $AM\perp BC$ . Prove that

Solution 5:

i) Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD²… (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, $DC=\frac{BC}{2}$ , we obtain

$AD^{2}+\left ( \frac{BC}{2} \right )^{2}+2MD\frac{BC}{2}=AC^{2}$

$AD^{2}+\left ( \frac{BC}{2} \right )^{2}+MD\times BC=AC^{2}$

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB² = AM² + MB²

= (AD² − DM²) + MB²

= (AD² − DM²) + (BD − MD)²

= AD² − DM² + BD² + MD² − 2BD × MD

= AD² + BD² − 2BD × MD

$AD^{2}+\left ( \frac{BC}{2} \right )^{2}-2\times \frac{BC}{2}\times MD$

$AD^{2}+\left ( \frac{BC}{2} \right )^{2}-BC\times MD$

(iii)Applying Pythagoras theorem in ΔABM, we obtain

AM² + MB² = AB² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC² … (2)

Adding equations (1) and (2), we obtain

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD − DM)² + (MD + DC)² = AB² + AC²

2AM²+BD² + DM² − 2BD.DM + MD² + DC² + 2MD.DC = AB²+ AC²

2AM² + 2MD² + BD² + DC² + 2MD (− BD + DC) = AB² + AC²

Comments

Tasneem

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 6:

Let ABCD be a parallelogram.

Let us draw perpendicular DE on extended side AB, and AF on side DC.

Applying Pythagoras theorem in ΔDEA, we obtain

DE² + EA² = DA² … (i)

Applying Pythagoras theorem in ΔDEB, we obtain

DE² + EB² = DB²

DE² + (EA + AB)² = DB²

(DE² + EA²) + AB² + 2EA × AB = DB²

DA² + AB² + 2EA × AB = DB² … (ii)

Applying Pythagoras theorem in ΔADF, we obtain

AD² = AF² + FD²

Applying Pythagoras theorem in ΔAFC, we obtain

AC² = AF² + FC²

= AF² + (DC − FD)²

= AF² + DC²+ FD² − 2DC × FD

= (AF² + FD²) + DC² − 2DC × FD

AC² = AD² + DC² − 2DC × FD … (iii)

Since ABCD is a parallelogram,

AB = CD … (iv)

And, BC = AD … (v)

In ΔDEA and ΔADF,

∠DEA = ∠AFD (Both 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common)

∴ ΔEAD ΔFDA (AAS congruence criterion)

⇒ EA = DF … (vi)

Adding equations (i) and (iii), we obtain

DA² + AB² + 2EA × AB + AD² + DC² − 2DC × FD = DB² + AC²

DA² + AB² + AD² + DC² + 2EA × AB − 2DC × FD = DB² + AC²

BC² + AB² + AD² + DC² + 2EA × AB − 2AB × EA = DB² + AC²

[Using equations (iv) and (vi)]

AB² + BC² + CD² + DA² = AC² + BD²

Comments

Tasneem

Question 7:

In the figure, ABC is a triangle in which $\angle ABC< 90^{\circ}$ and $AD\perp CB$ . Prove that $[AC^{2}=AB^{2}+BC^{2}-2BC.BD]$

Solution 7:

Applying Pythagoras theorem in ΔADB, we obtain

AD² + DB² = AB²

⇒ AD² = AB² − DB²… (1)

Applying Pythagoras theorem in ΔADC, we obtain

AD² + DC² = AC²

AB² − BD² + DC² = AC² [Using equation (1)]

AB² − BD² + (BC − BD)² = AC²

AC² = AB² − BD² + BC² + BD² −2BC × BD

= AB² + BC² − 2BC × BD

Comments

Tasneem

Question 8:

In the figure, PS is the bisector of $\angle QPR$ of $\triangle PQR$ . Prove that $\frac{OS}{SR}=\frac{PQ}{PR}$

Solution 8:

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given that, PS is the angle bisector of ∠QPR.

∠QPS = ∠SPR … (1)

By construction,

∠SPR = ∠PRT (As PS || TR) … (2)

∠QPS = ∠QTR (As PS || TR) … (3)

Using these equations, we obtain

∠PRT = ∠QTR

∴ PT = PR

By construction,

PS || TR

By using basic proportionality theorem for ΔQTR,

$\frac{QS}{SR} = \frac{QP}{PT}$

Comments

Tasneem

Question 9:

In the figure, ABC is a triangle in which $[\angle ABC> 90^{\circ}]$ and $AD\perp CB$ produced. Prove that $AC^{2}=AB^{2}+BC^{2}-2BC.BD$ .

Solution 9:

Applying Pythagoras theorem in ΔADB, we obtain

AB² = AD² + DB² … (1)

Applying Pythagoras theorem in ΔACD, we obtain

AC² = AD²+ DC²

AC² = AD² + (DB + BC)²

AC² = AD² + DB² + BC² + 2DB × BC

AC² = AB² + BC² + 2DB × BC [Using equation (1)]

Comments

Tasneem

All Exercises - Chapter 6 - Triangles

Exercise 6.6

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Learn Exercise 6.6 with Free Lessons & Tips

All Exercises - Chapter 6 - Triangles

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