# NCERT Solutions for Class 10 Maths Chapter 6 Triangles

## NCERT Solutions for Class 10 Maths Chapter 6 â€“ Download Free PDF

NCERT Solutions for Class 10 Maths Chapter 6 Triangles are provided here, which is considered to be one of the most important study materials for the students studying in CBSE Class 10. Chapter 6 of NCERT Solutions for Class 10 Maths is well structured in accordance with the CBSE Syllabus for 2023-24. It covers a vast topic, including a number of rules and theorems. Students often tend to get confused about which theorem to use while solving a variety of questions.

The solutions provided at BYJUâ€™S are designed in such a way that every step is explained clearly and in detail. The Solutions for NCERT Class 10 Maths are prepared by the subject experts to help students prepare better for their board exams. These solutions will be helpful not only for exam preparations, but also in solving homework and assignments.

The CBSE Class 10 examination often asks questions, either directly or indirectly, from the NCERT textbooks. Thus, the NCERT Solutions for Chapter 6 Triangles of Class 10 Maths is one of the best resources to prepare, and equip oneself to solve any type of questions in the exam, from the chapter. It is highly recommended that the students practise these NCERT Solutions on a regular basis to excel in the Class 10 board examinations.

## Exercise 6.1 Page: 122

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

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

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

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

(b) Proportional

2. Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures

Solution:

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

Solution:

From the given two figures, we can see their corresponding angles are different or unequal. Therefore, they are not similar.

## Exercise 6.2 Page: 128

1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:

(i) Given, inÂ â–³ ABC, DEâˆ¥BC

âˆ´ AD/DB = AE/EC [Using Basic proportionality theorem]

â‡’1.5/3 = 1/EC

â‡’EC = 3/1.5

EC = 3Ã—10/15 = 2 cm

Hence, EC = 2 cm.

(ii) Given, inÂ â–³ ABC, DEâˆ¥BC

âˆ´ AD/DB = AE/EC [Using Basic proportionality theorem]

â‡’ AD/7.2 = 1.8 / 5.4

â‡’ AD = 1.8 Ã—7.2/5.4 = (18/10)Ã—(72/10)Ã—(10/54) = 24/10

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

Solution:

Given, in Î”PQR, E and F are two points on side PQ and PR, respectively. See the figure below;

Â

(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm

Therefore, by using Basic proportionality theorem, we get,

PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3

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

So, we get, PE/EQ â‰  PF/FR

Hence, EF is not parallel to QR.

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

Therefore, by using Basic proportionality theorem, we get,

PE/QE = 4/4.5 = 40/45 = 8/9

And, PF/RF = 8/9

So, we get here,

PE/QE = PF/RF

Hence, EF is parallel to QR.

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

From the figure,

EQ = PQ â€“ PE = 1.28 â€“ 0.18 = 1.10 cm

And, FR = PR â€“ PF = 2.56 â€“ 0.36 = 2.20 cm

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55â€¦â€¦â€¦â€¦. (i)

And, PE/FR = 0.36/2.20 = 36/220 = 9/55â€¦â€¦â€¦â€¦ (ii)

So, we get here,

PE/EQ = PF/FR

Hence, EF is parallel to QR.

3. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD

Solution:

In the given figure, we can see, LM || CB,

By using basic proportionality theorem, we get,

AM/AB = AL/ACâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

Similarly, given, LN || CD and using basic proportionality theorem,

From equationÂ (i)Â andÂ (ii), we get,

Hence, proved.

4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC

Solution:

In Î”ABC, given as, DE || AC

Thus, by using Basic Proportionality Theorem, we get,

âˆ´BD/DA = BE/EC â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

In Â Î”BAE, given as, DF || AE

Thus, by using Basic Proportionality Theorem, we get,

âˆ´BD/DA = BF/FE â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

From equationÂ (i)Â andÂ (ii), we get

BE/EC = BF/FE

Hence, proved.

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

Solution:

Given,

In Î”PQO, DE || OQ

So by using Basic Proportionality Theorem,

PD/DO = PE/EQâ€¦â€¦â€¦â€¦â€¦â€¦ ..(i)

Again given, in Î”POR, DF || OR,

So by using Basic Proportionality Theorem,

PD/DO = PF/FRâ€¦â€¦â€¦â€¦â€¦â€¦â€¦ (ii)

From equationÂ (i)Â andÂ (ii), we get,

PE/EQ = PF/FR

Therefore, by converse of Basic Proportionality Theorem,

EFÂ ||Â QR, in Î”PQR.

6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:

Given here,

In Î”OPQ, AB || PQ

By using Basic Proportionality Theorem,

OA/AP = OB/BQâ€¦â€¦â€¦â€¦â€¦.(i)

Also given,

In Î”OPR, AC || PR

By using Basic Proportionality Theorem

âˆ´ OA/AP = OC/CRâ€¦â€¦â€¦â€¦â€¦(ii)

From equationÂ (i)Â andÂ (ii), we get,

OB/BQ = OC/CR

Therefore, by converse of Basic Proportionality Theorem,

In Î”OQR, BC ||Â QR.

7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution:

Given, in Î”ABC, D is the midpoint of AB such that AD=DB.

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

We have to prove that E is the mid point of AC.

Since, D is the mid-point of AB.

In Î”ABC, DE || BC,

By using Basic Proportionality Theorem,

From equation (i), we can write,

â‡’ 1 = AE/EC

âˆ´ AE = EC

Hence, proved, E is the midpoint 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).

Solution:

Given, in Î”ABC, D and E are the mid points of AB and AC, respectively, such that,

We have to prove that: DE || BC.

Since, D is the midpoint of AB

Â

Also given, E is the mid-point of AC.

âˆ´ AE=EC

â‡’ AE/EC = 1

From equationÂ (i)Â andÂ (ii), we get,

By converse of Basic Proportionality Theorem,

DE || BC

Hence, proved.

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.

Solution:

Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.

We have to prove, AO/BO = CO/DO

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

In Î”ADC, we have OE || DC

Therefore, by using Basic Proportionality Theorem

AE/ED = AO/CO â€¦â€¦â€¦â€¦â€¦..(i)

Now, In Î”ABD, OE || AB

Therefore, by using Basic Proportionality Theorem

DE/EA = DO/BOâ€¦â€¦â€¦â€¦â€¦.(ii)

From equationÂ (i)Â andÂ (ii), we get,

AO/CO = BO/DO

â‡’AO/BO = CO/DO

Hence, proved.

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

Solution:

Given, Quadrilateral ABCD where AC and BD intersect each other at O such that,

AO/BO = CO/DO.

We have to prove here, ABCD is a trapezium

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

In Î”DAB, EO || AB

Therefore, by using Basic Proportionality Theorem

DE/EA = DO/OB â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

Also, given,

AO/BO = CO/DO

â‡’ AO/CO = BO/DO

â‡’ CO/AO = DO/BO

â‡’DO/OB = CO/AO â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

From equationÂ (i)Â andÂ (ii), we get

DE/EA = CO/AO

Therefore, by using converse of Basic Proportionality Theorem,

EO || DC also EO || AB

â‡’ AB || DC.

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

## Exercise 6.3 Page: 138

1. State which pairs of triangles in the figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Solution:

(i) Given, in Î”ABC and Î”PQR,

âˆ A =Â âˆ P = 60Â°

âˆ B = âˆ Q = 80Â°

âˆ C = âˆ R = 40Â°

Therefore, by AAA similarity criterion,

âˆ´ Î”ABC ~ Î”PQR

(ii) Given, in Â Î”ABC and Î”PQR,

AB/QR = 2/4 = 1/2,

BC/RP = 2.5/5 = 1/2,

CA/PA = 3/6 = 1/2

By SSS similarity criterion,

Î”ABC ~ Î”QRP

(iii) Given, in Î”LMP and Î”DEF,

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 â‰  LM/EF

Therefore, Î”LMP and Î”DEF are not similar.

(iv) In Î”MNL and Î”QPR, it is given,

MN/QP = ML/QR = 1/2

âˆ M = âˆ Q = 70Â°

Therefore, by SAS similarity criterion

âˆ´ Î”MNL ~ Î”QPR

(v) In Î”ABC and Î”DEF, given that,

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

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

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

â‡’ âˆ B â‰  âˆ F

Hence, Î”ABC and Î”DEF are not similar.

(vi) In Î”DEF, by sum of angles of triangles, we know that,

âˆ DÂ +Â âˆ EÂ +Â âˆ F = 180Â°

â‡’ 70Â°Â + 80Â°Â + âˆ F = 180Â°

â‡’ âˆ F = 180Â° â€“ 70Â° â€“ 80Â°

â‡’ âˆ F = 30Â°

Similarly, In Î”PQR,

âˆ PÂ +Â âˆ QÂ +Â âˆ R = 180 (Sum of angles of Î”)

â‡’ âˆ PÂ + 80Â°Â + 30Â° = 180Â°

â‡’ âˆ P = 180Â° â€“ 80Â° -30Â°

â‡’ âˆ P = 70Â°

Now, comparing both the triangles, Î”DEF and Î”PQR, we have

âˆ D = âˆ P = 70Â°

âˆ F = âˆ Q = 80Â°

âˆ F = âˆ R = 30Â°

Therefore, by AAA similarity criterion,

Hence, Î”DEF ~ Î”PQR

2.Â Â In figure 6.35, Î”ODC ~ Î”OBA, âˆ  BOC = 125Â° and âˆ  CDO = 70Â°. Find âˆ  DOC, âˆ  DCO and âˆ  OAB.

Solution:

As we can see from the figure, DOB is a straight line.

Therefore, âˆ DOC + âˆ  COB = 180Â°

â‡’ âˆ DOC = 180Â° â€“ 125Â° (Given, âˆ  BOC = 125Â°)

= 55Â°

In Î”DOC, sum of the measures of the angles of a triangle is 180Âº

Therefore, âˆ DCO + âˆ  CDO + âˆ  DOC = 180Â°

â‡’ âˆ DCO + 70Âº + 55Âº = 180Â°(Given, âˆ  CDO = 70Â°)

â‡’ âˆ DCO = 55Â°

It is given that, Î”ODC ~ Î”OBA,

Therefore, Î”ODC ~ Î”OBA.

Hence, corresponding angles are equal in similar triangles

âˆ OAB = âˆ OCD

â‡’ âˆ  OAB = 55Â°

âˆ OAB = âˆ OCD

â‡’ âˆ OAB = 55Â°

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

Solution:

In Î”DOC and Î”BOA,

AB || CD, thus alternate interior angles will be equal,

âˆ´âˆ CDO = âˆ ABO

Similarly,

âˆ DCO = âˆ BAO

Also, for the two triangles Î”DOC and Î”BOA, vertically opposite angles will be equal;

âˆ´âˆ DOC = âˆ BOA

Hence, by AAA similarity criterion,

Î”DOC ~ Î”BOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

â‡’OA/OC = OB/OD

Hence, proved.

Â

4.Â In the fig.6.36, QR/QS = QT/PR andÂ âˆ 1 =Â âˆ 2. Show thatÂ Î”PQS ~Â Î”TQR.

Solution:

In Î”PQR,

âˆ PQR = âˆ PRQ

âˆ´ PQ = PR â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

Given,

QR/QS = QT/PRUsingÂ equation (i), we get

QR/QS = QT/QPâ€¦â€¦â€¦â€¦â€¦â€¦.(ii)

In Î”PQS and Î”TQR, by equation (ii),

QR/QS = QT/QP

âˆ Q = âˆ Q

âˆ´ Î”PQS ~ Î”TQR [By SAS similarity criterion]

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

Solution:

Given, S and T are point on sides PR and QR of Î”PQR

And âˆ P = âˆ RTS.

In Î”RPQ and Î”RTS,

âˆ RTS = âˆ QPS (Given)

âˆ R = âˆ R (Common angle)

âˆ´ Î”RPQ ~ Î”RTS (AA similarity criterion)

6. In the figure, if Î”ABE â‰… Î”ACD, show that Î”ADE ~ Î”ABC.

Solution:

Given, Î”ABE â‰… Î”ACD.

âˆ´ AB = AC [By CPCT] â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)

And, AD = AE [By CPCT] â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

In Î”ADE and Î”ABC, dividing eq.(ii) by eq(i),

âˆ A = âˆ A [Common angle]

âˆ´ Î”ADE ~ Î”ABC [SAS similarity criterion]

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

(i) Î”AEP ~ Î”CDP
(ii) Î”ABD ~ Î”CBE
(iv) Î”PDC ~ Î”BEC

Solution:

Given, altitudes AD and CE of Î”ABC intersect each other at the point P.

(i) In Î”AEP and Î”CDP,

âˆ AEP = âˆ CDP (90Â° each)

âˆ APE = âˆ CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

Î”AEP ~ Î”CDP

(ii) In Î”ABD and Î”CBE,

âˆ ADB = âˆ CEB ( 90Â° each)

âˆ ABD = âˆ CBE (Common Angles)

Hence, by AA similarity criterion,

Î”ABD ~ Î”CBE

âˆ AEP = âˆ ADB (90Â° each)

âˆ PAE = âˆ DAB (Common Angles)

Hence, by AA similarity criterion,

(iv) In Î”PDC and Î”BEC,

âˆ PDC = âˆ BEC (90Â° each)

âˆ PCD = âˆ BCE (Common angles)

Hence, by 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.

Solution:

Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,

In Î”ABE and Î”CFB,

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

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

âˆ´ Î”ABE ~ Î”CFB (AA similarity criterion)

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

(i) Î”ABC ~ Î”AMP

(ii) CA/PA = BC/MP

Solution:

Given, ABC and AMP are two right triangles, right angled at B and M, respectively.

(i) In Î”ABC and Î”AMP, we have,

âˆ CAB = âˆ MAP (common angles)

âˆ ABC = âˆ AMP = 90Â° (each 90Â°)

âˆ´ Î”ABC ~ Î”AMP (AA similarity criterion)

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

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

Hence, CA/PA = BC/MP

10. CD and GH are respectively the bisectors of âˆ ACB and âˆ 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
(ii) Î”DCBÂ ~Â Î”HGE
(iii) Î”DCA ~ Î”HGF

Solution:

Given, CD and GH are respectively the bisectors of âˆ ACB and âˆ EGF such that D and H lie on sides AB and FE of Î”ABC and Î”EFG, respectively.

(i) From the given condition,

Î”ABC ~ Î”FEG.

âˆ´ âˆ A = âˆ F, âˆ B = âˆ E, and âˆ ACB = âˆ FGE

Since, âˆ ACB = âˆ FGE

âˆ´ âˆ ACD = âˆ FGH (Angle bisector)

And, âˆ DCB = âˆ HGE (Angle bisector)

In Î”ACD and Î”FGH,

âˆ A = âˆ F

âˆ ACD = âˆ FGH

âˆ´ Î”ACD ~ Î”FGH (AA similarity criterion)

â‡’CD/GH = AC/FG

(ii) In Î”DCB and Î”HGE,

âˆ DCB = âˆ HGE (Already proved)

âˆ B = âˆ E (Already proved)

âˆ´ Î”DCB ~ Î”HGE (AA similarity criterion)

(iii) In Î”DCA and Î”HGF,

âˆ ACD = âˆ FGH (Already proved)

âˆ A = âˆ F (Already proved)

âˆ´ Î”DCA ~ Î”HGF (AA similarity criterion)

11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD âŠ¥ BC and EF âŠ¥ AC, prove that Î”ABD ~ Î”ECF.

Solution:

Given, ABC is an isosceles triangle.

âˆ´ AB = AC

â‡’ âˆ ABD = âˆ ECF

In Î”ABD and Î”ECF,

âˆ ADB = âˆ EFC (Each 90Â°)

âˆ´ Î”ABD ~ Î”ECF (using AA similarity criterion)

12.Â 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.

Solution:

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

We have to prove: Î”ABC ~ Î”PQR

As we know here,

â‡’AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)

â‡’ Î”ABD ~ Î”PQM [SSS similarity criterion]

âˆ´ âˆ ABD = âˆ PQM [Corresponding angles of two similar triangles are equal]

â‡’ âˆ ABC = âˆ PQR

In Î”ABC and Î”PQR

AB/PQ = BC/QR â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)

âˆ ABC = âˆ PQR â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

From equationÂ (i)Â andÂ (ii), we get,

Î”ABC ~ Î”PQR [SAS similarity criterion]

13. D is a point on the side BC of a triangle ABC such that âˆ ADC = âˆ BAC. Show that CA2Â = CB.CD

Solution:

Given, D is a point on the side BC of a triangle ABC such that âˆ ADC = âˆ BAC.

âˆ ACD = âˆ BCA (Common angles)

âˆ´ Î”ADC ~ Î”BAC (AA similarity criterion)

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

âˆ´ CA/CB = CD/CA

â‡’Â CA2Â = CB.CD.

Hence, proved.

14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show thatÂ Î”ABC ~ Î”PQR.

Solution:

Given: Two triangles Î”ABC and Î”PQR in which AD and PM are medians such that;

We have to prove, Î”ABC ~ Î”PQR

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

In Î”ABD and Î”CDE, we have

AD = DE Â [By Construction.]

BD = DC [Since, AP is the median]

and, âˆ ADB = âˆ CDE [Vertically opposite angles]

âˆ´ Î”ABDÂ â‰…Â Î”CDE [SAS criterion of congruence]

â‡’ AB = CE [By CPCT] â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

Also, in Î”PQM and Î”MNR,

PM = MN [By Construction.]

QM = MR [Since, PM is the median]

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

âˆ´ Î”PQM = Î”MNR [SAS criterion of congruence]

â‡’ PQ = RN [CPCT] â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

Now, AB/PQ = AC/PR = AD/PM

From equation (i)Â andÂ (ii),

â‡’ CE/RN = AC/PR = 2AD/2PM

â‡’ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]

âˆ´ Î”ACE ~ Î”PRN [SSS similarity criterion]

Therefore, âˆ 2 = âˆ 4

Similarly, âˆ 1 = âˆ 3

âˆ´ âˆ 1Â +Â âˆ 2Â =Â âˆ 3Â +Â âˆ 4

â‡’ âˆ A = âˆ P â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(iii)

Now, in Î”ABC and Î”PQR, we have

From equation (iii),

âˆ A = âˆ P

âˆ´ Î”ABC ~ Î”PQR [ SAS similarity criterion]

15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:

Given, Length of the vertical pole = 6m

Shadow of the pole = 4 m

Let Height of tower =Â hÂ m

Length of shadow of the tower = 28 m

In Î”ABC and Î”DEF,

âˆ C = âˆ E (angular elevation of sum)

âˆ B = âˆ F = 90Â°

âˆ´ Î”ABC ~ Î”DEF (AA similarity criterion)

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

âˆ´ 6/h = 4/28

â‡’h = (6Ã—28)/4

â‡’Â hÂ =Â 6 Ã— 7

â‡’Â hÂ = 42 m

Hence, the height of the tower is 42 m.

16. If AD and PM are medians of triangles ABC and PQR, respectively whereÂ Î”ABC ~Â Î”PQR prove that AB/PQ = AD/PM.

Solution:

Given, Î”ABC ~ Î”PQR

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

âˆ´AB/PQ = AC/PR = BC/QRâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

Also, âˆ A = âˆ P, âˆ B = âˆ Q, âˆ C = âˆ R â€¦â€¦â€¦â€¦.â€¦..(ii)

Since AD and PM are medians, they will divide their opposite sides.

âˆ´ BD = BC/2 and QM = QR/2 â€¦â€¦â€¦â€¦â€¦..â€¦â€¦â€¦â€¦.(iii)

From equationsÂ (i)Â andÂ (iii), we get

AB/PQ = BD/QM â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(iv)

In Î”ABD and Î”PQM,

From equation (ii), we have

âˆ B = âˆ Q

From equationÂ (iv), we have,

AB/PQ = BD/QM

âˆ´ Î”ABD ~ Î”PQM (SAS similarity criterion)

Exercise 6.4 Page: 143

1. Let Î”ABC ~ Î”DEF and their areas be, respectively, 64 cm2Â and 121 cm2. If EF = 15.4 cm, find BC.

Solution: Given, Î”ABC ~ Î”DEF,

Area of Î”ABC = 64 cm2

Area of Î”DEF =Â 121 cm2

EF = 15.4 cm

As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,

= AC2/DF2Â = BC2/EF2

âˆ´ 64/121 = BC2/EF2

â‡’ (8/11)2Â = (BC/15.4)2

â‡’ 8/11 = BC/15.4

â‡’ BC = 8Ã—15.4/11

â‡’ BC = 8 Ã— 1.4

â‡’ 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.

Solution:

Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

In Î”AOB and Î”COD, we have

âˆ 1 = âˆ 2 (Alternate angles)

âˆ 3 = âˆ 4 (Alternate angles)

âˆ 5 = âˆ 6 (Vertically opposite angle)

âˆ´ Î”AOB ~ Î”COD [AAA similarity criterion]

As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,

Area of (Î”AOB)/Area of (Î”COD) = AB2/CD2

= (2CD)2/CD2Â [âˆ´ AB = 2CD]

âˆ´ Area of (Î”AOB)/Area of (Î”COD)

= 4CD2/CD2 = 4/1

Hence, the required ratio of the area of Î”AOB and Î”COD = 4:1

3. In the figure, 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.

Solution:

Given, ABC and DBC are two triangles on the same base BC. AD intersects BC at O.

We have to prove: Area (Î”ABC)/Area (Î”DBC) = AO/DO

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

We know that area of a triangle = 1/2Â Ã— BaseÂ Ã— Height

In Î”APO and Î”DMO,

âˆ APO = âˆ DMO (Each 90Â°)

âˆ AOP = âˆ DOM (Vertically opposite angles)

âˆ´ Î”APO ~ Î”DMO (AA similarity criterion)

âˆ´ AP/DM = AO/DO

â‡’ Area (Î”ABC)/Area (Î”DBC) = AO/DO.

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

Solution:

Say Î”ABC and Î”PQR are two similar triangles and equal in area

Now let us prove Î”ABC â‰… Î”PQR.

Since, Î”ABC ~ Î”PQR

âˆ´ Area of (Î”ABC)/Area of (Î”PQR) = BC2/QR2

â‡’ BC2/QR2Â =1 [Since, Area(Î”ABC) = (Î”PQR)

â‡’ BC2/QR2

â‡’ BC = QR

Similarly, we can prove that

AB = PQ and AC = PR

Thus, Î”ABC â‰… Î”PQR [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.

Solution:

D, E, and F are the mid-points ofÂ Î”ABC
âˆ´Â DE || AC and
DE = (1/2) ACÂ (Midpoint theorem) â€¦. (1)
InÂ Â Î”BEDÂ andÂ Î”BCA
âˆ BEDÂ =Â âˆ BCAÂ (Corresponding angles)
âˆ BDEÂ =Â âˆ BACÂ (Corresponding angles)
âˆ EBDÂ =Â âˆ CBA (Common angles)
âˆ´Î”BEDâˆ¼Î”BCAÂ (AAA similarity criterion)
ar (Î”BED) / ar (Î”BCA)=(DE/AC)2
â‡’ar (Î”BED) / ar (Î”BCA)Â = (1/4)Â [From (1)]
â‡’ar (Î”BED)Â = (1/4) ar (Î”BCA)
Similarly,
ar (Î”CFE)Â = (1/4) ar (CBA)Â and ar (Î”ADF)Â = (1/4) ar (Î”ADF)Â = (1/4) ar (Î”ABC)
Also,
ar (Î”DEF)Â = ar (Î”ABC)Â âˆ’Â [ar (Î”BED)Â + ar (Î”CFE)Â + ar (Î”ADF)]
â‡’arÂ (Î”DEF)Â = arÂ (Î”ABC)Â âˆ’ (3/4) arÂ (Î”ABC)Â = (1/4)Â ar (Î”ABC)
â‡’ar (Î”DEF) /Â ar (Î”ABC)Â = (1/4)

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

Solution:

Given: AM and DN are the medians of triangles ABC and DEF respectively and Î”ABC ~ Î”DEF.

We have to prove: Area(Î”ABC)/Area(Î”DEF) = AM2/DN2

Since, Î”ABC ~ Î”DEF (Given)

âˆ´ Area(Î”ABC)/Area(Î”DEF) = (AB2/DE2) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

and, AB/DE = BC/EF = CA/FD â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

In Î”ABM and Î”DEN,

Since Î”ABC ~ Î”DEF

âˆ´ âˆ B = âˆ E

AB/DE = BM/EN [Already Proved in equationÂ (i)]

âˆ´ Î”ABC ~ Î”DEF [SAS similarity criterion]

â‡’ AB/DE = AM/DN â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

âˆ´ Î”ABM ~ Î”DEN

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

âˆ´ area(Î”ABC)/area(Î”DEF) = AB2/DE2Â = AM2/DN2

Hence, proved.

7.Â 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.

Solution:

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.

Area(Î”BQC) = Â½ Area(Î”APC)

Since, Î”APC and Î”BQC are both equilateral triangles, as per given,

âˆ´ Î”APC ~ Î”BQC [AAA similarity criterion]

âˆ´ area(Î”APC)/area(Î”BQC) = (AC2/BC2) = AC2/BC2

Since, Diagonal = âˆš2 side = âˆš2 BC = AC

â‡’ area(Î”APC) = 2Â Ã— area(Î”BQC)

â‡’ area(Î”BQC) = 1/2area(Î”APC)

Hence, proved.

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

Solution:

Given, Î”ABC and Î”BDE are two equilateral triangle. D is the midpoint of BC.

Â

âˆ´ BD = DC = 1/2BC

Let each side of triangle is 2a.

As, Î”ABC ~ Î”BDE

âˆ´ Area(Î”ABC)/Area(Î”BDE) = AB2/BD2Â = (2a)2/(a)2Â = 4a2/a2Â = 4/1 = 4:1

Hence, the correct answer is (C).

9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81

Solution:

Given, Sides of two similar triangles are in the ratio 4 : 9.

Let ABC and DEF are two similar triangles, such that,

Î”ABC ~ Î”DEF

And AB/DE = AC/DF = BC/EF = 4/9

As, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,

âˆ´ Area(Î”ABC)/Area(Î”DEF) = AB2/DE2Â

âˆ´ Area(Î”ABC)/Area(Î”DEF) =Â (4/9)2Â = 16/81 = 16:81

Hence, the correct answer is (D).

## Exercise 6.5 Page: 150

1. Â Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

(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

Solution:

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

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

49 + 576 = 625

(7)2Â + (24)2Â = (25)2

Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given, 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.

Clearly, 9 + 36 â‰  64

Or, 32Â + 62Â â‰  82

Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

Hence, the given triangle does not satisfies Pythagoras theorem.

(iii)Â Given, sides of triangleâ€™s are 50 cm, 80 cm, and 100 cm.

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

However, 2500 + 6400 â‰  10000

Or, 502Â + 802Â â‰  1002

As you can see, 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 does not satisfies Pythagoras theorem.

Hence, it is not a right triangle.

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

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

Thus, 144 +25 = 169

Or, 122Â + 52Â = 132

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

Hence, length of the hypotenuse of this triangle is 13 cm.

Â

2. PQR is a triangle right angled at P and M is a point on QR such that PM âŠ¥ QR. Show that PM2Â = QMÂ Ã—Â MR.

Solution:

Given, Î”PQR is right angled at P is a point on QR such that PMÂ âŠ¥QR

We have to prove, PM2Â = QMÂ Ã—Â MR

In Î”PQM, by Pythagoras theorem

PQ2Â = PM2Â + QM2

Or, PM2Â = PQ2Â â€“ QM2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

In Î”PMR, by Pythagoras theorem

PR2Â = PM2Â + MR2

Or, PM2Â = PR2Â â€“ MR2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

AddingÂ equation, (i)Â andÂ (ii), we get,

2PM2Â =Â (PQ2Â + PM2) â€“ (QM2Â + MR2)

= QR2Â â€“ QM2Â â€“ MR2 Â  Â  Â Â Â [âˆ´ QR2Â = PQ2Â + PR2]

= (QMÂ + MR)2Â â€“ QM2Â â€“ MR2

= 2QM Ã— MR

âˆ´ PM2Â = QMÂ Ã—Â MR

3. In Figure, ABD is a triangle right angled at AÂ and AC âŠ¥ BD. Show that
(i) AB2Â = BC Ã—Â BD
(ii) AC2Â = BC Ã—Â DC
(iii) AD2Â = BDÂ Ã—Â CD

Â

Solution:

âˆ DAB = âˆ ACB (Each 90Â°)

âˆ ABD = âˆ CBA (Common angles)

âˆ´ Î”ADB ~ Î”CAB [AA similarity criterion]

â‡’ AB/CB = BD/AB

â‡’ AB2Â = CB Ã—Â BD

(ii) Let âˆ CAB = x

In Î”CBA,

âˆ CBA = 180Â° â€“ 90Â° â€“ x

âˆ CBA = 90Â° â€“ x

âˆ CAD = 90Â° â€“ âˆ CBA

= 90Â° â€“Â x

âˆ CDA = 180Â° â€“ 90Â° â€“ (90Â° â€“Â x)

âˆ CDA =Â x

In Î”CBA and Î”CAD, we have

âˆ CAB = âˆ CDA

âˆ ACB = âˆ DCA (Each 90Â°)

âˆ´ Î”CBA ~ Î”CAD [AAA similarity criterion]

â‡’ AC/DC = BC/AC

â‡’ AC2Â = Â DC Ã— BC

(iii) In Î”DCA and Î”DAB,

âˆ DCA = âˆ DAB (Each 90Â°)

âˆ CDA = âˆ ADB (common angles)

âˆ´ Î”DCA ~ Î”DAB [AA similarity criterion]

â‡’ DC/DA = DA/DA

â‡’ AD2Â = BDÂ Ã—Â CD

4. ABC is an isosceles triangle right angled at C. Prove that AB2Â = 2AC2Â .

Solution:

Given, Î”ABC is an isosceles triangle right angled at C.

In Î”ACB, âˆ C = 90Â°

AC = BC (By isosceles triangle property)

AB2Â = AC2Â + BC2Â [By Pythagoras theorem]

= AC2Â +Â AC2Â [Since, AC = BC]

AB2Â = 2AC2

5.Â ABC is an isosceles triangle with AC = BC. IfÂ AB2Â = 2AC2, prove that ABC is a right triangle.

Solution:

Given, Î”ABC is an isosceles triangle having AC = BC and AB2Â = 2AC2

In Î”ACB,

AC = BC

AB2Â = 2AC2

AB2Â = AC2Â + AC2

= AC2Â + BC2Â [Since, AC = BC]

Hence, by Pythagoras theoremÂ Î”ABC is right angle triangle.

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

Solution:

Given, ABC is an equilateral triangle of side 2a.

AB = AC

Hence, BD = DC [by CPCT]

â‡’Â AD2 = 4a2Â â€“Â a2

7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Solution:

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

We have to prove, as per the question,

AB2Â + BC2Â + CD2Â + AD2Â = AC2Â + BD2

Since, the diagonals of a rhombus bisect each other at right angles.

Therefore, AO = CO and BO = DO

In Î”AOB,

âˆ AOB =Â 90Â°

AB2Â = AO2Â + BO2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â (i)Â [By Pythagoras theorem]

Similarly,

AD2Â = AO2Â + DO2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â (ii)

DC2Â = DO2Â + CO2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â (iii)

BC2Â = CO2Â + BO2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â (iv)

Adding equationsÂ (i) + (ii)Â + (iii)Â + (iv), we get,

AB2Â + AD2Â +Â DC2Â +Â BC2Â = 2(AO2Â + BO2Â + DO2Â + CO2)

= 4AO2Â + 4BO2Â [Since, AO = CO and BO =DO]

= (2AO)2Â + (2BO)2Â = AC2Â + BD2

AB2Â + AD2Â +Â DC2Â +Â BC2Â = AC2Â + BD2

Hence, proved.

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

ABC, OD âŠ¥ BC, OEÂ âŠ¥Â AC and OF âŠ¥ AB. Show that:
(i) OA2Â + OB2Â + OC2Â â€“ OD2Â â€“ OE2Â â€“ OF2Â = AF2Â + BD2Â + CE2Â ,
(ii) AF2Â + BD2Â + CE2Â = AE2Â + CD2Â + BF2.

Solution:

Given, in Î”ABC, O is a point in the interior of a triangle.

And OD âŠ¥ BC, OEÂ âŠ¥Â AC and OF âŠ¥ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in Î”AOF, we have

OA2Â = OF2Â +Â AF2

Similarly, inÂ Î”BOD

OB2Â = OD2Â + BD2

Similarly,Â inÂ Î”COE

OC2Â = OE2Â + EC2

OA2Â + OB2Â + OC2Â = OF2Â + AF2Â + OD2Â + BD2Â + OE2Â + EC2

OA2Â + OB2Â + OC2Â â€“ OD2Â â€“ OE2Â â€“ OF2Â = AF2Â + BD2Â + CE2.

(ii) AF2Â + BD2Â + EC2Â = (OA2Â â€“ OE2)Â + (OC2Â â€“ OD2)Â + (OB2Â â€“ OF2)

âˆ´ AF2Â + BD2Â + CE2Â = AE2Â + CD2Â + BF2.

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.

Solution:

Given, a ladder 10 m long reaches a window 8 m above the ground.

Let BA be the wall and AC be the ladder,

Therefore, by Pythagoras theorem,

AC2Â =Â AB2Â + BC2

102Â = 82Â + BC2

BC2Â = 100 â€“ 64

BC2Â = 36

BCÂ = 6m

Therefore, the distance of the foot of the ladder from the base of the wall isÂ 6 m.

10.Â 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?

Solution:

Given, 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.

Let AB be the pole and AC be the wire.

By Pythagoras theorem,

AC2Â =Â AB2Â + BC2

242Â = 182Â + BC2

BC2Â = 576 â€“ 324

BC2Â = 252

BCÂ = 6âˆš7m

Therefore, the distance from the base is 6âˆš7m.

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
Â hours?

Solution:

Given,

Speed of first aeroplane = 1000 km/hr

Distance covered by first aeroplane flying due north in
Â Â hours (OA) = 1000 Ã— 3/2 km = 1500 km

Speed of second aeroplane = 1200 km/hr

Distance covered by second aeroplane flying due west in
Â Â hours (OB) = 1200 Ã— 3/2 km = 1800 km

In right angle Î”AOB, by Pythagoras Theorem,

AB2Â =Â AO2Â + OB2

â‡’ AB2Â =Â (1500)2Â + (1800)2

â‡’ AB = âˆš(2250000Â + 3240000)

= âˆš5490000

â‡’ AB = 300âˆš61Â km

Hence, the distance between two aeroplanes will be 300âˆš61Â km.

12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Solution:

Given, Two poles of heights 6 m and 11 m stand on a plane ground.

And distance between the feet of the poles is 12 m.

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

Therefore, CP = 11 â€“ 6 = 5m

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

By Pythagoras theorem for Î”APC, we get,

AP2Â =Â PC2Â + AC2

(12m)2Â + (5m)2Â = (AC)2

AC2Â = (144+25) m2Â = 169 m2

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 AE2Â + BD2Â = AB2Â + DE2.

Solution:

Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

By Pythagoras theorem in Î”ACE, we get

AC2Â +Â CE2Â = AE2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)

In Î”BCD, by Pythagoras theorem, we get

BC2Â +Â CD2Â = BD2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

From equationsÂ (i)Â andÂ (ii), we get,

AC2Â +Â CE2Â + BC2Â +Â CD2Â = AE2Â + BD2Â â€¦â€¦â€¦â€¦..(iii)

In Î”CDE, by Pythagoras theorem, we get

DE2Â =Â CD2Â + CE2

In Î”ABC, by Pythagoras theorem, we get

AB2Â =Â AC2Â + CB2

Putting the above two values in equationÂ (iii), we get

DE2Â + AB2Â = AE2Â + BD2.

14. The perpendicular from A on side BC of aÂ Î” ABC intersects BC at D such that DB = 3CDÂ (see Figure). Prove that 2AB2Â = 2AC2Â + BC2.

Solution:

Given, the perpendicular from A on side BC of aÂ Î” ABC intersects BC at D such that;

DB = 3CD.

In Î” ABC,

AD âŠ¥BC and BD = 3CD

AB2Â =Â AD2Â + BD2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)

AC2Â =Â AD2Â + DC2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

Subtracting equationÂ (ii)Â from equationÂ (i), we get

AB2Â â€“ AC2Â = BD2Â â€“ DC2

= 9CD2Â â€“ CD2Â [Since, BD = 3CD]

= 8CD2

= 8(BC/4)2Â [Since, BC = DBÂ + CD = 3CDÂ + CD = 4CD]

Therefore, AB2Â â€“ AC2Â = BC2/2

â‡’ 2(AB2Â â€“ AC2) = BC2

â‡’ 2AB2Â â€“ 2AC2Â = BC2

âˆ´ 2AB2Â = 2AC2Â + BC2.

15. Â In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2Â = 7AB2.

Solution:

Given, ABC is an equilateral triangle.

And D is a point on side BC such that BD = 1/3BC

Let the side of the equilateral triangle beÂ a, and AE be the altitude of Î”ABC.

âˆ´ BE = EC = BC/2 = a/2

And, AE =Â aâˆš3/2

Given, BD = 1/3BC

âˆ´ BD =Â a/3

DE = BE â€“ BD =Â a/2 â€“Â a/3 =Â a/6

â‡’ 9 AD2Â = 7 AB2

16.Â 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.

Solution:

Given, an equilateral triangle say ABC,

Let the sides of the equilateral triangle be of length a, and AE be the altitude of Î”ABC.

âˆ´ BE = EC = BC/2 =Â a/2

In Î”ABE, by Pythagoras Theorem, we get

AB2Â = AE2Â + BE2

4AE2Â = 3a2

â‡’ 4 Ã— (Square of altitude) = 3 Ã— (Square of one side)

Hence, proved.

17. Tick the correct answer and justify: In Î”ABC, AB = 6âˆš3Â cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120Â°

(B) 60Â°
(C) 90Â°Â

(D) 45Â°

Solution:

Given, in Î”ABC, AB = 6âˆš3Â cm, AC = 12 cm and BC = 6 cm.

We can observe that,

AB2Â = 108

AC2Â = 144

And, BC2Â = 36

AB2Â + BC2Â = AC2

The given triangle, Î”ABC, is satisfying Pythagoras theorem.

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

âˆ´ âˆ B = 90Â°

Hence, the correct answer is (C).

## Exercise 6.6 Page: 152

1. In Figure, PS is the bisector of âˆ  QPR of âˆ† PQR. Prove that QS/PQ = SR/PR

Solution:

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

Given, PS is the angle bisector of âˆ QPR. Therefore,

âˆ QPS = âˆ SPRâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

As per the constructed figure,

âˆ SPR=âˆ PRT(Since, PS||TR)â€¦â€¦â€¦â€¦â€¦(ii)

âˆ QPS = âˆ QRT(Since, PS||TR) â€¦â€¦â€¦â€¦..(iii)

From the above equations, we get,

âˆ PRT=âˆ QTR

Therefore,

PT=PR

In â–³QTR, by basic proportionality theorem,

QS/SR = QP/PT

Since, PT=TR

Therefore,

QS/SR = PQ/PR

Hence, proved.

2. In Fig. 6.57, D is a point on hypotenuse AC of âˆ†ABC, such that BD âŠ¥AC, DM âŠ¥ BC and DN âŠ¥ AB. Prove that: (i) DM2 = DN . MC (ii) DN2 = DM . AN.

Solution:

1. Let us join Point D and B.

Given,

BD âŠ¥AC, DM âŠ¥ BC and DN âŠ¥ AB

Now from the figure we have,

DN || CB, DM || AB and âˆ B = 90 Â°

Therefore, DMBN is a rectangle.

So, DN = MB and DM = NB

The given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC.

âˆ´ âˆ CDB = 90Â° â‡’ âˆ 2 + âˆ 3 = 90Â° â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

In âˆ†CDM, âˆ 1 + âˆ 2 + âˆ DMC = 180Â°

â‡’ âˆ 1 + âˆ 2 = 90Â° â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (ii)

In âˆ†DMB, âˆ 3 + âˆ DMB + âˆ 4 = 180Â°

â‡’ âˆ 3 + âˆ 4 = 90Â° â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (iii)

From equation (i) and (ii), we get

âˆ 1 = âˆ 3

From equation (i) and (iii), we get

âˆ 2 = âˆ 4

In âˆ†DCM and âˆ†BDM,

âˆ 1 = âˆ 3 (Already Proved)

âˆ 2 = âˆ 4 (Already Proved)

âˆ´ âˆ†DCM âˆ¼ âˆ†BDM (AA similarity criterion)

BM/DM = DM/MC

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

â‡’ DM2 = DN Ã— MC

Hence, proved.

(ii) In right triangle DBN,

âˆ 5 + âˆ 7 = 90Â° â€¦â€¦â€¦â€¦â€¦â€¦.. (iv)

In right triangle DAN,

âˆ 6 + âˆ 8 = 90Â° â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (v)

D is the point in triangle, which is foot of the perpendicular drawn from B to AC.

âˆ´ âˆ ADB = 90Â° â‡’ âˆ 5 + âˆ 6 = 90Â° â€¦â€¦â€¦â€¦.. (vi)

From equation (iv) and (vi), we get,

âˆ 6 = âˆ 7

From equation (v) and (vi), we get,

âˆ 8 = âˆ 5

In âˆ†DNA and âˆ†BND,

âˆ 6 = âˆ 7 (Already proved)

âˆ 8 = âˆ 5 (Already proved)

âˆ´ âˆ†DNA âˆ¼ âˆ†BND (AA similarity criterion)

AN/DN = DN/NB

â‡’ DN2 = AN Ã— NB

â‡’ DN2 = AN Ã— DM (Since, NB = DM)

Hence, proved.

3. In Figure, ABC is a triangle in which âˆ ABC > 90Â° and AD âŠ¥ CB produced. Prove that

AC2= AB2+ BC2+ 2 BC.BD.

Solution:

By applying Pythagoras Theorem in âˆ†ADB, we get,

AB2 = AD2 + DB2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (i)

Again, by applying Pythagoras Theorem in âˆ†ACD, we get,

AC2 = AD2 + (DB + BC) 2

AC2 = AD2 + DB2 + BC2 + 2DB Ã— BC

From equation (i), we can write,

AC2 = AB2 + BC2 + 2DB Ã— BC

Hence, proved.

4. In Figure, ABC is a triangle in which âˆ  ABC < 90Â° and AD âŠ¥ BC. Prove that

AC2= AB2+ BC2 â€“ 2 BC.BD.

Solution:

By applying Pythagoras Theorem in âˆ†ADB, we get,

We can write it as;

â‡’ AD2 = AB2 âˆ’ DB2 â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

By applying Pythagoras Theorem in âˆ†ADC, we get,

From equation (i),

AB2 âˆ’ BD2 + DC2 = AC2

AB2 âˆ’ BD2 + (BC âˆ’ BD) 2 = AC2

AC2 = AB2 âˆ’ BD2 + BC2 + BD2 âˆ’2BC Ã— BD

AC2 = AB2 + BC2 âˆ’ 2BC Ã— BD

Hence, proved.

5. In Figure, AD is a median of a triangle ABC and AM âŠ¥ BC. Prove that :

(i) AC2 = AD2 + BC.DM + 2 (BC/2) 2

(ii) AB2 = AD2 â€“ BC.DM + 2 (BC/2) 2

(iii) AC2 + AB2 = 2 AD2 + Â½ BC2

Solution:

(i) By applying Pythagoras Theorem in âˆ†AMD, we get,

AM2 + MD2 = AD2 â€¦â€¦â€¦â€¦â€¦â€¦. (i)

Again, by applying Pythagoras Theorem in âˆ†AMC, we get,

AM2 + MC2 = AC2

AM2 + (MD + DC) 2 = AC2

(AM2 + MD2 ) + DC2 + 2MD.DC = AC2

From equation(i), we get,

AD2 + DC2 + 2MD.DC = AC2

Since, DC=BC/2, thus, we get,

AD2 + (BC/2) 2 + 2MD.(BC/2) 2 = AC2

AD2 + (BC/2) 2 + 2MD Ã— BC = AC2

Hence, proved.

(ii) By applying Pythagoras Theorem in âˆ†ABM, we get;

AB2 = AM2 + MB2

= (AD2 âˆ’ DM2) + MB2

= (AD2 âˆ’ DM2) + (BD âˆ’ MD) 2

= AD2 âˆ’ DM2 + BD2 + MD2 âˆ’ 2BD Ã— MD

= AD2 + BD2 âˆ’ 2BD Ã— MD

= AD2 + (BC/2)2 â€“ 2(BC/2) MD

= AD2 + (BC/2)2 â€“ BC MD

Hence, proved.

(iii) By applying Pythagoras Theorem in âˆ†ABM, we get,

AM2 + MB2 = AB2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦.â€¦ (i)

By applying Pythagoras Theorem in âˆ†AMC, we get,

AM2 + MC2 = AC2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦..â€¦ (ii)

Adding both the equations (i) and (ii), we get,

2AM2 + MB2 + MC2 = AB2 + AC2

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

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

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

2(AM2+ MD2) + (BC/2) 2 + (BC/2) 2 + 2MD (-BC/2 + BC/2) 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.

Solution:

Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.

By applying Pythagoras Theorem in âˆ†DEA, we get,

DE2 + EA2 = DA2 â€¦â€¦â€¦â€¦â€¦â€¦.â€¦ (i)

By applying Pythagoras Theorem in âˆ†DEB, we get,

DE2 + EB2 = DB2

DE2 + (EA + AB) 2 = DB2

(DE2 + EA2) + AB2 + 2EA Ã— AB = DB2

DA2 + AB2 + 2EA Ã— AB = DB2 â€¦â€¦â€¦â€¦â€¦. (ii)

By applying Pythagoras Theorem in âˆ†ADF, we get,

Again, applying Pythagoras theorem in âˆ†AFC, we get,

AC2 = AF2 + FC2 = AF2 + (DC âˆ’ FD) 2

= AF2 + DC2 + FD2 âˆ’ 2DC Ã— FD

= (AF2 + FD2) + DC2 âˆ’ 2DC Ã— FD AC2

AC2= AD2 + DC2 âˆ’ 2DC Ã— FD â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iii)

Since ABCD is a parallelogram,

AB = CD â€¦â€¦â€¦â€¦â€¦â€¦â€¦.â€¦(iv)

And BC = AD â€¦â€¦â€¦â€¦â€¦â€¦. (v)

âˆ DEA = âˆ AFD (Each 90Â°)

âˆ´ âˆ†EAD â‰… âˆ†FDA (AAS congruence criterion)

â‡’ EA = DF â€¦â€¦â€¦â€¦â€¦â€¦ (vi)

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

DA2 + AB2 + 2EA Ã— AB + AD2 + DC2 âˆ’ 2DC Ã— FD = DB2 + AC2

DA2 + AB2 + AD2 + DC2 + 2EA Ã— AB âˆ’ 2DC Ã— FD = DB2 + AC2

From equation (iv) and (vi),

BC2 + AB2 + AD2 + DC2 + 2EA Ã— AB âˆ’ 2AB Ã— EA = DB2 + AC2

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

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

(i) âˆ†APC ~ âˆ† DPB

(ii) AP . PB = CP . DP

Solution:

Firstly, let us join CB, in the given figure.

(i) In âˆ†APC and âˆ†DPB,

âˆ APC =Â âˆ DPB (Vertically opposite angles)

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

Therefore,

âˆ†APCÂ âˆ¼Â âˆ†DPB (AA similarity criterion)

(ii) In the above, we have proved that âˆ†APCÂ âˆ¼Â âˆ†DPB

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

âˆ´ AP/DP = PC/PB = CA/BD

â‡’AP/DP = PC/PB

âˆ´AP. PB = PC. DP

Hence, proved.

8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

(i) âˆ† PAC ~ âˆ† PDB

(ii) PA . PB = PC . PD.

Solution:

(i) In âˆ†PAC and âˆ†PDB,

âˆ P = âˆ P (Common Angles)

As we know, exterior angle of a cyclic quadrilateral is âˆ PCA and âˆ PBD is opposite interior angle, which are both equal.

âˆ PAC = âˆ PDB

Thus, âˆ†PAC âˆ¼ âˆ†PDB(AA similarity criterion)

(ii) We have already proved above,

âˆ†APC âˆ¼ âˆ†DPB

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

Therefore,

AP/DP = PC/PB = CA/BD

AP/DP = PC/PB

âˆ´ AP. PB = PC. DP

9. In Figure, D is a point on side BC of âˆ† ABC such that BD/CD = AB/AC. Prove that AD is the bisector of âˆ  BAC.

Solution:

In the given figure, let us extend BA to P such that;

AP = AC.

Now join PC.

Given, BD/CD = AB/AC

â‡’ BD/CD = AP/AC

By using the converse of basic proportionality theorem, we get,

âˆ BAD = âˆ APC (Corresponding angles) â€¦â€¦â€¦â€¦â€¦â€¦.. (i)

And, âˆ DAC = âˆ ACP (Alternate interior angles) â€¦â€¦.â€¦ (ii)

By the new figure, we have;

AP = AC

â‡’ âˆ APC = âˆ ACP â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (iii)

On comparing equations (i), (ii), and (iii), we get,

Therefore, AD is the bisector of the angle BAC.

Hence, proved.

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 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:

Let us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the

horizontal distance of the fly from the tip of the fishing rod. Therefore, AC is now the length of the string.

To find AC, we have to use Pythagoras theorem in âˆ†ABC, is such way;

AC2 = AB2+ BC2

AB2 = (1.8 m) 2 + (2.4 m) 2

AB2 = (3.24 + 5.76) m2

AB2 = 9.00 m2

âŸ¹ AB = âˆš9 m = 3m

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

As its given, she pulls the string at the rate of 5 cm per second.

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

Let us say now, the fly is at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC âˆ’ String pulled by Nazima in 12 seconds

= (3.00 âˆ’ 0.6) m

= 2.4 m

(1.8 m) 2 + BD2 = (2.4 m) 2

BD2 = (5.76 âˆ’ 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

## NCERT Solutions for Class 10 Maths Chapter 6 â€“ Triangles

NCERT Solutions Class 10 Maths Chapter 6, Triangles, is part of the Unit Geometry, which constitutes 15 marks of the total marks of 80. On the basis of the updated CBSE Class 10 Syllabus for 2023-24, this chapter belongs to the Unit-Geometry and has the second-highest weightage. Hence, having a clear understanding of the concepts, theorems and problem-solving methods in this chapter is mandatory to score well in the board examination of Class 10 Maths.

### Main topics covered in this chapter include:

#### 6.1 Introduction

From your earlier classes, you are familiar with triangles and many of their properties. In Class 9, you have studied congruence of triangles in detail. In this chapter, we shall study about those figures which have the same shape, but not necessarily the same size. Two figures having the same shape (and not necessarily the same size) are called similar figures. In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.

#### 6.2 Similar Figures

In Class 9, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent. The topic explains similarity of figures by performing the relevant activity. Similar figures are two figures having the same shape, but not necessarily the same size.

#### 6.3 Similarity of Triangles

The topic recalls triangles and its similarities. Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). It explains Basic Proportionality Theorem and different theorems are discussed performing various activities.

#### 6.4 Criteria for Similarity of Triangles

In the previous section, we stated that two triangles are similar (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). The topic discusses the criteria for similarity of triangles referring to the topics we have studied in earlier classes. It also contains different theorems explained with proper examples.

#### 6.5 Areas of Similar Triangles

You have learnt that in two similar triangles, the ratio of their corresponding sides is the same. The topic Areas of Similar Triangles consists of theorem and relatable examples to prove the theorem.

#### 6.6 Pythagoras Theorem

You are already familiar with the Pythagoras Theorem from your earlier classes. You have also seen proof of this theorem in Class 9. Now, we shall prove this theorem using the concept of similarity of triangles. Hence, the theorem is verified through some activities, and you can make use of it while solving certain problems.

#### 6.7 Summary

The summary contains the points you have studied in the chapter. Going through the points mentioned in the summary will help you to recollect all the important concepts and theorems of the chapter.

### List of Exercises in NCERT Class 10 Maths Chapter 6:

Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)

Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)

Exercise 6.3 Solutions 16 Questions (1 main question with 6 sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)

Exercise 6.4 Solutions 9 Questions (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.5 Solutions 17 Questions (15 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)

Triangle is one of the most interesting and exciting chapters of the unit Geometry as it takes us through the different aspects and concepts related to the geometrical figure triangle. A triangle is a plane figure that has three sides and three angles. This chapter covers various topics and sub-topics related to triangles, including a detailed explanation of similar figures, different theorems related to the similarities of triangles with proof, and the areas of similar triangles. The chapter concludes by explaining the Pythagoras theorem and the ways to use it in solving problems. Read and learn Chapter 6 of the Class 10 MathsÂ NCERT textbook to learn more about Triangles and the concepts covered in it. Ensure to learn the NCERT Solutions for Class 10 effectively to score high in the board examinations.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 6 â€“ Triangles

• Helps to ensure that the students use the concepts in solving the problems.
• Encourages the children to come up with diverse solutions to problems.
• Hints are given for those questions which are difficult to solve.
• Helps the students in checking if the solutions they gave for the questions are correct or not.

Disclaimer â€“Â

Dropped Topics â€“Â

6.5 Areas of similar triangles
6.6 Pythagoras theorem

## Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 6

Q1

### Why should we learn all the concepts present in NCERT Solutions for Class 10 Maths Chapter 6?

The concepts covered in Chapter 6 of NCERT Solution Maths provide questions based on the exam pattern and model question paper. So it is necessary to learn all the concepts present in NCERT Solutions for Class 10 Maths Chapter 6.
Q2

### List out the important topics present in NCERT Solutions for Class 10 Maths Chapter 6.

The topics covered in the chapters are Introduction to the Triangles, similar figures, the similarity of triangles, criteria for similarity of Triangles, areas of similar triangles and Pythagoras Theorem. These concepts are important from an exam perspective. It is strictly based on the latest syllabus of the CBSE for 2023-24 and also depends on the CBSE question paper design and marking scheme.
Q3

### How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 6?

There are 6 exercises in NCERT Solutions for Class 10 Maths Chapter 6. The first exercise contains 3 questions, the second exercise contains 10 questions, the third exercise has 16 questions, the fourth exercise has 9 questions, the fifth exercise has 17 questions, and the last or sixth exercise has ten questions based on the concepts of triangles.