# NCERT Solutions for Class 10 Maths Chapter 6 Triangles

## NCERT Solutions for Class 10 Maths Chapter 6 โ 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 the triangles and 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 one particular question.

The solutions provided at BYJUโS are in such a way that every step while solving a problem, is explained clearly and in detail. The **Solutions for NCERT Class 10 Maths** are prepared by the subject experts to help students in their board exam preparation in a better way. These solutions can 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** and, 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 for the students to practice these NCERT Solutions on a regular basis to excel in the Class 10 Board examination.Chapter 6 Triangles

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## Exercise 6.1 Page: 122

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

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

Answer: Similar

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

Answer: Similar

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

Answer: 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)**

Answer: (a) Equal

(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, in โณ ABC, DEโฅBC

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

โ1.5/3 = 1/EC

โEC = 3/1.5

EC = 3ร10/15 = 2 cm

Hence, EC = 2 cm.

(ii) Given, in โณ ABC, DEโฅBC

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

โ AD/7.2 = 1.8 / 5.4

โ AD = 1.8 ร7.2/5.4 = (18/10)ร(72/10)ร(10/54) = 24/10

โ AD = 2.4

Hence, AD = 2.4 cm.

**2. E and F are points on the sides PQ and PR respectively of a ฮPQR. 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 ฮPQR, E and F are two points on side PQ and PR respectively. See the figure below;

(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 โ 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 โ PE = 1.28 โ 0.18 = 1.10 cm

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

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55**โฆโฆโฆโฆ. (i)**

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

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

โดAN/AD = AL/AC**โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ(ii)**

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

AM/AB = AN/AD

Hence, proved.

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

**Solution:**

In ฮABC, given as, DE || AC

Thus, by using Basic Proportionality Theorem, we get,

โดBD/DA = BE/EC โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ**(i)**

In ฮABC, given as, DF || AE

Thus, by using Basic Proportionality Theorem, we get,

โดBD/DA = BF/FE โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ**(ii)**

From equation **(i)** and **(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 ฮPQO, DE || OQ

So by using Basic Proportionality Theorem,

PD/DO = PE/EQโฆโฆโฆโฆโฆโฆ **..(i)**

Again given, in ฮPOR, DF || OR,

So by using Basic Proportionality Theorem,

PD/DO = PF/FRโฆโฆโฆโฆโฆโฆโฆ **(ii)**

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

PE/EQ = PF/FR

Therefore, by converse of Basic Proportionality Theorem,

EF || QR, in ฮPQR.

**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 ฮOPQ, AB || PQ

By using Basic Proportionality Theorem,

OA/AP = OB/BQโฆโฆโฆโฆโฆ.**(i)**

Also given,

In ฮOPR, AC || PR

By using Basic Proportionality Theorem

โด OA/AP = OC/CRโฆโฆโฆโฆโฆ(ii)

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

OB/BQ = OC/CR

Therefore, by converse of Basic Proportionality Theorem,

In ฮOQR, BC || 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 ฮABC, 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.

โด AD=DB

โAD/DB = 1 โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ. **(i)**

In ฮABC, DE || BC,

By using Basic Proportionality Theorem,

Therefore, AD/DB = AE/EC

From equation (i), we can write,

โ 1 = AE/EC

โด 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 ฮABC, D and E are the mid points of AB and AC respectively, such that,

AD=BD and AE=EC.

We have to prove that: DE || BC.

Since, D is the midpoint of AB

โด AD=DB

โAD/BD = 1โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.. **(i)**

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

โด AE=EC

โ AE/EC = 1

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

AD/BD = AE/EC

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 ฮADC, we have OE || DC

Therefore, By using Basic Proportionality Theorem

AE/ED = AO/CO **โฆโฆโฆโฆโฆ..(i)**

Now, In ฮABD, OE || AB

Therefore, By using Basic Proportionality Theorem

DE/EA = DO/BO**โฆโฆโฆโฆโฆ.(ii)**

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

AO/CO = BO/DO

โAO/BO = CO/DO

Hence, proved.

**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.**

**Solution:**

Given, Quadrilateral ABCD where AC and BD intersects 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 ฮDAB, EO || AB

Therefore, By using Basic Proportionality Theorem

DE/EA = DO/OB โฆโฆโฆโฆโฆโฆโฆโฆ(i)

Also, given,

AO/BO = CO/DO

โ AO/CO = BO/DO

โ CO/AO = DO/BO

โDO/OB = CO/AO **โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..(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

โ AB || DC.

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

## Exercise 6.3 Page: 138

**1. State which pairs of triangles in 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 ฮABC and ฮPQR,

โ A = โ P = 60ยฐ

โ B = โ Q = 80ยฐ

โ C = โ R = 40ยฐ

Therefore by AAA similarity criterion,

โด ฮABC ~ ฮPQR

(ii) Given, in ฮABC and ฮPQR,

AB/QR = BC/RP = CA/PQ

By SSS similarity criterion,

ฮABC ~ ฮQRP

(iii) Given, in ฮLMP and ฮDEF,

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 โ LM/EF

Therefore, ฮLMP and ฮDEF are not similar.

(iv) In ฮMNL and ฮQPR, it is given,

MN/QP = ML/QR = 1/2

โ M = โ Q = 70ยฐ

Therefore, by SAS similarity criterion

โด ฮMNL ~ ฮQPR

(v) In ฮABC and ฮDEF, given that,

AB = 2.5, BC = 3, โ A = 80ยฐ, EF = 6, DF = 5, โ F = 80ยฐ

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

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

โ โ B โ โ F

Hence, ฮABC and ฮDEF are not similar.

(vi) In ฮDEF, by sum of angles of triangles, we know that,

โ D + โ E + โ F = 180ยฐ

โ 70ยฐ + 80ยฐ + โ F = 180ยฐ

โ โ F = 180ยฐ โ 70ยฐ โ 80ยฐ

โ โ F = 30ยฐ

Similarly, In ฮPQR,

โ P + โ Q + โ R = 180 (Sum of angles of ฮ)

โ โ P + 80ยฐ + 30ยฐ = 180ยฐ

โ โ P = 180ยฐ โ 80ยฐ -30ยฐ

โ โ P = 70ยฐ

Now, comparing both the triangles, ฮDEF and ฮPQR, we have

โ D = โ P = 70ยฐ

โ F = โ Q = 80ยฐ

โ F = โ R = 30ยฐ

Therefore, by AAA similarity criterion,

Hence, ฮDEF ~ ฮPQR

**2. In the figure, ฮODC โ ยผ ฮOBA, โ BOC = 125ยฐ and โ CDO = 70ยฐ. Find โ DOC, โ DCO and โ OAB.**

**Solution:**

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

Therefore, โ DOC + โ COB = 180ยฐ

โ โ DOC = 180ยฐ โ 125ยฐ (Given, โ BOC = 125ยฐ)

= 55ยฐ

In ฮDOC, sum of the measures of the angles of a triangle is 180ยบ

Therefore, โ DCO + โ CDO + โ DOC = 180ยฐ

โ โ DCO + 70ยบ + 55ยบ = 180ยฐ(Given, โ CDO = 70ยฐ)

โ โ DCO = 55ยฐ

It is given that, ฮODC โ ยผ ฮOBA,

Therefore, ฮODC ~ ฮOBA.

Hence, Corresponding angles are equal in similar triangles

โ OAB = โ OCD

โ โ OAB = 55ยฐ

โ OAB = โ OCD

โ โ OAB = 55ยฐ

**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 ฮDOC and ฮBOA,

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

โดโ CDO = โ ABO

Similarly,

โ DCO = โ BAO

Also, for the two triangles ฮDOC and ฮBOA, vertically opposite angles will be equal;

โดโ DOC = โ BOA

Hence, by AAA similarity criterion,

ฮDOC ~ ฮBOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

โOA/OC = OB/OD

Hence, proved.

**4. In the fig.6.36, QR/QS = QT/PR and โ 1 = โ 2. Show that ฮPQS ~ ฮTQR.**

**Solution:**

In ฮPQR,

โ PQR = โ PRQ

โด PQ = PR โฆโฆโฆโฆโฆโฆโฆโฆโฆ**(i)**

Given,

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

QR/QS = QT/QP**โฆโฆโฆโฆโฆโฆ.(ii)**

In ฮPQS and ฮTQR, by equation (ii),

QR/QS = QT/QP

โ Q = โ Q

โด ฮPQS ~ ฮTQR [By SAS similarity criterion]

**5. S and T are point on sides PR and QR of ฮPQR such that โ P = โ RTS. Show that ฮRPQ ~ ฮRTS.**

**Solution:**

Given, S and T are point on sides PR and QR of ฮPQR

And โ P = โ RTS.

In ฮRPQ and ฮRTS,

โ RTS = โ QPS (Given)

โ R = โ R (Common angle)

โด ฮRPQ ~ ฮRTS (AA similarity criterion)

**6. In the figure, if ฮABE โ
ฮACD, show that ฮADE ~ ฮABC.**

**Solution:**

Given, ฮABE โ ฮACD.

โด AB = AC [By CPCT] โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.**(i)**

And, AD = AE [By CPCT] โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ**(ii)**

In ฮADE and ฮABC, dividing eq.(ii) by eq(i),

AD/AB = AE/AC

โ A = โ A [Common angle]

โด ฮADE ~ ฮABC [SAS similarity criterion]

**7. In the figure, altitudes AD and CE of ฮABC intersect each other at the point P. Show that:**

**(i) ฮAEP ~ ฮCDP(ii) ฮABD ~ ฮCBE(iii) ฮAEP ~ ฮADB(iv) ฮPDC ~ ฮBEC**

**Solution:**

Given, altitudes AD and CE of ฮABC intersect each other at the point P.

(i) In ฮAEP and ฮCDP,

โ AEP = โ CDP (90ยฐ each)

โ APE = โ CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

ฮAEP ~ ฮCDP

(ii) In ฮABD and ฮCBE,

โ ADB = โ CEB ( 90ยฐ each)

โ ABD = โ CBE (Common Angles)

Hence, by AA similarity criterion,

ฮABD ~ ฮCBE

(iii) In ฮAEP and ฮADB,

โ AEP = โ ADB (90ยฐ each)

โ PAE = โ DAB (Common Angles)

Hence, by AA similarity criterion,

ฮAEP ~ ฮADB

(iv) In ฮPDC and ฮBEC,

โ PDC = โ BEC (90ยฐ each)

โ PCD = โ BCE (Common angles)

Hence, by AA similarity criterion,

ฮPDC ~ ฮBEC

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

**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 ฮABE and ฮCFB,

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

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

โด ฮABE ~ ฮCFB (AA similarity criterion)

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

**(i) ฮABC ~ ฮAMP**

**(ii) CA/PA = BC/MP**

**Solution:**

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

(i) In ฮABC and ฮAMP, we have,

โ CAB = โ MAP (common angles)

โ ABC = โ AMP = 90ยฐ (each 90ยฐ)

โด ฮABC ~ ฮAMP (AA similarity criterion)

(ii) As, ฮABC ~ ฮAMP (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 โ ACB and โ EGF such that D and H lie on sides AB and FE of ฮABC and ฮEFG respectively. If ฮABC ~ ฮFEG, Show that:**

**(i) CD/GH = AC/FG(ii) ฮDCB ~ ฮHGE(iii) ฮDCA ~ ฮHGF**

**Solution:**

Given, CD and GH are respectively the bisectors of โ ACB and โ EGF such that D and H lie on sides AB and FE of ฮABC and ฮEFG respectively.

(i) From the given condition,

ฮABC ~ ฮFEG.

โด โ A = โ F, โ B = โ E, and โ ACB = โ FGE

Since, โ ACB = โ FGE

โด โ ACD = โ FGH (Angle bisector)

And, โ DCB = โ HGE (Angle bisector)

In ฮACD and ฮFGH,

โ A = โ F

โ ACD = โ FGH

โด ฮACD ~ ฮFGH (AA similarity criterion)

โCD/GH = AC/FG

(ii) In ฮDCB and ฮHGE,

โ DCB = โ HGE (Already proved)

โ B = โ E (Already proved)

โด ฮDCB ~ ฮHGE (AA similarity criterion)

(iii) In ฮDCA and ฮHGF,

โ ACD = โ FGH (Already proved)

โ A = โ F (Already proved)

โด ฮDCA ~ ฮHGF (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 โฅ BC and EF โฅ AC, prove that ฮABD ~ ฮECF.**

**Solution:**

Given, ABC is an isosceles triangle.

โด AB = AC

โ โ ABD = โ ECF

In ฮABD and ฮECF,

โ ADB = โ EFC (Each 90ยฐ)

โ BAD = โ CEF (Already proved)

โด ฮABD ~ ฮECF (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 ฮPQR (see Fig 6.41). Show that ฮABC ~ ฮPQR.**

**Solution:**

Given, ฮABC and ฮPQR, AB, BC and median AD of ฮABC are proportional to sides PQ, QR and median PM of ฮPQR

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

We have to prove: ฮABC ~ ฮPQR

As we know here,

AB/PQ = BC/QR = AD/PM

โAB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)

โ ฮABD ~ ฮPQM [SSS similarity criterion]

โด โ ABD = โ PQM [Corresponding angles of two similar triangles are equal]

โ โ ABC = โ PQR

In ฮABC and ฮPQR

AB/PQ = BC/QR โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.**(i)**

โ ABC = โ PQR โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ**(ii)**

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

ฮABC ~ ฮPQR [SAS similarity criterion]

**13. D is a point on the side BC of a triangle ABC such that โ ADC = โ BAC. Show that CA ^{2} = CB.CD**

**Solution:**

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

In ฮADC and ฮBAC,

โ ADC = โ BAC (Already given)

โ ACD = โ BCA (Common angles)

โด ฮADC ~ ฮBAC (AA similarity criterion)

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

โด CA/CB = CD/CA

โ CA^{2} = 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 that ฮABC ~ ฮPQR.**

**Solution:**

Given: Two triangles ฮABC and ฮPQR in which AD and PM are medians such that;

AB/PQ = AC/PR = AD/PM

We have to prove, ฮABC ~ ฮPQR

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 ฮABD and ฮCDE, we have

AD = DE [By Construction.]

BD = DC [Since, AP is the median]

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

โด ฮABD โ ฮCDE [SAS criterion of congruence]

โ AB = CE [By CPCT] โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..**(i)**

Also, in ฮPQM and ฮMNR,

PM = MN [By Construction.]

QM = MR [Since, PM is the median]

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

โด ฮPQM = ฮMNR [SAS criterion of congruence]

โ PQ = RN [CPCT] โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ**(ii)**

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

From equation **(i)** and **(ii)**,

โCE/RN = AC/PR = AD/PM

โ CE/RN = AC/PR = 2AD/2PM

โ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]

โด ฮACE ~ ฮPRN [SSS similarity criterion]

Therefore, โ 2 = โ 4

Similarly, โ 1 = โ 3

โด โ 1 + โ 2 = โ 3 + โ 4

โ โ A = โ P โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.**(iii)**

Now, in ฮABC and ฮPQR, we have

AB/PQ = AC/PR (Already given)

From equation (iii),

โ A = โ P

โด ฮABC ~ ฮPQR [ 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 = *h* m

Length of shadow of the tower = 28 m

In ฮABC and ฮDEF,

โ C = โ E (angular elevation of sum)

โ B = โ F = 90ยฐ

โด ฮABC ~ ฮDEF (AA similarity criterion)

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

โด 6/h = 4/28

โh = (6ร28)/4

โ *h* = 6 ร 7

โ *h *= 42 m

Hence, the height of the tower is 42 m.

**16. If AD and PM are medians of triangles ABC and PQR, respectively where ฮABC ~ ฮPQR prove that AB/PQ = AD/PM.**

**Solution:**

Given, ฮABC ~ ฮPQR

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

โดAB/PQ = AC/PR = BC/QR**โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ(i**)

Also, โ A = โ P, โ B = โ Q, โ C = โ R โฆโฆโฆโฆ.โฆ..**(ii)**

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

โด BD = BC/2 and QM = QR/2 โฆโฆโฆโฆโฆ..โฆโฆโฆโฆ.**(iii)**

From equations **(i)** and **(iii)**, we get

AB/PQ = BD/QM **โฆโฆโฆโฆโฆโฆโฆโฆโฆ.(iv)**

In ฮABD and ฮPQM,

From equation (ii), we have

โ B = โ Q

From equation **(iv), we have,**

AB/PQ = BD/QM

โด ฮABD ~ ฮPQM (SAS similarity criterion)

โAB/PQ = BD/QM = AD/PM

Exercise 6.4 Page: 143

**1. Let ฮABC ~ ฮDEF and their areas be, respectively, 64 cm ^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC.**

**Solution: **Given, ฮABC ~ ฮDEF,

Area of ฮABC = 64 cm^{2}

Area of ฮDEF = 121 cm^{2}

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,

= AC^{2}/DF^{2} = BC^{2}/EF^{2}

โด 64/121 = BC^{2}/EF^{2}

โ (8/11)^{2} = (BC/15.4)^{2}

โ 8/11 = BC/15.4

โ BC = 8ร15.4/11

โ BC = 8 ร 1.4

โ 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 ฮAOB and ฮCOD, we have

โ 1 = โ 2 (Alternate angles)

โ 3 = โ 4 (Alternate angles)

โ 5 = โ 6 (Vertically opposite angle)

โด ฮAOB ~ ฮCOD [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 (ฮAOB)/Area of (ฮCOD) = AB^{2}/CD^{2}

= (2CD)^{2}/CD^{2} [โด AB = 2CD]

โด Area of (ฮAOB)/Area of (ฮCOD)

= 4CD^{2}/CD^{2} = 4/1

Hence, the required ratio of the area of ฮAOB and ฮCOD = 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 (ฮABC)/area (ฮDBC) = 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 (ฮABC)/Area (ฮDBC) = AO/DO

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

We know that area of a triangle = 1/2 ร Base ร Height

In ฮAPO and ฮDMO,

โ APO = โ DMO (Each 90ยฐ)

โ AOP = โ DOM (Vertically opposite angles)

โด ฮAPO ~ ฮDMO (AA similarity criterion)

โด AP/DM = AO/DO

โ Area (ฮABC)/Area (ฮDBC) = AO/DO.

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

**Solution:**

Say ฮABC and ฮPQR are two similar triangles and equal in area

Now let us prove ฮABC โ ฮPQR.

Since, ฮABC ~ ฮPQR

โด Area of (ฮABC)/Area of (ฮPQR) = BC^{2}/QR^{2}

โ BC^{2}/QR^{2} =1 [Since, Area(ฮABC) = (ฮPQR)

โ BC^{2}/QR^{2}

โ BC = QR

Similarly, we can prove that

AB = PQ and AC = PR

Thus, ฮABC โ ฮPQR [SSS criterion of congruence]

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

**Solution:**

Given, D, E and F are respectively the mid-points of sides AB, BC and CA of ฮABC.

In ฮABC,

F is the mid-point of AB (Already given)

E is the mid-point of AC (Already given)

So, by the mid-point theorem, we have,

FE || BC and FE = 1/2BC

โ FE || BC and FE || BD [BD = 1/2BC]

Since, opposite sides of parallelogram are equal and parallel

โด BDEF is parallelogram.

Similarly, in ฮFBD and ฮDEF, we have

FB = DE (Opposite sides of parallelogram BDEF)

FD = FD (Common sides)

BD = FE (Opposite sides of parallelogram BDEF)

โด ฮFBD โ ฮDEF

Similarly, we can prove that

ฮAFE โ ฮDEF

ฮEDC โ ฮDEF

As we know, if triangles are congruent, then they are equal in area.

So,

Area(ฮFBD) = Area(ฮDEF) โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ**(i)**

Area(ฮAFE) = Area(ฮDEF) โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.**(ii)**

and,

Area(ฮEDC) = Area(ฮDEF) โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.**(iii)**

Now,

Area(ฮABC) = Area(ฮFBD) + Area(ฮDEF) + Area(ฮAFE) + Area(ฮEDC) โฆโฆโฆ**(iv)**

Area(ฮABC) = Area(ฮDEF) + Area(ฮDEF) + Area(ฮDEF) + Area(ฮDEF)

From equation **(i)**, **(ii)** and **(iii)**,

โ Area(ฮDEF) = (1/4)Area(ฮABC)

โ Area(ฮDEF)/Area(ฮABC) = 1/4

Hence, Area(ฮDEF): Area(ฮABC) = 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.**

**Solution:**

Given: AM and DN are the medians of triangles ABC and DEF respectively and ฮABC ~ ฮDEF.

We have to prove: Area(ฮABC)/Area(ฮDEF) = AM^{2}/DN^{2}

Since, ฮABC ~ ฮDEF (Given)

โด Area(ฮABC)/Area(ฮDEF) = (AB^{2}/DE^{2}) โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ**(i)**

and, AB/DE = BC/EF = CA/FD โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ**(ii)**

In ฮABM and ฮDEN,

Since ฮABC ~ ฮDEF

โด โ B = โ E

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

โด ฮABC ~ ฮDEF [SAS similarity criterion]

โ AB/DE = AM/DN โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..**(iii)**

โด ฮABM ~ ฮDEN

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

โด area(ฮABC)/area(ฮDEF) = AB^{2}/DE^{2} = AM^{2}/DN^{2}

Hence, proved.

^{}**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.**

**Solution:**

Given, ABCD is a square whose one diagonal is AC. ฮAPC and ฮBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

Area(ฮBQC) = ยฝ Area(ฮAPC)

Since, ฮAPC and ฮBQC are both equilateral triangles, as per given,

โด ฮAPC ~ ฮBQC [AAA similarity criterion]

โด area(ฮAPC)/area(ฮBQC) = (AC^{2}/BC^{2}) = AC^{2}/BC^{2}

Since, Diagonal = โ2 side = โ2 BC = AC

โ area(ฮAPC) = 2 ร area(ฮBQC)

โ area(ฮBQC) = 1/2area(ฮAPC)

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**, **ฮABC and ฮBDE are two equilateral triangle. D is the midpoint of BC.

โด BD = DC = 1/2BC

Let each side of triangle is 2*a*.

As, ฮABC ~ ฮBDE

โด Area(ฮABC)/Area(ฮBDE) = AB^{2}/BD^{2} = (2*a*)^{2}/(*a*)^{2} = 4*a*^{2}/*a*^{2} = 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,

ฮABC ~ ฮDEF

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,

โด Area(ฮABC)/Area(ฮDEF) = AB^{2}/DE^{2 }

โด Area(ฮABC)/Area(ฮDEF) = (4/9)^{2 }= 16/81 = 16:81

Hence, the correct answer is (D).

## Exercise 6.5 Page: 150

**1. 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)^{2} + (24)^{2} = (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 โ 64

Or, 3^{2} + 6^{2} โ 8^{2}

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) Given, sides of triangleโs are 50 cm, 80 cm, and 100 cm.

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

However, 2500 + 6400 โ 10000

Or, 50^{2} + 80^{2} โ 100^{2}

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, 12^{2} + 5^{2} = 13^{2}

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.

**2. PQR is a triangle right angled at P and M is a point on QR such that PM โฅ QR. Show that PM ^{2} = QM ร MR.**

**Solution:**

Given, ฮPQR is right angled at P is a point on QR such that PM โฅQR

We have to prove, PM^{2} = QM ร MR

In ฮPQM, by Pythagoras theorem

PQ^{2} = PM^{2} + QM^{2}

Or, PM^{2} = PQ^{2} โ QM^{2} โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..**(i)**

In ฮPMR, by Pythagoras theorem

PR^{2} = PM^{2} + MR^{2}

Or, PM^{2} = PR^{2} โ MR^{2} โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..**(ii)**

Adding equation, **(i)** and **(ii)**, we get,

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

= QR^{2} โ QM^{2} โ MR^{2 } [โด QR^{2} = PQ^{2} + PR^{2}]

= (QM + MR)^{2} โ QM^{2} โ MR^{2}

= 2QM ร MR

โด PM^{2} = QM ร MR

**3. In Figure, ABD is a triangle right angled at A and AC โฅ BD. Show that(i) AB ^{2} = BC ร BD(ii) AC^{2} = BC ร DC(iii) AD^{2} = BD ร CD**

**Solution:**

(i) In ฮADB and ฮCAB,

โ DAB = โ ACB (Each 90ยฐ)

โ ABD = โ CBA (Common angles)

โด ฮADB ~ ฮCAB [AA similarity criterion]

โ AB/CB = BD/AB

โ AB^{2} = CB ร BD

(ii) Let โ CAB = x

In ฮCBA,

โ CBA = 180ยฐ โ 90ยฐ โ x

โ CBA = 90ยฐ โ x

Similarly, in ฮCAD

โ CAD = 90ยฐ โ โ CBA

= 90ยฐ โ* *x

โ CDA = 180ยฐ โ 90ยฐ โ (90ยฐ โ x)

โ CDA = x

In ฮCBA and ฮCAD, we have

โ CBA = โ CAD

โ CAB = โ CDA

โ ACB = โ DCA (Each 90ยฐ)

โด ฮCBA ~ ฮCAD [AAA similarity criterion]

โ AC/DC = BC/AC

โ AC^{2} = DC ร BC

(iii) In ฮDCA and ฮDAB,

โ DCA = โ DAB (Each 90ยฐ)

โ CDA = โ ADB (common angles)

โด ฮDCA ~ ฮDAB [AA similarity criterion]

โ DC/DA = DA/DA

โ AD^{2} = BD ร CD

**4. ABC is an isosceles triangle right angled at C. Prove that AB ^{2} = 2AC^{2} .**

**Solution:**

Given, ฮABC is an isosceles triangle right angled at C.

In ฮACB, โ C = 90ยฐ

AC = BC (By isosceles triangle property)

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

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

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

**5. ABC is an isosceles triangle with AC = BC. If AB ^{2} = 2AC^{2}, prove that ABC is a right triangle.**

**Solution:**

Given, ฮABC is an isosceles triangle having AC = BC and AB^{2} = 2AC^{2}

In ฮACB,

AC = BC

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

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

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

Hence, by Pythagoras theorem ฮABC is right angle triangle.

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

**Solution:**

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

Draw, AD โฅ BC

In ฮADB and ฮADC,

AB = AC

AD = AD

โ ADB = โ ADC [Both are 90ยฐ]

Therefore, ฮADB โ ฮADC by RHS congruence.

Hence, BD = DC [by CPCT]

In right angled ฮADB,

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

(2*a*)^{2} = AD^{2 }+ *a*^{2 }

โ AD^{2 =} 4*a*^{2} โ *a*^{2}

โ AD^{2 =} 3*a*^{2}

โ AD^{ =} โ3a

**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,

AB^{2 }+ BC^{2 }+ CD^{2} + AD^{2 }= AC^{2 }+ BD^{2}

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

Therefore, AO = CO and BO = DO

In ฮAOB,

โ AOB = 90ยฐ

AB^{2} = AO^{2 }+ BO^{2 }โฆโฆโฆโฆโฆโฆโฆโฆ.. **(i)** [By Pythagoras theorem]

Similarly,

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

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

BC^{2} = CO^{2 }+ BO^{2 }โฆโฆโฆโฆโฆโฆโฆโฆ.. **(iv)**

Adding equations **(i) + (ii) + (iii) + (iv)**, we get,

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

= 4AO^{2 }+ 4BO^{2 }[Since, AO = CO and BO =DO]

= (2AO)^{2 }+ (2BO)^{2} = AC^{2 }+ BD^{2}

AB^{2 }+ AD^{2 }+^{ }DC^{2 }+^{ }BC^{2} = AC^{2 }+ BD^{2}

Hence, proved.

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

**ABC, OD โฅ BC, OE โฅ AC and OF โฅ AB. Show that:(i) OA ^{2} + OB^{2} + OC^{2} โ OD^{2} โ OE^{2} โ OF^{2} = AF^{2} + BD^{2} + CE^{2} ,(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.**

**Solution:**

Given, in ฮABC, O is a point in the interior of a triangle.

And OD โฅ BC, OE โฅ AC and OF โฅ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in ฮAOF, we have

OA^{2} = OF^{2} + AF^{2}

Similarly, in ฮBOD

OB^{2} = OD^{2} + BD^{2}

Similarly, in ฮCOE

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} โ OD^{2} โ OE^{2} โ OF^{2} = AF^{2} + BD^{2} + CE^{2}.

(ii) AF^{2} + BD^{2} + EC^{2} = (OA^{2} โ OE^{2}) + (OC^{2} โ OD^{2}) + (OB^{2} โ OF^{2})

โด 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.**

**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,

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

10^{2} = 8^{2} + BC^{2}

BC^{2 }= 100 โ 64

BC^{2 }= 36

BC^{ }= 6m

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

**10. 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,

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

24^{2} = 18^{2} + BC^{2}

BC^{2 }= 576 โ 324

BC^{2 }= 252

BC^{ }= 6โ7m

Therefore, the distance from the base is 6โ7m.

**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?**

**Solution:**

Given,

Speed of first aeroplane = 1000 km/hr

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

Speed of second aeroplane = 1200 km/hr

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

In right angle ฮAOB, by Pythagoras Theorem,

AB^{2} =^{ }AO^{2} + OB^{2}

โ AB^{2} =^{ }(1500)^{2} + (1800)^{2}

โ AB = โ(2250000 + 3240000)

= โ5490000

โ AB = 300โ61 km

Hence, the distance between two aeroplanes will be 300โ61 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 โ 6 = 5m

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

By Pythagoras theorem for ฮAPC, we get,

AP^{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}.**

**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 ฮACE, we get

AC^{2} +^{ }CE^{2} = AE^{2} โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.**(i)**

In ฮBCD, by Pythagoras theorem, we get

BC^{2} +^{ }CD^{2} = BD^{2} โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..**(ii)**

From equations **(i)** and **(ii)**, we get,

AC^{2} +^{ }CE^{2} + BC^{2} +^{ }CD^{2} = AE^{2} + BD^{2} โฆโฆโฆโฆ..**(iii)**

In ฮCDE, by Pythagoras theorem, we get

DE^{2} =^{ }CD^{2} + CE^{2}

In ฮABC, by Pythagoras theorem, we get

AB^{2} =^{ }AC^{2} + CB^{2}

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

DE^{2} + AB^{2} = AE^{2} + BD^{2}.

**14. The perpendicular from A on side BC of a ฮ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB ^{2} = 2AC^{2} + BC^{2}.**

**Solution:**

Given, the perpendicular from A on side BC of a ฮ ABC intersects BC at D such that;

DB = 3CD.

In ฮ ABC,

AD โฅBC and BD = 3CD

In right angle triangle, ADB and ADC, by Pythagoras theorem,

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

AC^{2} =^{ }AD^{2} + DC^{2} โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..**(ii)**

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

AB^{2} โ AC^{2} = BD^{2} โ DC^{2}

= 9CD^{2} โ CD^{2} [Since, BD = 3CD]

= 8CD^{2}

= 8(BC/4)^{2 }[Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB^{2} โ AC^{2} = BC^{2}/2

โ 2(AB^{2} โ AC^{2}) = BC^{2}

โ 2AB^{2} โ 2AC^{2} = BC^{2}

โด 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}.**

**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 be *a*, and AE be the altitude of ฮABC.

โด BE = EC = BC/2 = a/2

And, AE = aโ3/2

Given, BD = 1/3BC

โด BD = a/3

DE = BE โ BD = a/2 โ a/3 = a/6

In ฮADE, by Pythagoras theorem,

AD^{2} = AE^{2} + DE^{2 }

โ 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.**

**Solution:**

Given, an equilateral triangle say ABC,

Let the sides of the equilateral triangle be of length a, and AE be the altitude of ฮABC.

โด BE = EC = BC/2 = a/2

In ฮABE, by Pythagoras Theorem, we get

AB^{2} = AE^{2} + BE^{2}

4AE^{2} = 3a^{2}

โ 4 ร (Square of altitude) = 3 ร (Square of one side)

Hence, proved.^{}

**17. Tick the correct answer and justify: In ฮABC, AB = 6โ3 cm, AC = 12 cm and BC = 6 cm.The angle B is:(A) 120ยฐ**

**(B) 60ยฐ(C) 90ยฐ **

**(D) 45ยฐ**

**Solution:**

Given, in ฮABC, AB = 6โ3 cm, AC = 12 cm and BC = 6 cm.

We can observe that,

AB^{2} = 108

AC^{2} = 144

And, BC^{2} = 36

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

The given triangle, ฮABC, is satisfying Pythagoras theorem.

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

โด โ B = 90ยฐ

Hence, the correct answer is (C).

## Exercise 6.6 Page: 152

**1. In Figure, PS is the bisector of โ QPR of โ 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 โ QPR. Therefore,

โ QPS = โ SPRโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..(i)

As per the constructed figure,

โ SPR=โ PRT(Since, PS||TR)โฆโฆโฆโฆโฆ(ii)

โ QPS = โ QRT(Since, PS||TR) โฆโฆโฆโฆ..(iii)

From the above equations, we get,

โ PRT=โ QTR

Therefore,

PT=PR

In โณQTR, 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 โABC, such that BD โฅAC, DM โฅ BC and DN โฅ AB. Prove that: (i) DM ^{2} = DN . MC (ii) DN^{2} = DM . AN.**

**Solution:**

- Let us join Point D and B.

Given,

BD โฅAC, DM โฅ BC and DN โฅ AB

Now from the figure we have,

DN || CB, DM || AB and โ B = 90 ยฐ

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.

โด โ CDB = 90ยฐ โ โ 2 + โ 3 = 90ยฐ โฆโฆโฆโฆโฆโฆโฆโฆ. (i)

In โCDM, โ 1 + โ 2 + โ DMC = 180ยฐ

โ โ 1 + โ 2 = 90ยฐ โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.. (ii)

In โDMB, โ 3 + โ DMB + โ 4 = 180ยฐ

โ โ 3 + โ 4 = 90ยฐ โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.. (iii)

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

โ 1 = โ 3

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

โ 2 = โ 4

In โDCM and โBDM,

โ 1 = โ 3 (Already Proved)

โ 2 = โ 4 (Already Proved)

โด โDCM โผ โBDM (AA similarity criterion)

BM/DM = DM/MC

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

โ DM^{2} = DN ร MC

Hence, proved.

(ii) In right triangle DBN,

โ 5 + โ 7 = 90ยฐ โฆโฆโฆโฆโฆโฆ.. (iv)

In right triangle DAN,

โ 6 + โ 8 = 90ยฐ โฆโฆโฆโฆโฆโฆโฆ (v)

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

โด โ ADB = 90ยฐ โ โ 5 + โ 6 = 90ยฐ โฆโฆโฆโฆ.. (vi)

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

โ 6 = โ 7

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

โ 8 = โ 5

In โDNA and โBND,

โ 6 = โ 7 (Already proved)

โ 8 = โ 5 (Already proved)

โด โDNA โผ โBND (AA similarity criterion)

AN/DN = DN/NB

โ DN^{2} = AN ร NB

โ DN^{2} = AN ร DM (Since, NB = DM)

Hence, proved.

**3. In Figure, ABC is a triangle in which โ ABC > 90ยฐ and AD โฅ CB produced. Prove that**

**AC ^{2}= AB^{2}+ BC^{2}+ 2 BC.BD.**

**Solution:**

By applying Pythagoras Theorem in โADB, we get,

AB^{2} = AD^{2} + DB^{2} โฆโฆโฆโฆโฆโฆโฆโฆโฆ (i)

Again, by applying Pythagoras Theorem in โACD, we get,

AC^{2} = AD^{2} + DC^{2}

AC^{2} = AD^{2} + (DB + BC) ^{2}

AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2DB ร BC

From equation (i), we can write,

AC^{2} = AB^{2} + BC^{2} + 2DB ร BC

Hence, proved.

**4. In Figure, ABC is a triangle in which โ ABC < 90ยฐ and AD โฅ BC. Prove that**

**AC ^{2}= AB^{2}+ BC^{2} โ 2 BC.BD.**

**Solution:**

By applying Pythagoras Theorem in โADB, we get,

AB^{2} = AD^{2} + DB^{2}

We can write it as;

โ AD^{2} = AB^{2} โ DB^{2} โฆโฆโฆโฆโฆโฆ.. (i)

By applying Pythagoras Theorem in โADC, we get,

AD^{2} + DC^{2} = AC^{2}

From equation (i),

AB^{2} โ BD^{2} + DC^{2} = AC^{2}

AB^{2} โ BD^{2} + (BC โ BD)^{ 2} = AC^{2}

AC^{2} = AB^{2} โ BD^{2} + BC^{2} + BD^{2} โ2BC ร BD

AC^{2 }= AB^{2} + BC^{2} โ 2BC ร BD

Hence, proved.

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

**(i) AC ^{2} = AD^{2} + BC.DM + 2 (BC/2)^{ 2}**

**(ii) AB ^{2} = AD^{2} โ BC.DM + 2 (BC/2)^{ 2}**

**(iii) AC ^{2} + AB^{2} = 2 AD^{2} + ยฝ BC^{2}**

**Solution:**

(i) By applying Pythagoras Theorem in โAMD, we get,

AM^{2} + MD^{2} = AD^{2} โฆโฆโฆโฆโฆโฆ. (i)

Again, by applying Pythagoras Theorem in โAMC, we get,

AM^{2} + MC^{2} = AC^{2}

AM^{2} + (MD + DC)^{ 2} = AC^{2}

(AM^{2} + MD^{2} ) + DC^{2} + 2MD.DC = AC^{2}

From equation(i), we get,

AD^{2} + DC^{2} + 2MD.DC = AC^{2}

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

AD^{2 }+ (BC/2)^{ 2 }+ 2MD.(BC/2)^{ 2} = AC^{2}

AD^{2 }+ (BC/2)^{ 2 }+ 2MD ร BC = AC^{2}

Hence, proved.

(ii) By applying Pythagoras Theorem in โABM, we get;

AB^{2} = AM^{2} + MB^{2}

= (AD^{2} โ DM^{2}) + MB^{2}

= (AD^{2} โ DM^{2}) + (BD โ MD)^{ 2}

= AD^{2} โ DM^{2} + BD^{2} + MD^{2} โ 2BD ร MD

= AD^{2} + BD^{2} โ 2BD ร MD

= AD^{2 }+ (BC/2)^{2 }โ 2(BC/2) MD

= AD^{2 }+ (BC/2)^{2 }โ BC MD

Hence, proved.

(iii) By applying Pythagoras Theorem in โABM, we get,

AM^{2} + MB^{2} = AB^{2} โฆโฆโฆโฆโฆโฆโฆ.โฆ (i)

By applying Pythagoras Theorem in โAMC, we get,

AM^{2} + MC^{2} = AC^{2} โฆโฆโฆโฆโฆโฆโฆ..โฆ (ii)

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

2AM^{2} + MB^{2} + MC^{2} = AB^{2} + AC^{2}

2AM^{2} + (BD โ DM)^{ 2} + (MD + DC)^{ 2} = AB^{2} + AC^{2}

2AM^{2}+BD^{2} + DM^{2} โ 2BD.DM + MD^{2} + DC^{2} + 2MD.DC = AB^{2} + AC^{2}

2AM^{2} + 2MD^{2} + BD^{2} + DC^{2} + 2MD (โ BD + DC) = AB^{2} + AC^{2}

2(AM^{2}+ MD^{2}) + (BC/2)^{ 2} + (BC/2)^{ 2} + 2MD (-BC/2 + BC/2)^{ 2} = AB^{2} + AC^{2}

2AD^{2 }+ BC^{2}/2 = AB^{2} + AC^{2}

**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 โDEA, we get,

DE^{2} + EA^{2} = DA^{2} โฆโฆโฆโฆโฆโฆ.โฆ (i)

By applying Pythagoras Theorem in โDEB, we get,

DE^{2} + EB^{2} = DB^{2}

DE^{2} + (EA + AB)^{ 2} = DB^{2}

(DE^{2} + EA^{2}) + AB^{2} + 2EA ร AB = DB^{2}

DA^{2} + AB^{2} + 2EA ร AB = DB^{2} โฆโฆโฆโฆโฆ. (ii)

By applying Pythagoras Theorem in โADF, we get,

AD^{2} = AF^{2} + FD^{2}

Again, applying Pythagoras theorem in โAFC, we get,

AC^{2} = AF^{2} + FC^{2} = AF^{2} + (DC โ FD)^{ 2}

= AF^{2} + DC^{2} + FD^{2} โ 2DC ร FD

= (AF^{2} + FD^{2}) + DC^{2} โ 2DC ร FD AC^{2}

AC^{2}= AD^{2} + DC^{2} โ 2DC ร FD โฆโฆโฆโฆโฆโฆโฆ (iii)

Since ABCD is a parallelogram,

AB = CD โฆโฆโฆโฆโฆโฆโฆ.โฆ(iv)

And BC = AD โฆโฆโฆโฆโฆโฆ. (v)

In โDEA and โADF,

โ DEA = โ AFD (Each 90ยฐ)

โ EAD = โ ADF (EA || DF)

AD = AD (Common Angles)

โด โEAD โ โFDA (AAS congruence criterion)

โ EA = DF โฆโฆโฆโฆโฆโฆ (vi)

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

DA^{2} + AB^{2} + 2EA ร AB + AD^{2} + DC^{2} โ 2DC ร FD = DB^{2} + AC^{2}

DA^{2} + AB^{2} + AD^{2} + DC^{2} + 2EA ร AB โ 2DC ร FD = DB^{2} + AC^{2}

From equation (iv) and (vi),

BC^{2} + AB^{2} + AD^{2} + DC^{2} + 2EA ร AB โ 2AB ร EA = DB^{2} + AC^{2}

AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

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

**(i) โAPC ~ โ DPB**

**(ii) AP . PB = CP . DP**

**Solution:**

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

(i) In โAPC and โDPB,

โ APC = โ DPB (Vertically opposite angles)

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

Therefore,

โAPC โผ โDPB (AA similarity criterion)

(ii) In the above, we have proved that โAPC โผ โDPB

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

โด AP/DP = PC/PB = CA/BD

โAP/DP = PC/PB

โดAP. 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) โ PAC ~ โ PDB**

**(ii) PA . PB = PC . PD.**

**Solution:**

(i) In โPAC and โPDB,

โ P = โ P (Common Angles)

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

โ PAC = โ PDB

Thus, โPAC โผ โPDB(AA similarity criterion)

(ii) We have already proved above,

โAPC โผ โDPB

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

Therefore,

AP/DP = PC/PB = CA/BD

AP/DP = PC/PB

โด AP. PB = PC. DP

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

**Solution:**

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

AP = AC.

Now join PC.

Given, BD/CD = AB/AC

โ BD/CD = AP/AC

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

AD || PC

โ BAD = โ APC (Corresponding angles) โฆโฆโฆโฆโฆโฆ.. (i)

And, โ DAC = โ ACP (Alternate interior angles) โฆโฆ.โฆ (ii)

By the new figure, we have;

AP = AC

โ โ APC = โ ACP โฆโฆโฆโฆโฆโฆโฆโฆ. (iii)

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

โ BAD = โ APC

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 โABC, is such way;

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

AB^{2 }= (1.8 m)^{ 2} + (2.4 m)^{ 2}

AB^{2 }= (3.24 + 5.76) m^{2}

AB^{2} = 9.00 m^{2}

โน AB = โ9 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 ร 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 โ String pulled by Nazima in 12 seconds

= (3.00 โ 0.6) m

= 2.4 m

In โADB, by Pythagoras Theorem,

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

(1.8 m) ^{2} + BD^{2} = (2.4 m)^{ 2}

BD^{2} = (5.76 โ 3.24) m^{2} = 2.52 m^{2}

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- 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 CBSE Class 10 syllabus, this chapter belongs to the unit that has the second-highest weightage. Hence, having a clear understanding of the concepts, theorems and the 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 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 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 the 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 Maths 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 examination.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 6 โ 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.

Also Access |

NCERT Exemplar for Class 10 Maths Chapter 6 Triangle |

CBSE Notes for Class 10 Maths Chapter 6 Triangle |

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

### 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.

### 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, 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 board, and also depends on the CBSE question paper design and marking scheme.

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

There are 6 exercises are there 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.