**NCERT Solutions for Class 10 Maths Chapter 6 Triangles. Question Answer.**

Page No: 122

**Exercise 6.1**

**1. Fill in the blanks using correct word given in the brackets:-**

(i) All circles are __________. (congruent, similar)

► Similar

(ii) All squares are __________. (similar, congruent)

► Similar

(iii) All __________ triangles are similar. (isosceles, equilateral)

► Equilateral

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

► (a) Equal, (b) Proportional(iii) All __________ triangles are similar. (isosceles, equilateral)

► Equilateral

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

2. Give two different examples of pair of

(i) Similar figures

(ii) Non-similar figures

**Answer**

(i) Two twenty-rupee notes, Two two rupees coins.

(ii) One rupee coin and five rupees coin, One rupee not and ten rupees note.

3. State whether the following quadrilaterals are similar or not:

**Answer**

The given two figures are not similar because their corresponding angles are
not equal.

Page No: 128

**Exercise 6.2**

**Answer**

(i) In △ ABC, DE∥BC (Given)

∴ AD/DB = AE/EC [By using Basic proportionality theorem]

⇒ 1.5/3 = 1/EC
⇒ Σ EC = 3/1.5

EC = 3×10/15 = 2 cmHence, EC = 2 cm.

(ii) In △ ABC, DE∥BC (Given)

∴ AD/DB = AE/EC [By using Basic proportionality theorem]

⇒ AD/7.2 = 1.8/5.4

⇒ AD = 1.8×7.2/5.4 = 18/10 × 72/10 × 10/54 = 24/10

⇒ AD = 2.4

Hence, AD = 2.4 cm.

2. E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR.

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

**Answer**

In ΔPQR, E and F are two points on side PQ and PR respectively.

(i) PE = 3.9 cm, EQ = 3 cm (Given)

PF = 3.6 cm, FR = 2,4 cm (Given)

∴ PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3 [By using Basic proportionality theorem]

And, PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5

So, PE/EQ ≠ PF/FR

Hence, EF is not parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8cm, RF = 9cm

∴ PE/QE = 4/4.5 = 40/45 = 8/9 [By using Basic proportionality theorem]

And, PF/RF = 8/9

So, PE/QE = PF/RF

Hence, EF is parallel to QR.

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm (Given)

Here, EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm

And, FR = PR - PF = 2.56 - 0.36 = 2.20 cm

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55 ...

**(i)**
And, PE/FR = 0.36/2.20 = 36/220 = 9/55 ...

**(ii)**
∴ PE/EQ = PF/FR.

Hence, EF is parallel to QR.

**Answer**

In the given figure, LM || CB

By using basic proportionality theorem, we get,
AM/MB = AL/LC ...

Similarly, LN || CD

∴ AN/AD = AL/LC ...

From **(i)**Similarly, LN || CD

∴ AN/AD = AL/LC ...

**(ii)****(i)**and

**(ii)**, we get

AM/MB = AN/AD

4. In the fig 6.19, DE||AC and DF||AE. Prove that

BF/FE = BE/EC

**Answer**

∴ BD/DA = BE/EC ...

**(i)**[By using Basic Proportionality Theorem]
In ΔABC, DF || AE (Given)

∴ BD/DA = BF/FE ...**(ii)**[By using Basic Proportionality Theorem]

From equation

**(i)**and

**(ii)**, we get

BE/EC = BF/FE

5. In the fig 6.20, DE||OQ and DF||OR, show that EF||QR.

**Answer**

In ΔPQO, DE || OQ (Given)

∴ PD/DO = PE/EQ ...

**(i)**[By using Basic Proportionality Theorem]
In ΔPQO, DE || OQ (Given)

∴ PD/DO = PF/FR ...

**(ii)**[By using Basic Proportionality Theorem]
From equation

**(i)**and**(ii)**, we get
PE/EQ = PF/FR

In ΔPQR, EF || QR. [By converse of Basic Proportionality Theorem]

**Answer**

In ΔOPQ, AB || PQ (Given)

∴ OA/AP = OB/BQ ...

**(i)**[By using Basic Proportionality Theorem]
In ΔOPR, AC || PR (Given)

∴ OA/AP = OC/CR ...

**(ii)**[By using Basic Proportionality Theorem]
From equation

**(i)**and**(ii)**, we get
OB/BQ = OC/CR

In ΔOQR, BC || QR. [By converse of Basic Proportionality Theorem].

**Answer**

Given: ΔABC in which D is the mid point of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE
|| BC.

To Prove: E is the mid point of AC.

Proof: D is the mid-point of AB.

∴ AD=DB

⇒ AD/BD = 1 ... **(i)**

In ΔABC, DE || BC,

Therefore, AD/DB = AE/EC [By using Basic Proportionality Theorem]⇒1 = AE/EC [From equation

**(i)**]

∴ AE =EC

Hence, E is the mid point of AC.

8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

**Answer**

To Prove: DE || BC

Proof: D is the mid point of AB (Given)

∴ AD=DB

⇒ AD/BD = 1 ... **(i)**

Also, E is the mid-point of AC (Given)

∴ AE=EC

⇒AE/EC = 1 [From equation ∴ AE=EC

**(i)**]

From equation

**(i)**and

**(ii)**, we get

AD/BD = AE/EC

Hence, DE || BC [By converse of Basic Proportionality Theorem]

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

**Answer**

Given: ABCD is a trapezium in which AB || DC in which diagonals AC and BD intersect each other at O.

To Prove: AO/BO = CO/DO

Construction: Through O, draw EO || DC || AB

Proof: In ΔADC, we have

OE || DC (By Construction)

∴ AE/ED = AO/CO ...

**(i)**[By using Basic Proportionality Theorem]

In ΔABD, we have

OE || AB (By Construction)

∴ DE/EA = DO/BO ...

**(ii)**[By using Basic Proportionality Theorem]

From equation

**(i)**and

**(ii)**, we get

AO/CO = BO/DO

⇒ AO/BO = CO/DO

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

**Answer**

Given: Quadrilateral ABCD in which diagonals AC and BD intersects each other
at O such that AO/BO = CO/DO.

To Prove: ABCD is a trapezium

Construction: Through O, draw line EO, where EO || AB, which meets AD at E.

Proof: In ΔDAB, we have

EO || AB

∴ DE/EA = DO/OB ...

**(i)**[By using Basic Proportionality Theorem]
Also, AO/BO = CO/DO (Given)

⇒ AO/CO = BO/DO

⇒ CO/AO = BO/DO

⇒ DO/OB = CO/AO ...

**(ii)**
From equation

**(i)**and**(ii)**, we get
DE/EA = CO/AO

Therefore, By using converse of Basic Proportionality Theorem, EO || DC also
EO || AB

⇒ AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.

Page No: 138

**Exercise 6.3**

**Answer**

(i) In ΔABC and ΔPQR, we have

∠A = ∠P = 60° (Given)

∠B = ∠Q = 80° (Given)∠C = ∠R = 40° (Given)

∴ ΔABC ~ ΔPQR (AAA similarity criterion)

(ii) In ΔABC and ΔPQR, we have

AB/QR = BC/RP = CA/PQ

∴ ΔABC ~ ΔQRP (SSS similarity criterion)

(iii) In ΔLMP and ΔDEF, we have

LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6

MP/DE = 2/4 = 1/2

PL/DF = 3/6 = 1/2

LM/EF= 2.7/5 = 27/50

Here, MP/DE = PL/DF ≠ LM/EF

Hence, ΔLMP and ΔDEF are not similar.

(iv) In ΔMNL and ΔQPR, we have

MN/QP = ML/QR = 1/2

∠M = ∠Q = 70°

∴ ΔMNL ~ ΔQPR (SAS similarity criterion)

(v) In ΔABC and ΔDEF, we have

AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°

Here, AB/DF = 2.5/5 = 1/2

And, BC/EF = 3/6 = 1/2

⇒ ∠B ≠ ∠F

Hence, ΔABC and ΔDEF are not similar.

(vi) In ΔDEF,we have

∠D + ∠E + ∠F = 180° (sum of angles of a triangle)

⇒ 70° + 80° + ∠F = 180°

⇒ ∠F = 180° - 70° - 80°

⇒ ∠F = 30°

In PQR, we have

∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)

⇒ ∠P + 80° + 30° = 180°

⇒ ∠P = 180° - 80° -30°

⇒ ∠P = 70°

In ΔDEF and ΔPQR, we have

∠D = ∠P = 70°

∠F = ∠Q = 80°

∠F = ∠R = 30°

Hence, ΔDEF ~ ΔPQR (AAA similarity criterion)

Page No: 139

2. In the fig 6.35, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

**Answer**

DOB is a straight line.

Therefore, ∠DOC + ∠ COB = 180°

⇒ ∠DOC = 180° - 125°

= 55°

In ΔDOC,

∠DCO + ∠ CDO + ∠ DOC = 180°

(Sum of the measures of the angles of a triangle is 180º.)

⇒ ∠DCO + 70º + 55º = 180°

⇒ ∠DCO = 55°

It is given that ΔODC ~ ΔOBA.

∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]∠DCO + ∠ CDO + ∠ DOC = 180°

(Sum of the measures of the angles of a triangle is 180º.)

⇒ ∠DCO + 70º + 55º = 180°

⇒ ∠DCO = 55°

It is given that ΔODC ~ ΔOBA.

⇒ ∠ OAB = 55°

∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]

⇒ ∠OAB = 55°

3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD

**Answer**

∠CDO = ∠ABO [Alternate interior angles as AB || CD]

∠DCO = ∠BAO [Alternate interior angles as AB || CD]

∠DOC = ∠BOA [Vertically opposite angles]

∴ ΔDOC ~ ΔBOA [AAA similarity criterion]

∴ DO/BO = OC/OA [ Corresponding sides are proportional]

⇒ OA/OC = OB/OD

4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

**Answer**

In ΔPQR, ∠PQR = ∠PRQ

∴ PQ = PR ...

**(i)**

Given,QR/QS = QT/PR

Using

**(i)**, we get

QR/QS = QT/QP ...

**(ii)**

In ΔPQS and ΔTQR,

QR/QS = QT/QP [using

**(ii)**]

∠Q = ∠Q

∴ ΔPQS ~ ΔTQR [SAS similarity criterion]

5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

**Answer**

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ΔRPQ ~ ΔRTS (By AA similarity criterion)

6. In the fig 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

**Answer**

∴ AB = AC [By cpct] ...

**(i)**

And, AD = AE [By cpct] ...

**(ii)**

In ΔADE and ΔABC,

AD/AB = AE/AC [Dividing equation

∠A = ∠A [Common angle]**(ii)**by**(i)**]∴ ΔADE ~ ΔABC [By SAS similarity criterion]

7. In the fig 6.38, altitudes AD and CE of ΔABC intersect each other
at the point P. Show that:

(ii) ΔABD ~ ΔCBE

(iii) ΔAEP ~ ΔADB

(iv) ΔPDC ~ ΔBEC

**Answer**

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Each 90°)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by using AA similarity criterion,

ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)

Hence, by using AA similarity criterion,

ΔABD ~ ΔCBE

Hence, by using AA similarity criterion,

ΔABD ~ ΔCBE

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (Each 90°)

∠PAE = ∠DAB (Common)

Hence, by using AA similarity criterion,

ΔAEP ~ ΔADB

Hence, by using AA similarity criterion,

ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion,

ΔPDC ~ ΔBEC

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

**Answer**

∠A = ∠C (Opposite angles of a parallelogram)

∠AEB = ∠CBF (Alternate interior angles as AE || BC)

∴ ΔABE ~ ΔCFB (By AA similarity criterion)

9. In the fig 6.39, ABC and AMP are two right triangles, right angled
at B and M respectively, prove that:

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D
and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG,
Show that:

(i) CD/GH = AC/FG
(i) ΔABC ~ ΔAMP

(ii) CA/PA = BC/MP

**Answer**
(i) In ΔABC and ΔAMP, we have

∠A = ∠A (common angle)

∠ABC = ∠AMP = 90° (each 90°)

∴ ΔABC ~ ΔAMP (By AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (By AA similarity criterion)

If two triangles are similar then the corresponding sides are equal,

Hence, CA/PA = BC/MP

(ii) ΔDCB ~ ΔHGE

(iii) ΔDCA ~ ΔHGF

**Answer**

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)

In ΔACD and ΔFGH,

∠A = ∠F (Proved above)

∠ACD = ∠FGH (Proved above)

∴ ΔACD ~ ΔFGH (By AA similarity criterion)

⇒ CD/GH = AC/FG

(ii) In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Proved above)

∠B = ∠E (Proved above)

∴ ΔDCB ~ ΔHGE (By AA similarity criterion)

∠DCB = ∠HGE (Proved above)

∠B = ∠E (Proved above)

∴ ΔDCB ~ ΔHGE (By AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

∴ ΔDCA ~ ΔHGF (By AA similarity criterion)

∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

∴ ΔDCA ~ ΔHGF (By AA similarity criterion)

Page No: 141

11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

**Answer**

It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Proved above)

∴ ΔABD ~ ΔECF (By using AA similarity criterion)

12. Sides AB and BC and median AD of a triangle ABC are respectively
proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show
that ΔABC ~ ΔPQR.

**Answer**

Given: ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides
PQ, QR and median PM of ΔPQR

i.e., AB/PQ = BC/QR = AD/PM

To Prove: ΔABC ~ ΔPQR

Proof: AB/PQ = BC/QR = AD/PM

⇒ AB/PQ = BC/QR = AD/PM (D is the mid-point of BC. M is the mid point of QR)

⇒ ΔABD ~ ΔPQM [SSS similarity criterion]

∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]

⇒ ∠ABC = ∠PQR

In ΔABC and ΔPQR

AB/PQ = BC/QR ...

**(i)**
∠ABC = ∠PQR ...

**(ii)**
From equation

**(i)**and**(ii)**, we get
ΔABC ~ ΔPQR [By SAS similarity criterion]

^{2}= CB.CD

**Answer**

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ΔADC ~ ΔBAC (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.

∴ CA/CB =CD/CA

⇒ CA

^{2}= CB.CD.**Answer**

Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that
AB/PQ = AC/PR = AD/PM

To Prove: ΔABC ~ ΔPQR

Construction: Produce AD to E so that AD = DE. Join CE, Similarly produce PM
to N such that PM = MN, also Join RN.

Proof: In ΔABD and ΔCDE, we have

AD = DE [By Construction]

BD = DC [∴ AP is the median]

and, ∠ADB = ∠CDE [Vertically opp. angles]

∴ ΔABD ≅ ΔCDE [By SAS criterion of congruence]

⇒ AB = CE [CPCT] ...

**(i)**
Also, in ΔPQM and ΔMNR, we have

PM = MN [By Construction]

QM = MR [∴ PM is the median]

and, ∠PMQ = ∠NMR [Vertically opposite angles]

∴ ΔPQM = ΔMNR [By SAS criterion of congruence]

⇒ PQ = RN [CPCT] ...

**(ii)**
Now, AB/PQ = AC/PR = AD/PM

⇒ CE/RN = AC/PR = AD/PM ...[From

**(i)**and**(ii)**]
⇒ CE/RN = AC/PR = 2AD/2PM

⇒ CE/RN = AC/PR = AE/PN [∴ 2AD = AE and 2PM = PN]

∴ ΔACE ~ ΔPRN [By SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠A = ∠P ...

**(iii)**
Now, In ΔABC and ΔPQR, we have

AB/PQ = AC/PR (Given)

∠A = ∠P [From

**(iii)**]
∴ ΔABC ~ ΔPQR [By SAS similarity criterion]

**Answer**

Length of the vertical pole = 6m (Given)

Shadow of the pole = 4 m (Given)

Let Height of tower =

*h*m
Length of shadow of the tower = 28 m (Given)

In ΔABC and ΔDEF,

∠C = ∠E (angular elevation of sum)

∠B = ∠F = 90°

∴ ΔABC ~ ΔDEF (By AA similarity criterion)

∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)

∴ 6/

*h*= 4/28
⇒

*h*= 6×28/4
⇒

*h*= 6 × 7
⇒

*h*= 42 m
Hence, the height of the tower is 42 m.

**Answer**

We know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR ...

**(i)**

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …

**(ii)**

Since AD and PM are medians, they will divide their opposite sides.∴ BD = BC/2 and QM = QR/2 ...

**(iii)**

From equations

**(i)**and

**(iii)**, we get

AB/PQ = BD/QM ...

**(iv)**

In ΔABD and ΔPQM,

∠B = ∠Q [Using equation

**(ii)**]

AB/PQ = BD/QM [Using equation

**(iv)**]

∴ ΔABD ~ ΔPQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/PM.

Page No: 143

**Exercise 6.4**

1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm

^{2}and 121 cm

^{2}. If EF = 15.4 cm, find BC.

**Answer**

It is given that,

Area of ΔABC = 64 cm

^{2}

Area of ΔDEF = 121 cm

^{2}

EF = 15.4 cm

and, ΔABC ~ ΔDEF∴ Area of ΔABC/Area of ΔDEF = AB

^{2}/DE

^{2}

= AC

^{2}/DF

^{2}= BC

^{2}/EF

^{2}...

**(i)**

[If two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides]

∴ 64/121 = BC

^{2}/EF

^{2}

⇒ (8/11)

^{2}= (BC/15.4)

^{2}

⇒ 8/11 = BC/15.4

⇒ BC = 8×15.4/11

⇒ BC = 8 × 1.4

⇒ BC = 11.2 cm

2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

**Answer**

In ΔAOB and ΔCOD, we have

∠1 = ∠2 (Alternate angles)

∠3 = ∠4 (Alternate angles)

∠5 = ∠6 (Vertically opposite angle)

∴ ΔAOB ~ ΔCOD [By AAA similarity criterion]

Now, Area of (ΔAOB)/Area of (ΔCOD)

= AB

^{2}/CD

^{2}[If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides]

= (2CD)

^{2}/CD

^{2}[∴ AB = CD]

∴ Area of (ΔAOB)/Area of (ΔCOD)

= 4CD

^{2}/CD = 4/1

Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

3. In the fig 6.53, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.

**Answer**

Given: ABC and DBC are triangles on the same base BC. Ad intersects BC at O.

To Prove: area (ΔABC)/area (ΔDBC) = AO/DO.

Construction: Let us draw two perpendiculars AP and DM on line BC.

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each equals to 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ~ ΔDMO (By AA similarity criterion)∴ AP/DM = AO/DO

⇒ area (ΔABC)/area (ΔDBC) = AO/DO.

4. If the areas of two similar triangles are equal, prove that they are congruent.

**Answer**

Given: ΔABC and ΔPQR are similar and equal in area.

To Prove: ΔABC ≅ ΔPQR

Proof: Since, ΔABC ~ ΔPQR

∴ Area of (ΔABC)/Area of (ΔPQR) = BC

^{2}/QR

^{2}

⇒ BC

^{2}/QR

^{2}=1 [Since, Area(ΔABC) = (ΔPQR)

⇒ BC

^{2}/QR

^{2}

⇒ BC = QR

Similarly, we can prove that

AB = PQ and AC = PR

Thus, ΔABC ≅ ΔPQR [BY SSS criterion of congruence]

5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

**Answer**

To Find: area(ΔDEF) and area(ΔABC)

Solution: In ΔABC, we have

F is the mid point of AB (Given)

E is the mid-point of AC (Given)

So, by the mid-point theorem, we have

FE || BC and FE = 1/2BC

⇒ FE || BC and FE || BD [BD = 1/2BC]

∴ BDEF is parallelogram [Opposite sides of parallelogram are equal and parallel]

Similarly in ΔFBD and ΔDEF, we have

FB = DE (Opposite sides of parallelogram BDEF)

FD = FD (Common)

BD = FE (Opposite sides of parallelogram BDEF)

∴ ΔFBD ≅ ΔDEF

Similarly, we can prove that

ΔAFE ≅ ΔDEF

ΔEDC ≅ ΔDEF

If triangles are congruent,then they are equal in area.

So, area(ΔFBD) = area(ΔDEF) ...

**(i)**

area(ΔAFE) = area(ΔDEF) ...

**(ii)**

and, area(ΔEDC) = area(ΔDEF) ...

**(iii)**

Now, area(ΔABC) = area(ΔFBD) + area(ΔDEF) + area(ΔAFE) + area(ΔEDC) ...

**(iv)**

area(ΔABC) = area(ΔDEF) + area(ΔDEF) + area(ΔDEF) + area(ΔDEF)

⇒ area(ΔDEF) = 1/4area(ΔABC) [From

**(i)**,

**(ii)**and

**(iii)**]

⇒ area(ΔDEF)/area(ΔABC) = 1/4

Hence, area(ΔDEF):area(ΔABC) = 1:4

6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

**Answer**

Given: AM and DN are the medians of triangles ABC and DEF respectively and
ΔABC ~ ΔDEF.

To Prove: area(ΔABC)/area(ΔDEF) = AM

^{2}/DN^{2}Proof: ΔABC ~ ΔDEF (Given)

∴ area(ΔABC)/area(ΔDEF) = (AB

^{2}/DE

^{2}) ...

**(i)**

and, AB/DE = BC/EF = CA/FD ...

**(ii)**

In ΔABM and ΔDEN, we have

∠B = ∠E [Since ΔABC ~ ΔDEF]

AB/DE = BM/EN [Prove in

**(i)**]

∴ ΔABC ~ ΔDEF [By SAS similarity criterion]

⇒ AB/DE = AM/DN ...

**(iii)**

∴ ΔABM ~ ΔDEN

As the areas of two similar triangles are proportional to the squares of the corresponding sides.

∴ area(ΔABC)/area(ΔDEF) = AB

^{2}/DE

^{2}= AM

^{2}/DN

^{2}

^{}7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

**Answer**

Given: ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

To Prove: area(ΔBQC) = 1/2area(ΔAPC)

Proof: ΔAPC and ΔBQC are both equilateral triangles (Given)

∴ ΔAPC ~ ΔBQC [AAA similarity criterion]

∴ area(ΔAPC)/area(ΔBQC) = (AC

^{2}/BC

^{2}) = AC

^{2}/BC

^{2}

⇒ area(ΔBQC) = 1/2area(ΔAPC)

**Tick the correct answer and justify:**

8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is

(A) 2 : 1

(B) 1 : 2

(C) 4 : 1

(D) 1 : 4

**Answer**

ΔABC and ΔBDE are two equilateral triangle. D is the mid point of BC.

∴ BD = DC = 1/2BC

Let each side of triangle is 2

*a*.
As, ΔABC ~ ΔBDE

∴ area(ΔABC)/area(ΔBDE) = AB

^{2}/BD^{2}= (2*a*)^{2}/(*a*)^{2}= 4*a*^{2}/*a*^{2}= 4/1 = 4:1
Hence, the correct option is (C).

(A) 2 : 3

(B) 4 : 9

(C) 81 : 16

(D) 16 : 81

**Answer**

Let ABC and DEF are two similarity triangles ΔABC ~ ΔDEF (Given)

and, AB/DE = AC/DF = BC/EF = 4/9 (Given)

∴ area(ΔABC)/area(ΔDEF) = AB

^{2}/DE^{2 }[the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides]
∴ area(ΔABC)/area(ΔDEF) = (4/9)

^{2 }= 16/81 = 16:81
Hence, the correct option is (D).

Page No: 150

**Exercise 6.5**

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

**Answer**

(i) Given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of these sides, we will get 49, 576, and 625.

49 + 576 = 625

(7)

^{2}+ (24)

^{2}= (25)

^{2}

The sides of the given triangle are satisfying Pythagoras theorem.Hence, it is right angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will get 9, 64, and 36.

However, 9 + 36 ≠ 64

Or, 3

^{2}+ 6

^{2}≠ 8

^{2}

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

(iii) Given that sides are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will get 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50

^{2}+ 80

^{2}≠ 100

^{2}

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

Hence, it is not a right triangle.

(iv) Given that sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will get 169, 144, and 25.

Clearly, 144 +25 = 169

Or, 12

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

Length of the hypotenuse of this triangle is 13 cm.

^{2}+ 5^{2}= 13^{2}The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

Length of the hypotenuse of this triangle is 13 cm.

2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM

^{2}= QM × MR.

**Answer**

Given: ΔPQR is right angled at P is a point on QR such that PM ⊥QR.

To prove: PM

^{2}= QM × MR
Proof: In ΔPQM, we have

PQ

^{2}= PM^{2}+ QM^{2}[By Pythagoras theorem]
Or, PM

^{2}= PQ^{2}- QM^{2}...**(i)**
In ΔPMR, we have

PR

^{2}= PM^{2}+ MR^{2}[By Pythagoras theorem]
Or, PM

^{2}= PR^{2}- MR^{2}...**(ii)**
Adding

**(i)**and**(ii)**, we get
2PM

^{2}= (PQ^{2}+ PM^{2}) - (QM^{2}+ MR^{2})
= QR

^{2}- QM^{2}- MR^{2 }[∴ QR^{2}= PQ^{2}+ PR^{2}]
= (QM + MR)

^{2}- QM^{2}- MR^{2}
= 2QM × MR

∴ PM

^{2}= QM × MR(i) AB

^{2}= BC × BD

(ii) AC

^{2}= BC × DC

(iii) AD

^{2}= BD × CD

**Answer**

(i) In ΔADB and ΔCAB, we have

∠DAB = ∠ACB (Each equals to 90°)

∠ABD = ∠CBA (Common angle)

∴ ΔADB ~ ΔCAB [AA similarity criterion]

⇒ AB/CB = BD/AB

⇒ AB

^{2}= CB × BD

(ii) Let ∠CAB =

*x*

In ΔCBA,

∠CBA = 180° - 90° -

*x*

∠CBA = 90° -

*x*

Similarly, in ΔCAD

∠CAD = 90° - ∠CBA = 90° -

*x*

∠CDA = 180° - 90° - (90° -

*x*)

∠CDA =

*x*

In ΔCBA and ΔCAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA (Each equals to 90°)

∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]

⇒ AC/DC = BC/AC

⇒ AC

^{2}= DC × BC

(iii) In ΔDCA and ΔDAB, we have

∠DCA = ∠DAB (Each equals to 90°)

∠CDA = ∠ADB (common angle)

∴ ΔDCA ~ ΔDAB [By AA similarity criterion]

⇒ DC/DA = DA/DA

⇒ AD

^{2}= BD × CD

4. ABC is an isosceles triangle right angled at C. Prove that AB

^{2}= 2AC

^{2}.

**Answer**

Given that ΔABC is an isosceles triangle right angled at C.

In ΔACB, ∠C = 90°

AC = BC (Given)

AB

^{2}= AC^{2}+ BC^{2}([By using Pythagoras theorem]
= AC

^{2}+ AC^{2}[Since, AC = BC]
AB

^{2}= 2AC^{2}^{}

^{2}= 2AC

^{2}, prove that ABC is a right triangle.

**Answer**

Given that ΔABC is an isosceles triangle having AC = BC and
AB

^{2}= 2AC^{2}
In ΔACB,

AC = BC (Given)

AB

^{2}= 2AC^{2}(Given)
AB

^{2}= AC^{2 }+ AC^{2}
= AC

^{2}+ BC^{2 }[Since, AC = BC]
Hence, By Pythagoras theorem ΔABC is right angle triangle.

6. ABC is an equilateral triangle of side 2a. Find each of its
altitudes.

**Answer**

ABC is an equilateral triangle of side 2a.

Draw, AD ⊥ BC

In ΔADB and ΔADC, we have

AB = AC [Given]

AD = AD [Given]

∠ADB = ∠ADC [equal to 90°]

Therefore, ΔADB ≅ ΔADC by RHS congruence.

Hence, BD = DC [by CPCT]

In right angled ΔADB,

AB

^{2}= AD^{2 }+ BD^{2}
(2

*a*)^{2}= AD^{2 }+*a*^{2 }
⇒ AD

^{2 }= 4*a*^{2}-*a*^{2}
⇒ AD

^{2 }= 3*a*^{2}
⇒ AD

^{ }= √3a
7. Prove that the sum of the squares of the sides of rhombus is equal to
the sum of the squares of its diagonals.

**Answer**

ABCD is a rhombus whose diagonals AC and BD intersect at O. [Given]

We have to prove that,

AB

Since, the diagonals of a rhombus bisect each other at right angles.^{2 }+ BC^{2 }+ CD^{2}+ AD^{2 }= AC^{2 }+ BD^{2}
Therefore, AO = CO and BO = DO

In ΔAOB,

∠AOB = 90°

AB

^{2}= AO^{2 }+ BO^{2 }...**(i)**[By Pythagoras]
Similarly,

AD

^{2}= AO^{2 }+ DO^{2 }...**(ii)**
DC

^{2}= DO^{2 }+ CO^{2 }...**(iii)**
BC

^{2}= CO^{2 }+ BO^{2 }...**(iv)**
Adding equations

**(i) + (ii) + (iii) + (iv)**we get,
AB

= 4AO^{2 }+ AD^{2 }+^{ }DC^{2 }+^{ }BC^{2}= 2(AO^{2 }+ BO^{2 }+ DO^{2 }+ CO^{2 })^{2 }+ 4BO

^{2 }[Since, AO = CO and BO =DO]

= (2AO)

^{2}

^{ }+ (2BO)

^{2}= AC

^{2 }+ BD

^{2}

Page No: 151

8. In Fig. 6.54, O is a point in the interior of a triangle

(i) OA

^{2}+ OB

^{2}+ OC

^{2}– OD

^{2}– OE

^{2}– OF

^{2}= AF

^{2}+ BD

^{2}+ CE

^{2},

(ii) AF

^{2}+ BD

^{2}+ CE

^{2}= AE

^{2}+ CD

^{2}+ BF

^{2}.

**Answer**

Join OA, OB and OC

(i) Applying Pythagoras theorem in ΔAOF, we have

OA

Similarly, in ΔBOD^{2}= OF^{2}+ AF^{2}OB

^{2}= OD

^{2}+ BD

^{2}

Similarly, in ΔCOE

OC

^{2}= OE

^{2}+ EC

^{2}

Adding these equations,

OA

^{2}+ OB

^{2}+ OC

^{2}= OF

^{2}+ AF

^{2}+ OD

^{2}+ BD

^{2}+ OE

^{2 }+ EC

^{2}

OA

^{2}+ OB

^{2}+ OC

^{2}– OD

^{2}– OE

^{2}– OF

^{2}= AF

^{2}+ BD

^{2}+ CE

^{2}.

(ii) AF

^{2}+ BD

^{2}+ EC

^{2}= (OA

^{2}- OE

^{2}) + (OC

^{2}- OD

^{2}) + (OB

^{2}- OF

^{2})

∴ AF

^{2}+ BD

^{2}+ CE

^{2}= AE

^{2}+ CD

^{2}+ BF

^{2}.

9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

**Answer**

Let BA be the wall and Ac be the ladder,

Therefore, by Pythagoras theorem,we have

AC

^{2}=^{ }AB^{2}+ BC^{2}
10

^{2}= 8^{2}+ BC^{2}
BC

^{2 }= 100 - 64
BC

^{2 }= 36
BC

Therefore, the distance of the foot of the ladder from the base of the wall
is 6 m.^{ }= 6m10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?

**Answer**

By Pythagoras theorem,

AC

24

BC

BC

BC

Therefore, the distance from the base is 6√7m.

^{2}=^{ }AB^{2}+ BC^{2}24

^{2}= 18^{2}+ BC^{2}BC

^{2 }= 576 - 324BC

^{2 }= 252BC

^{ }= 6√7mTherefore, the distance from the base is 6√7m.

11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after hours?

**Answer**

Speed of first aeroplane = 1000 km/hr

Distance covered by first aeroplane due north in hours (OA) = 100 × 3/2 km = 1500 km

Speed of second aeroplane = 1200 km/hr

Distance covered by second aeroplane due west in hours (OB) = 1200 × 3/2 km = 1800 km

In right angle ΔAOB, we have

AB

^{2}=^{ }AO^{2}+ OB^{2}
⇒ AB

^{2}=^{ }(1500)2 + (1800)2
⇒ AB = √2250000 + 3240000

= √5490000

⇒ AB = 300√61 km

Hence, the distance between two aeroplanes will be 300√61 km.

**Answer**

Let CD and AB be the poles of height 11 m and 6 m.

Therefore, CP = 11 - 6 = 5 m

From the figure, it can be observed that AP = 12m

Applying Pythagoras theorem for ΔAPC, we get

AP

(12m)

AC

AC = 13m

Therefore, the distance between their tops is 13 m.

^{2}=^{ }PC^{2}+ AC^{2}(12m)

^{2}+ (5m)^{2}= (AC)^{2}AC

^{2}= (144+25)m^{2}= 169 m^{2}AC = 13m

Therefore, the distance between their tops is 13 m.

13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE

^{2}+ BD

^{2}= AB

^{2}+ DE

^{2}.

**Answer**

Applying Pythagoras theorem in ΔACE, we get

AC

^{2}+^{ }CE^{2}= AE^{2}....**(i)**
Applying Pythagoras theorem in ΔBCD, we get

BC

^{2}+^{ }CD^{2}= BD^{2}....**(ii)**
Using equations

**(i)**and**(ii)**, we get
AC

^{2}+^{ }CE^{2}+ BC^{2}+^{ }CD^{2}= AE^{2}+ BD^{2}...**(iii)**
Applying Pythagoras theorem in ΔCDE, we get

DE

^{2}=^{ }CD^{2}+ CE^{2}
Applying Pythagoras theorem in ΔABC, we get

AB

Putting these values in equation ^{2}=^{ }AC^{2}+ CB^{2}**(iii)**, we get

DE

^{2}+ AB

^{2}= AE

^{2}+ BD

^{2}.

14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Fig. 6.55). Prove that 2AB

^{2}= 2AC

^{2}+ BC

^{2}.

**Answer**

Given that in ΔABC, we have

AD ⊥BC and BD = 3CD

In right angle triangles ADB and ADC, we have

AB

^{2}=^{ }AD^{2}+ BD^{2}...**(i)**
AC

^{2}=^{ }AD^{2}+ DC^{2}...**(ii)**[By Pythagoras theorem]
Subtracting equation

**(ii)**from equation**(i)**, we get
AB

= 9CD^{2}- AC^{2}= BD^{2}- DC^{2}^{2}- CD

^{2}[∴ BD = 3CD]

= 9CD

^{2}= 8(BC/4)

^{2 }[Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB

^{2}- AC

^{2}= BC

^{2}/2

⇒ 2(AB

^{2}- AC

^{2}) = BC

^{2}

⇒ 2AB

^{2}- 2AC

^{2}= BC

^{2}

∴ 2AB

^{2}= 2AC

^{2}+ BC

^{2}.

15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD

^{2}= 7AB

^{2}.

**Answer**

*a*, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 =

*a*/2
And, AE =

Given that, BD = 1/3BC

∴ BD = *a*√3/2Given that, BD = 1/3BC

*a*/3

DE = BE - BD =

Applying Pythagoras theorem in ΔADE, we get*a*/2 -*a*/3 =*a*/6AD

^{2}= AE

^{2}+ DE

^{2 }

⇒ 9 AD

^{2}= 7 AB

^{2}

^{}

^{}16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

**Answer**

∴ BE = EC = BC/2 =

*a*/2

Applying Pythagoras theorem in ΔABE, we get

AB

^{2}= AE

^{2}+ BE

^{2}

4AE

^{2}= 3a

^{2}

⇒ 4 × (Square of altitude) = 3 × (Square of one side)

^{}

17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.

The angle B is:

(A) 120°

(B) 60°

(C) 90°

(C) 90°

(D) 45°

**Answer**

We can observe that

AB

^{2}= 108

AC

^{2}= 144

And, BC

^{2}= 36

AB

^{2}+ BC

^{2}= AC

^{2}

The given triangle, ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

∴ ∠B = 90°

Hence, the correct option is (C).

Page No. 152

**Exercise 6.6**

1. In Fig. 6.56, PS is the bisector of ∠QPR of ΔPQR. Prove that QS/SR = PQ/PR

**Answer**

Given, in figure, PS is the bisector of ∠QPR of ∆PQR.

Now, draw RT SP || to meet QP produced in T.

Proof:

∵ RT SP || and transversal PR intersects them

∴ ∠1 = ∠2

**(Alternate interior angle)…(i)**
∴ RT SP || and transversalQT intersects them

∴ ∠3 = ∠4

**(Corresponding angle) …(ii)**
But ∠1 = ∠3

**(Given)**
∴ ∠2 = ∠4

**[From Eqs. (i) and (ii)]**
∴ PT = PR …(iii)

**(∵ Sides opposite to equal angles of a triangle are equal)**
Now, in ∆QRT,

PS || RT

**(By construction)**
∴ QS/SR = PQ/PT

**(By basic proportionally theorem)**
⇒ QS/SR = PQ/PR

**[From Eq. (iii)]**2. In Fig. 6.57, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC and

DN ⊥ AB. Prove that :

(i) DM

^{2}= DN.MC

(ii) DN

^{2}= DM.AN

**Answer**

Given that, D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC and DN ⊥ AB.

Now, join NM.

Let BD and NM intersect at O.

(i) In ∆DMC and ∆NDM,

∠DMC = ∠NDM

**(Each equal to 90°)**
∠MCD = ∠DMN

Let MCD = ∠1

Then, ∠MDC = 90° − (90°-∠1)

= ∠1

**(∵∠MCD + ∠MDC + ∠DMC = 180°)**
∴ ∠ODM = 90° − (90° − ∠1)

= ∠1

⇒ ∠DMN = ∠1

∴ ∆DMO ~ ∆NDM

**(AA similarity criterion)**
∴ DM/ND = MC/DM

**(Corresponding sides of the similar triangles are proportional)**

⇒ DM

^{2}= MC ND
(ii) In ∆DNM and ∆NAD,

∠NDM = ∠AND

**(Each equal to 90°)**
∠DNM = ∠NAD

Let ∠NAD = ∠2

Then, ∠NDA = 90° − ∠2

∵∠NDA + ∠DAN + ∠DNA = 180°

∴ ∠ODN = 90° − (90° − ∠2) = ∠2

∴ ∠DNO = ∠2

∴ ∆DNM ~ ∆NAD

**(AA similarity criterion)**
∴ DN/NA = DM/ND

⇒ DN/AN = DM/DN

⇒ DN

^{2}= DM×AN3. In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC

^{2}= AB

^{2}+ BC

^{2}+ 2 BC.BD.

**Answer**

Given that, in figure, ABC is a triangle in which ∠ABC> 90° and AD
⊥ CB produced.

Proof :

In right ADC,

∠D = 90°

AC

^{2}= AD^{2}+ DC^{2}**(By Pythagoras theorem)**
= AD

^{2}+ (BD + BC)^{2}**[∵DC = DB + BC]**
= (AD

^{2}+ DB^{2}) + BC^{2}+ 2BD.BC**[∵ (a + b)**^{2}= a^{2}+ b^{2}+ 2ab]
= AB

^{2}+ BC^{2}+ 2BC.BD**[∵In right ADB with ∠D = 90°, AB**

^{2}= AD^{2}+ DB^{2}] (By Pythagoras theorem)4. In Fig. 6.59, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC

^{2}= AB

^{2}+ BC

^{2}– 2BC.BD.

**Answer**

Given that, in figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC.

Proof:

In right △ADC,

∠D = 90°

AC

= AD^{2}= AD^{2}+ DC^{2}**(By Pythagoras theorem)**^{2}+ (BC - BD)

^{2}

**[∵BC = BD + DC]**

= AD

^{2}+ (BC - BD)

^{2}

**(BC = BD + DC)**

= AD

^{2}+ BC

^{2}+ BD

^{2}- 2BC.BD

**[∵ (a + b)**

^{2}= a^{2}+ b^{2}+ 2ab]= (AD

^{2}+ BD

^{2}) + BC

^{2}- 2BC . BD

= AB

^{2}+ BC

^{2}- 2BC . BD

**{In right △ADB with ∠D = 90°, AB**

^{2}= AD^{2}+ BD^{2}} (By Pythagoras theorem)5. In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

(i) AC

^{2}= AD

^{2}+ BC.DM + (BC/2)

^{2}

(ii) AB

^{2}= AD

^{2}– BC.DM + (BC/2)

^{2}

(iii) AC

^{2}+ AB

^{2}= 2 AD

^{2}+ 1/2 BC

^{2}

**Answer**

Given that, in figure, AD is a median of a ∆ABC and AM ⊥ BC.

Proof:

(i) In right ∆AMC,

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**

Given that, ABCD is a parallelogram whose diagonals are AC and BD.

Now, draw AM⊥DC and BN⊥D

**(produced)**.
Proof:

In right ∆AMD and ∆BNC,

AD = BC (Opposite sides of a parallelogram)AM = BN (Both are altitudes of the same parallelogram to the same base) ,

△AMD ⩭ △BNC (RHS congruence criterion)

MD = NC (CPCT) ---

**(i)**

In right △BND,

∠N = 90°

BD

^{2}= BN

^{2}+ DN

^{2}

**(By Pythagoras theorem)**

= BN

^{2}+ (DC + CN)

^{2}

**(∵ DN = DC + CN)**

= BN

^{2}+ DC

^{2}+ CN

^{2}+ 2DC.CN

**[∵ (a + b)**

^{2}= a^{2}+ b^{2}+ 2ab]= (BN

^{2}+ CN

^{2}) + DC

^{2}+ 2DC.CN

= BC

^{2}+ DC

^{2}+ 2DC.CN

**--- (ii)**

**(∵In right △BNC with ∠N = 90°)**

BN

^{2}+ CN

^{2}= BC

^{2}(By Pythagoras theorem)

In right △AMC,

∠M = 90°

AC

^{2}= AM

^{2}+ MC

^{2}

**(∵MC = DC - DM)**

= AM

^{2}+ (DC - DM)

^{2}

**[∵ (a + b)**

^{2}= a^{2}+ b^{2}+ 2ab]= AM

^{2}+ DC

^{2}+ DM

^{2}- 2DC.DM

(AM

^{2}+ DM

^{2}) + DC

^{2}- 2DC.DM

= AD

^{2}+ DC

^{2}- 2DC.DM

**[∵ In right triangle AMD with ∠M = 90°, AD**

^{2}= AM^{2}+ DM^{2}(By Pythagoras theorem)]= AD

^{2}+ AB

^{2}= 2DC.CN

**--- (iii)**

**[∵ DC = AB, opposite sides of parallelogram and BM = CN from eq (i)]**

Now, on adding Eqs. (iii) and (ii), we get

AC

^{2}+ BD

^{2}= (AD

^{2}+ AB

^{2}) + (BC

^{2}+ DC

^{2})

= AB

^{2}+ BC

^{2}+ CD

^{2}+ DA

^{2}

7. In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that :

(i) Δ APC ~ Δ DPB

(ii) AP . PB = CP . DP

**Answer**

Given that, in figure, two chords AB and CD intersects each other at the point P.

Proof:

(i) ∆APC and ∆DPB

∠APC = ∠DPB

**(Vertically opposite angles)**

∠CAP = ∠BDP

**(Angles in the same segment)**

∴ ∆APC ~ ∆DPB

**(AA similarity criterion)**

(ii) ∆APC ~ ∆DPB

**[Proved in (i)]**

∴ AP/DP = CP/BP

(∴ Corresponding sides of two similar triangles are proportional)

⇒ AP.BP = CP.DP

⇒ AP.PB = CP.DP

(i) Δ PAC ~ Δ PDB

(ii) PA.PB = PC.PD

**Answer**

Given that, in figure, two chords AB and CD of a circle intersect each other at the point P (when produced) out the circle.

Proof:

(i) We know that, in a cyclic quadrilaterals, the exterior angle is equal to the interior opposite angle.

Therefore, ∠PAC = ∠PDB

**…(i)**

and ∠PCA = ∠PBD

**…(ii)**

In view of Eqs. (i) and (ii), we get

∆PAC ~ ∆PDB

**(∵ AA similarity criterion)**

(ii) ∆PAC ~ ∆PDB

**[Proved in (i)]**

∴ PA/PD = PC/PB

**(∵ Corresponding sides of the similar triangles are proportional)**

⇒ PA.PB = PC.PD

9. In Fig. 6.63, D is a point on side BC of ΔABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠BAC.

**Answer**

Given that,D is a point on side BC of ∆ABC such that BD/CD = AB/AC

Now, from BA produce cut off AE = A. JoinCE.

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 (see Fig. 6.64)? 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**

Length of the string that she has out