Solve the followings Questions.

1. In figure, PS is the bisector of ∠QPR of Δ PQR. Prove that QS/SR = PQ/PR .

Answer:

Given:

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,

⇒ QS/SR = QP/PT

⇒ QS/SR = PQ/PR (∴PT = PR)

2. In figure, D is a point on hypotenuse AC of Δ ABC, BD ⊥ AC, DM ⊥ BC and DN ⊥AB. Prove that:

(i) DM²= DN.MC

(ii) DN²= DM.AN

Answer: (i) Let us join DB

We have, DN || CB, DM || AB, and ∠B = 90°

∴ DMBN is a rectangle.

∴ DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.

∴ ∠CDB = 90°

⇒ ∠2 + ∠3 = 90° … (1)

In ΔCDM,

∠1 + ∠2 + ∠DMC = 180°

⇒ ∠1 + ∠2 = 90° … (2)

In ΔDMB,

∠3 + ∠DMB + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° … (3)

From equation (1) and (2), we obtain

∠1 = ∠3

From equation (1) and (3), we obtain

∠2 = ∠4

In ΔDCM and ΔBDM,

∠1 = ∠3 (Proved above)

∠2 = ∠4 (Proved above)

∴ ΔDCM ∼ ΔBDM (AA similarity criterion)

⇒ (BM)/(DM) = (DM)/(MC)

⇒ (DN)/(DM) = (DM)/(MC)         ∴ (BM = DN)

⇒ DM² = DN × MC

(ii) In right triangle DBN,

∠5 + ∠7 = 90° … (4)

In right triangle DAN,

∠6 + ∠8 = 90° … (5)

D is the foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90°

⇒ ∠5 + ∠6 = 90° … (6)

From equation (4) and (6), we obtain

∠6 = ∠7

From equation (5) and (6), we obtain

∠8 = ∠5

In ΔDNA and ΔBND,

∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

∴ ΔDNA ∼ ΔBND (AA similarity criterion)

=> AN/DN = DN/NB

⇒ DN² = AN × NB

⇒DN² = AN × DM (As NB = DM)

3. In figure, ABC is a triangle in which ∠ABC > 90⁰ and AD ⊥ CB produced. Prove that:

AC² = AB² + BC.BD

Answer

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 x BC

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

4. In figure, ABC is a triangle in which ∠ABC < 90⁰ and AD ⊥ BC produced. Prove that: AC² =AB² + BC² – 2BC.BD

Answer:

Applying Pythagoras theorem in ΔADB, we obtain

AD² + DB² = AB²

⇒ AD² =  AB² – DB² …(i)

Applying Pythagoras theorem in ΔADC, we obtain

AD² + DC² = AC²

AB² – BD² + DC² = AC² …. Using eqn.(i)

AB² – BD² + (BC – BD)² = AC²

⇒ AB² – BD²  + BC² + BD² – 2BC x BD

 = AB² + BC² – 2BC x BD

5. In figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i) AC² = AD² + BC.DM + (BC/2)²

(ii) AC² = AD² – BC.DM + (BC/2)²

(iii) AC² + AB² = 2AD² + 1/2(BC)²

Answer:

(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 = BC/2, we obtain

AD² + (BC/2)² + 2MD.(BC/2) = AC²

AD² + (BC/2)² + MC.BC = AC²

(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 x MD

= AD² + (BC/2)² – 2(BC/2) x MD

= AD² + (BC/2)² – BC x 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

2(AM² + MD²)  + (BC/2)² + (BC/2)² + 2MD(-BC/2 + BC/2) = AB² + AC²

2AD² + BC²/2 = AB² + AC²

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

Answer:

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 x AB = DB²

DA² + AB² + 2EA x AB = DB²  ….(ii)

Applying Pythagoras theorem in ΔADF, we obtain

AD² = AF² + FD²

Applying Pythagoras theorem in ΔAFC, we obtain

AC² = AF² + FC²

AC² = AF² + (DC – FD)²

AC² =AF² + DC² + FD² – 2DC x FD

AC² =(AF² + FD²) + DC² – 2DC x FD

AC² =AD² + DC² – 2DC x FD….(iii)

Since ABCD is a parallelogram,

AB = CD … (iv)

AB = CD … (v)

In ΔDEA and ΔADF,

∠DEA = ∠AFD (Both 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common)

∴ ΔEADΔFDA (AAS congruence criterion)

⇒ EA = DF … (vi)

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

DA² + AB² + 2EA x AB + AD² + DC² – 2DC x FD = DB² + AC²

DA² + AB² + AD² + DC² + 2EA x AB – 2DC x FD = DB² + AC²

BC² + AB² + AD² + DC² + 2EA x AB – 2AB x EA = DB² + AC²

[Using equations (iv) & (vi)

AB² + BC² + CD² + DA² = AC² + BD²

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

(i) ΔAPC ~ ΔDPB

(ii) AP.PB = CP.DP

Answer:

(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.

∴AP/DP = PC/PB = CA/BD

⇒AP/DP = PC/PB

∴ AP. PB = PC. DP

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

(i) ΔPAC ~ ΔPDB

(ii) PA.PB = PC.PD

Answer:

(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.

∴ PA/PD = AC/DB = PC/PB

⇒PA/PD = PC/PB

∴ PA.PB = PC.PD

9. In figure, D is appointing on side BC of ΔABC such that BD/CD = AB/AC Prove that AD is the bisector of ∠BAC.

Answer:

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

It is given that,

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

⇒ BD/CD = 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.

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 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 be the horizontal distance of the fly from her after 12 seconds?

Answer:

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 m²

⇒AB = √9 m = 3 m.

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.

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2