## NCERT Solutions for Class 10th Maths Chapter 6 Triangles

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
⇒ Î£ 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 Î”PQR. 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 Î”PQR, 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 Î”ABC, 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 Î”PQO, DE || OQ (Given)

∴ PD/DO = PE/EQ ...

**(i)**[By using Basic Proportionality Theorem]
In Î”PQO, 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 Î”PQR, EF || QR. [By converse of Basic Proportionality Theorem]

**Answer**

In Î”OPQ, AB || PQ (Given)

∴ OA/AP = OB/BQ ...

**(i)**[By using Basic Proportionality Theorem]
In Î”OPR, 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 Î”OQR, BC || QR. [By converse of Basic Proportionality Theorem].

**Answer**

Given: Î”ABC 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 Î”ABC, 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 Î”ADC, we have

OE || DC (By Construction)

∴ AE/ED = AO/CO ...

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

In Î”ABD, 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 Î”DAB, 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 Î”ABC and Î”PQR, we have

∠A = ∠P = 60° (Given)

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

∴ Î”ABC ~ Î”PQR (AAA similarity criterion)

(ii) In Î”ABC and Î”PQR, we have

AB/QR = BC/RP = CA/PQ

∴ Î”ABC ~ Î”QRP (SSS similarity criterion)

(iii) In Î”LMP and Î”DEF, 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, Î”LMP and Î”DEF are not similar.

(iv) In Î”MNL and Î”QPR, we have

MN/QP = ML/QR = 1/2

∠M = ∠Q = 70°

∴ Î”MNL ~ Î”QPR (SAS similarity criterion)

(v) In Î”ABC and Î”DEF, 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, Î”ABC and Î”DEF are not similar.

(vi) In Î”DEF,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 Î”)

⇒ ∠P + 80° + 30° = 180°

⇒ ∠P = 180° - 80° -30°

⇒ ∠P = 70°

In Î”DEF and Î”PQR, we have

∠D = ∠P = 70°

∠F = ∠Q = 80°

∠F = ∠R = 30°

Hence, Î”DEF ~ Î”PQR (AAA similarity criterion)

Page No: 139

2. In the fig 6.35, Î”ODC ∝ ¼ Î”OBA, ∠ 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 Î”DOC,

∠DCO + ∠ CDO + ∠ DOC = 180°

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

⇒ ∠DCO + 70Âº + 55Âº = 180°

⇒ ∠DCO = 55°

It is given that Î”ODC ~ Î”OBA.

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

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

⇒ ∠DCO + 70Âº + 55Âº = 180°

⇒ ∠DCO = 55°

It is given that Î”ODC ~ Î”OBA.

⇒ ∠ 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]

∴ Î”DOC ~ Î”BOA [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 Î”PQS ~ Î”TQR.

**Answer**

In Î”PQR, ∠PQR = ∠PRQ

∴ PQ = PR ...

**(i)**

Given,QR/QS = QT/PR

Using

**(i)**, we get

QR/QS = QT/QP ...

**(ii)**

In Î”PQS and Î”TQR,

QR/QS = QT/QP [using

**(ii)**]

∠Q = ∠Q

∴ Î”PQS ~ Î”TQR [SAS similarity criterion]

5. S and T are point on sides PR and QR of Î”PQR such that ∠P = ∠RTS. Show that Î”RPQ ~ Î”RTS.

**Answer**

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ Î”RPQ ~ Î”RTS (By AA similarity criterion)

6. In the fig 6.37, if Î”ABE ≅ Î”ACD, show that Î”ADE ~ Î”ABC.

**Answer**

∴ AB = AC [By cpct] ...

**(i)**

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

**(ii)**

In Î”ADE and Î”ABC,

AD/AB = AE/AC [Dividing equation

∠A = ∠A [Common angle]**(ii)**by**(i)**]∴ Î”ADE ~ Î”ABC [By SAS similarity criterion]

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

(ii) Î”ABD ~ Î”CBE

(iii) Î”AEP ~ Î”ADB

(iv) Î”PDC ~ Î”BEC

**Answer**

(i) In Î”AEP and Î”CDP,

∠AEP = ∠CDP (Each 90°)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by using AA similarity criterion,

Î”AEP ~ Î”CDP

(ii) In Î”ABD and Î”CBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)

Hence, by using AA similarity criterion,

Î”ABD ~ Î”CBE

Hence, by using AA similarity criterion,

Î”ABD ~ Î”CBE

(iii) In Î”AEP and Î”ADB,

∠AEP = ∠ADB (Each 90°)

∠PAE = ∠DAB (Common)

Hence, by using AA similarity criterion,

Î”AEP ~ Î”ADB

Hence, by using AA similarity criterion,

Î”AEP ~ Î”ADB

(iv) In Î”PDC and Î”BEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion,

Î”PDC ~ Î”BEC

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Î”ABE ~ Î”CFB.

**Answer**

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

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

∴ Î”ABE ~ Î”CFB (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 Î”ABC and Î”EFG respectively. If Î”ABC ~ Î”FEG, Show that:

(i) CD/GH = AC/FG
(i) Î”ABC ~ Î”AMP

(ii) CA/PA = BC/MP

**Answer**
(i) In Î”ABC and Î”AMP, we have

∠A = ∠A (common angle)

∠ABC = ∠AMP = 90° (each 90°)

∴ Î”ABC ~ Î”AMP (By AA similarity criterion)

(ii) As, Î”ABC ~ Î”AMP (By AA similarity criterion)

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

Hence, CA/PA = BC/MP

(ii) Î”DCB ~ Î”HGE

(iii) Î”DCA ~ Î”HGF

**Answer**

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)

In Î”ACD and Î”FGH,

∠A = ∠F (Proved above)

∠ACD = ∠FGH (Proved above)

∴ Î”ACD ~ Î”FGH (By AA similarity criterion)

⇒ CD/GH = AC/FG

(ii) In Î”DCB and Î”HGE,

∠DCB = ∠HGE (Proved above)

∠B = ∠E (Proved above)

∴ Î”DCB ~ Î”HGE (By AA similarity criterion)

∠DCB = ∠HGE (Proved above)

∠B = ∠E (Proved above)

∴ Î”DCB ~ Î”HGE (By AA similarity criterion)

(iii) In Î”DCA and Î”HGF,

∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

∴ Î”DCA ~ Î”HGF (By AA similarity criterion)

∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

∴ Î”DCA ~ Î”HGF (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 Î”ABD ~ Î”ECF.

**Answer**

It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In Î”ABD and Î”ECF,

∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Proved above)

∴ Î”ABD ~ Î”ECF (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 Î”PQR (see Fig 6.41). Show that Î”ABC ~ Î”PQR.

**Answer**

Given: Î”ABC and Î”PQR, AB, BC and median AD of Î”ABC are proportional to sides PQ, QR and median PM of Î”PQR

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

To Prove: Î”ABC ~ Î”PQR

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)

⇒ Î”ABD ~ Î”PQM [SSS similarity criterion]

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

⇒ ∠ABC = ∠PQR

In Î”ABC and Î”PQR

AB/PQ = BC/QR ...

**(i)**
∠ABC = ∠PQR ...

**(ii)**
From equation

**(i)**and**(ii)**, we get
Î”ABC ~ Î”PQR [By SAS similarity criterion]

^{2}= CB.CD

**Answer**

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ Î”ADC ~ Î”BAC (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 Î”ABC and Î”PQR in which AD and PM are medians such that AB/PQ = AC/PR = AD/PM

To Prove: Î”ABC ~ Î”PQR

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 Î”ABD and Î”CDE, we have

AD = DE [By Construction]

BD = DC [∴ AP is the median]

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

∴ Î”ABD ≅ Î”CDE [By SAS criterion of congruence]

⇒ AB = CE [CPCT] ...

**(i)**
Also, in Î”PQM and Î”MNR, we have

PM = MN [By Construction]

QM = MR [∴ PM is the median]

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

∴ Î”PQM = Î”MNR [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]

∴ Î”ACE ~ Î”PRN [By SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠A = ∠P ...

**(iii)**
Now, In Î”ABC and Î”PQR, we have

AB/PQ = AC/PR (Given)

∠A = ∠P [From

**(iii)**]
∴ Î”ABC ~ Î”PQR [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 Î”ABC and Î”DEF,

∠C = ∠E (angular elevation of sum)

∠B = ∠F = 90°

∴ Î”ABC ~ Î”DEF (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 Î”ABD and Î”PQM,

∠B = ∠Q [Using equation

**(ii)**]

AB/PQ = BD/QM [Using equation

**(iv)**]

∴ Î”ABD ~ Î”PQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/PM.

Page No: 143

**Exercise 6.4**

1. Let Î”ABC ~ Î”DEF 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 Î”ABC = 64 cm

^{2}

Area of Î”DEF = 121 cm

^{2}

EF = 15.4 cm

and, Î”ABC ~ Î”DEF∴ Area of Î”ABC/Area of Î”DEF = 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 Î”AOB and Î”COD, we have

∠1 = ∠2 (Alternate angles)

∠3 = ∠4 (Alternate angles)

∠5 = ∠6 (Vertically opposite angle)

∴ Î”AOB ~ Î”COD [By AAA similarity criterion]

Now, Area of (Î”AOB)/Area of (Î”COD)

= 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 (Î”AOB)/Area of (Î”COD)

= 4CD

^{2}/CD = 4/1

Hence, the required ratio of the area of Î”AOB and Î”COD = 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 (Î”ABC)/area (Î”DBC) = AO/DO.

**Answer**

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

To Prove: area (Î”ABC)/area (Î”DBC) = AO/DO.

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

In Î”APO and Î”DMO,

∠APO = ∠DMO (Each equals to 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ Î”APO ~ Î”DMO (By AA similarity criterion)∴ AP/DM = AO/DO

⇒ area (Î”ABC)/area (Î”DBC) = AO/DO.

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

**Answer**

Given: Î”ABC and Î”PQR are similar and equal in area.

To Prove: Î”ABC ≅ Î”PQR

Proof: Since, Î”ABC ~ Î”PQR

∴ Area of (Î”ABC)/Area of (Î”PQR) = BC

^{2}/QR

^{2}

⇒ BC

^{2}/QR

^{2}=1 [Since, Area(Î”ABC) = (Î”PQR)

⇒ BC

^{2}/QR

^{2}

⇒ BC = QR

Similarly, we can prove that

AB = PQ and AC = PR

Thus, Î”ABC ≅ Î”PQR [BY SSS criterion of congruence]

5. D, E and F are respectively the mid-points of sides AB, BC and CA of Î”ABC. Find the ratio of the area of Î”DEF and Î”ABC.

**Answer**

To Find: area(Î”DEF) and area(Î”ABC)

Solution: In Î”ABC, 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 Î”FBD and Î”DEF, we have

FB = DE (Opposite sides of parallelogram BDEF)

FD = FD (Common)

BD = FE (Opposite sides of parallelogram BDEF)

∴ Î”FBD ≅ Î”DEF

Similarly, we can prove that

Î”AFE ≅ Î”DEF

Î”EDC ≅ Î”DEF

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

So, area(Î”FBD) = area(Î”DEF) ...

**(i)**

area(Î”AFE) = area(Î”DEF) ...

**(ii)**

and, area(Î”EDC) = area(Î”DEF) ...

**(iii)**

Now, area(Î”ABC) = area(Î”FBD) + area(Î”DEF) + area(Î”AFE) + area(Î”EDC) ...

**(iv)**

area(Î”ABC) = area(Î”DEF) + area(Î”DEF) + area(Î”DEF) + area(Î”DEF)

⇒ area(Î”DEF) = 1/4area(Î”ABC) [From

**(i)**,

**(ii)**and

**(iii)**]

⇒ area(Î”DEF)/area(Î”ABC) = 1/4

Hence, area(Î”DEF):area(Î”ABC) = 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 Î”ABC ~ Î”DEF.

To Prove: area(Î”ABC)/area(Î”DEF) = AM

^{2}/DN^{2}Proof: Î”ABC ~ Î”DEF (Given)

∴ area(Î”ABC)/area(Î”DEF) = (AB

^{2}/DE

^{2}) ...

**(i)**

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

**(ii)**

In Î”ABM and Î”DEN, we have

∠B = ∠E [Since Î”ABC ~ Î”DEF]

AB/DE = BM/EN [Prove in

**(i)**]

∴ Î”ABC ~ Î”DEF [By SAS similarity criterion]

⇒ AB/DE = AM/DN ...

**(iii)**

∴ Î”ABM ~ Î”DEN

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

∴ area(Î”ABC)/area(Î”DEF) = 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. Î”APC and Î”BQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

To Prove: area(Î”BQC) = 1/2area(Î”APC)

Proof: Î”APC and Î”BQC are both equilateral triangles (Given)

∴ Î”APC ~ Î”BQC [AAA similarity criterion]

∴ area(Î”APC)/area(Î”BQC) = (AC

^{2}/BC

^{2}) = AC

^{2}/BC

^{2}

⇒ area(Î”BQC) = 1/2area(Î”APC)

**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**

Î”ABC and Î”BDE are two equilateral triangle. D is the mid point of BC.

∴ BD = DC = 1/2BC

Let each side of triangle is 2

*a*.
As, Î”ABC ~ Î”BDE

∴ area(Î”ABC)/area(Î”BDE) = 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 Î”ABC ~ Î”DEF (Given)

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

∴ area(Î”ABC)/area(Î”DEF) = 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(Î”ABC)/area(Î”DEF) = (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: Î”PQR is right angled at P is a point on QR such that PM ⊥QR.

To prove: PM

^{2}= QM × MR
Proof: In Î”PQM, we have

PQ

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

^{2}= PQ^{2}- QM^{2}...**(i)**
In Î”PMR, 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 Î”ADB and Î”CAB, we have

∠DAB = ∠ACB (Each equals to 90°)

∠ABD = ∠CBA (Common angle)

∴ Î”ADB ~ Î”CAB [AA similarity criterion]

⇒ AB/CB = BD/AB

⇒ AB

^{2}= CB × BD

(ii) Let ∠CAB =

*x*

In Î”CBA,

∠CBA = 180° - 90° -

*x*

∠CBA = 90° -

*x*

Similarly, in Î”CAD

∠CAD = 90° - ∠CBA = 90° -

*x*

∠CDA = 180° - 90° - (90° -

*x*)

∠CDA =

*x*

In Î”CBA and Î”CAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA (Each equals to 90°)

∴ Î”CBA ~ Î”CAD [By AAA similarity criterion]

⇒ AC/DC = BC/AC

⇒ AC

^{2}= DC × BC

(iii) In Î”DCA and Î”DAB, we have

∠DCA = ∠DAB (Each equals to 90°)

∠CDA = ∠ADB (common angle)

∴ Î”DCA ~ Î”DAB [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 Î”ABC is an isosceles triangle right angled at C.

In Î”ACB, ∠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 Î”ABC is an isosceles triangle having AC = BC and AB

^{2}= 2AC^{2}
In Î”ACB,

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 Î”ABC 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 Î”ADB and Î”ADC, we have

AB = AC [Given]

AD = AD [Given]

∠ADB = ∠ADC [equal to 90°]

Therefore, Î”ADB ≅ Î”ADC by RHS congruence.

Hence, BD = DC [by CPCT]

In right angled Î”ADB,

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 Î”AOB,

∠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 Î”AOF, we have

OA

Similarly, in Î”BOD^{2}= OF^{2}+ AF^{2}OB

^{2}= OD

^{2}+ BD

^{2}

Similarly, in Î”COE

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 Î”AOB, 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**

Therefore, CP = 11 - 6 = 5 m

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

Applying Pythagoras theorem for Î”APC, 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 Î”ACE, we get

AC

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

DE

^{2}=^{ }CD^{2}+ CE^{2}
Applying Pythagoras theorem in Î”ABC, 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 Î” 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 Î”ABC, 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 Î”ABC.

∴ 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 Î”ADE, 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 Î”ABE, 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 Î”ABC, 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, Î”ABC, 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 Î”PQR. 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 Î”ABC, 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) Î” APC ~ Î” 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) Î” PAC ~ Î” 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 Î”ABC 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

**Go Back To NCERT Solutions for Class 10th Maths**