# NCERT solution for class 10 maths triangles ( Chapter 6)

#### Solution for Exercise 6.1

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)

(i) Similar
(ii) Similar
(iii)Equilateral
(iv)(a) Equal
(b) Proportional

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

(i)Two Equilateral Triangle and Two Rectangle

(ii) 1)Triangle and rhombus

2)Rectangle and circle

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

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

#### Solution for Exercise 6.2

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

(i) Given, in Triangle ABC, DE II BC
$${{AD} \over {DB}} = {{AE} \over {EC}}$$ [Using Basic proportionality theorem]
=$${{1.5} \over {3}} = {{1} \over {EC}}$$
=EC = $${{3} \over {1.5}}$$
EC = 3×$${{10} \over {15}}$$ = 2 cm
Hence, EC = 2 cm.

(ii) Given, in Triangle ABC, DE is parallel to BC
= $${{AD} \over {DB}} = {{AE} \over {EC}}$$ [Using Basic proportionality theorem]
= $${{AD} \over {7.2}}= {{1.8} \over {5.4}}$$
= $$AD = 1.8 ×{{7.2} \over {5.4}} = ({{18} \over {10}})×({{72} \over {10}})×({{10} \over {54}}) = {{24} \over {10}}$$

2. E and F are points on the sides PQ and PR respectively of a $$\triangle PQR$$. 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

Given, in $$\triangle PQR$$, E and F are two points on side PQ and PR respectively.
(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,
$${{PE} \over {EQ}} = {{3.9} \over {3}} = {{39} \over {30}} = {{13} \over {10}} = 1.3$$
And $${{PF} \over {FR}} = {{3.6} \over {2.4}} = {{36} \over {24}} = {{3} \over {2}} = 1.5$$
So, we get, $${{PE} \over {EQ}} \ne {{PF} \over {FR}}$$
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,
$${{PE} \over {QE}} = {{4} \over {4.5}} = {{40} \over {45}} = {{8} \over {9}}$$
And, $${{PF} \over {RF}} = {{8} \over {9}}$$
So, we get here,
$${{PE} \over {QE}} = {{PF} \over {RF}}$$
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 – PE = 1.28 – 0.18 = 1.10 cm
And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm
So, $${{PE} \over {EQ}} = {{0.18} \over {1.10}} = {{18} \over {110}} = {{9} \over {55}}$$………….(i)
And, $${{PE} \over {FR}} = {{0.36} \over {2.20}} = {{36} \over {220}} = {{9} \over {55}}$$…………(ii)
So, we get here,
$${{PE} \over {EQ}} = {{PF} \over {FR}}$$
Hence, EF is parallel to QR.

3. In the figure, if LM || CB and LN || CD, prove that $$\frac{AM}{AB} = \frac{AN}{AD}$$.

In the given figure, we can see, LM || CB,
By using basic proportionality theorem, we get,
$$\frac{AM}{AB} = \frac{AL}{AC}$$…………………….. (i)
Similarly, given, LN || CD and using basic proportionality theorem,
$$\frac{AN}{AD} = \frac{AL}{AC}$$……………………………(ii)
From equation (i) and (ii), we get,
$$\frac{AM}{AB} = \frac{AN}{AD}$$
Hence, proved.

4. In the figure, DE||AC and DF || AE. Prove that $$\frac{BF}{FE} = \frac{BE}{EC}$$.

In $$\triangle ABC$$, given as, DE || AC
Thus, by using Basic Proportionality Theorem, we get,
$$\frac{BD}{DA} = \frac{BE}{EC}$$ ………………………………………………(i)
In $$\triangle ABC$$, given as, DF || AE
Thus, by using Basic Proportionality Theorem, we get,
$$\frac{BD}{DA} = \frac{BF}{FE}$$ ………………………………………………(ii)
From equation (i) and (ii), we get
$$\frac{BE}{EC} = \frac{BF}{FE}$$
Hence, proved.

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

Given,
In $$\triangle PQO$$, DE || OQ
So by using Basic Proportionality Theorem,
$$\frac{PD}{DO} = \frac{PE}{EQ}$$ ………………. (i)
Again given, in $$\triangle PQO$$, DE || OQ ,
So by using Basic Proportionality Theorem,
$$\frac{PD}{DO} = \frac{PF}{FR}$$ ………………… (ii)
From equation (i) and (ii), we get,
$$\frac{PE}{EQ} = \frac{PF}{FR}$$
Therefore, by converse of Basic Proportionality Theorem,
EF || QR, in $$\triangle PQR$$.

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.

Given here,
In $$\triangle OPQ$$, AB || PQ
By using Basic Proportionality Theorem,
$$\frac{OA}{AP} = \frac{OB}{BQ}$$…………….(i)
Also given,
In $$\triangle OPR$$, AC || PR
By using Basic Proportionality Theorem
$$\frac{OA}{AP} = \frac{OC}{CR}$$……………(ii)
From equation (i) and (ii), we get,
$$\frac{OB}{BQ} = \frac{OC}{CR}$$
Therefore, by converse of Basic Proportionality Theorem,
In $$\triangle OQR$$, 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).

Given, in $$\triangle ABC$$, 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.
=>$$\frac{AD}{DB}$$ = 1 …………………………. (i)
In $$\triangle ABC$$, DE || BC,
By using Basic Proportionality Theorem,
Therefore, $$\frac{AD}{DB} = \frac{AE}{EC}$$
From equation (i), we can write,
=> 1 = $$\frac{AE}{EC}$$
=> 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).

Given, in $$\triangle ABC$$, 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
=>$$\frac{AD}{BD}$$ = 1……………………………….. (i)
Also given, E is the mid-point of AC.
AE=EC
=> $$\frac{AE}{EC}$$ = 1
From equation (i) and (ii), we get,
$$\frac{AD}{BD} = \frac{AE}{EC}$$
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 $$\frac{AO}{BO} = \frac{CO}{DO}$$.

Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.
We have to prove, $$\frac{AO}{BO} = \frac{CO}{DO}$$
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In $$\triangle ADC$$, we have OE || DC
Therefore, By using Basic Proportionality Theorem
$$\frac{AE}{ED} = \frac{AO}{CO}$$ ……………..(i)
Now, In $$\triangle ABD$$, OE || AB
Therefore, By using Basic Proportionality Theorem
$$\frac{DE}{EA} = \frac{DO}{BO}$$…………….(ii)
From equation (i) and (ii), we get,
$$\frac{AO}{CO} = \frac{BO}{DO}$$
=>$$\frac{AO}{BO} = \frac{CO}{DO}$$
Hence, proved.

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

Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,
$$\frac{AO}{BO} = \frac{CO}{DO}$$.
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 $$\triangle DAB$$, EO || AB
Therefore, By using Basic Proportionality Theorem
$$\frac{DE}{EA} = \frac{DO}{OB}$$ ……………………(i)
Also, given,
$$\frac{AO}{BO} = \frac{CO}{DO}$$
=> $$\frac{AO}{CO} = \frac{BO}{DO}$$
=> $$\frac{CO}{AO} = \frac{DO}{BO}$$
=> $$\frac{DO}{OB} = \frac{CO}{AO}$$ …………………………..(ii)
From equation (i) and (ii), we get
$$\frac{DE}{EA} = \frac{CO}{AO}$$
Therefore, By using converse of Basic Proportionality Theorem,
EO || DC also EO || AB
=> AB || DC.
Hence, quadrilateral ABCD is a trapezium with AB || CD.

#### Solution for Exercise 6.3

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:

i) Given, in $$\triangle$$ ABC and $$\triangle$$ PQR,
$$\angle$$A = $$\angle$$P = 60°
$$\angle$$B = $$\angle$$Q = 80°
$$\angle$$C = $$\angle$$R = 40°
Therefore by AAA similarity criterion,
$$\triangle$$ ABC is similar to $$\triangle$$ PQR

(ii) Given, in $$\triangle$$ ABC and $$\triangle$$ PQR,
AB/QR = BC/RP = CA/PQ
By SSS similarity criterion,
$$\triangle$$ ABC is similar to $$\triangle$$ QRP

(iii) Given, in $$\triangle$$ LMP and $$\triangle$$ 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 $$\ne$$ LM/EF
Therefore, $$\triangle$$ LMP and $$\triangle$$ DEF are not similar.

(iv) In $$\triangle$$ MNL and $$\triangle$$ QPR, it is given,
MN/QP = ML/QR = 1/2
$$\angle$$M = $$\angle$$Q = 70°
Therefore, by SAS similarity criterion
$$\triangle$$ MNL is similar to $$\triangle$$ QPR

(v) In $$\triangle$$ ABC and $$\triangle$$ DEF, given that,
AB = 2.5, BC = 3, $$\angle$$A = 80°, EF = 6, DF = 5, $$\angle$$F = 80°
Here , AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
$$\angle$$B $$\ne$$ $$\angle$$F
Hence, $$\triangle$$ ABC and $$\triangle$$ DEF are not similar.

(vi) In $$\triangle$$ DEF, by sum of angles of triangles, we know that,
$$\angle$$D + $$\angle$$E + $$\angle$$F = 180°
70° + 80° + $$\angle$$F = 180°
$$\angle$$F = 180° – 70° – 80°
$$\angle$$F = 30°
Similarly, In $$\triangle$$ PQR,
$$\angle$$P + $$\angle$$Q + $$\angle$$R = 180 (Sum of angles of $$\triangle$$ )
$$\angle$$P + 80° + 30° = 180°
$$\angle$$P = 180° – 80° -30°
$$\angle$$P = 70°
Now, comparing both the triangles, $$\triangle$$ DEF and $$\triangle$$ PQR, we have
$$\angle$$D = $$\angle$$P = 70°
$$\angle$$E = $$\angle$$Q = 80°
$$\angle$$F = $$\angle$$R = 30°
Therefore, by AAA similarity criterion,
Hence, $$\triangle$$ DEF is similar to $$\triangle$$ PQR

2. In the figure, $$\triangle$$ ODC $$\sim$$ $$\triangle$$ OBA, $$\angle$$ BOC = 125° and $$\angle$$ CDO = 70°. Find $$\angle$$ DOC, $$\angle$$ DCO and $$\angle$$ OAB.

As we can see from the figure, DOB is a straight line.
Therefore, $$\angle$$ DOC + $$\angle$$ COB = 180°
$$\angle$$ DOC = 180° – 125° (Given, $$\angle$$ BOC = 125°)
= 55°
In $$\triangle$$ DOC, Sum of the measures of the angles of a triangle is 180º
Therefore, $$\angle$$ DCO + $$\angle$$ CDO + $$\angle$$ DOC = 180°
$$\angle$$ DCO + 70º + 55º = 180°(Given, $$\angle$$ CDO = 70°)
$$\angle$$ DCO = 55°
It is given that, $$\triangle$$ ODC $$\sim$$ OBA,
Therefore, $$\triangle$$ ODC ~ OBA.
Hence, Corresponding angles are equal in similar triangles
$$\angle$$ OAB = $$\angle$$ OCD
$$\angle$$ OAB = 55°

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

In $$\triangle$$DOC and $$\triangle$$BOA,
AB || CD, thus alternate interior angles will be equal,
$$\angle$$CDO = $$\angle$$ABO
Similarly,
$$\angle$$DCO = $$\angle$$BAO
Also, for the two triangles $$\triangle$$DOC and $$\triangle$$BOA, vertically opposite angles will be equal;
$$\angle$$DOC = $$\angle$$BOA
Hence, by AAA similarity criterion,
$$\triangle$$DOC ~ $$\triangle$$BOA
Thus, the corresponding sides are proportional.
DO/BO = OC/OA
OA/OC = OB/OD
Hence, proved.

4. In the fig.6.36, QR/QS = QT/PR and $$\angle$$ 1 = $$\angle$$ 2. Show that $$\triangle$$ PQS ~ $$\triangle$$ TQR.

In $$\triangle$$ PQR,
$$\angle$$ PQR = $$\angle$$ PRQ
PQ = PR ………………………(i)
Given,
QR/QS = QT/PR
Using equation (i), we get
QR/QS = QT/QP……………….(ii)
In $$\triangle$$ PQS and $$\triangle$$ TQR, by equation (ii),
QR/QS = QT/QP
$$\angle$$ Q = $$\angle$$ Q
$$\triangle$$ PQS ~ $$\triangle$$ TQR [By SAS similarity criterion]

5. S and T are point on sides PR and QR of $$\triangle$$ PQR such that $$\angle$$ P = $$\angle$$ RTS. Show that $$\triangle$$ RPQ ~ $$\triangle$$ RTS.

Given, S and T are point on sides PR and QR of $$\triangle$$ PQR
And $$\angle$$ P = $$\angle$$ RTS.
In $$\triangle$$ RPQ and $$\triangle$$ RTS,
$$\angle$$ RTS = $$\angle$$ QPS (Given)
$$\angle$$ R = $$\angle$$ R (Common angle)
$$\triangle$$ RPQ ~ $$\triangle$$ RTS (AA similarity criterion)

6. In the figure, if $$\triangle$$ ABE $${\displaystyle \cong }$$ $$\triangle$$ ACD, show that $$\triangle$$ ADE ~ $$\triangle$$ ABC.

Given, $$\triangle$$ ABE $${\displaystyle \cong }$$ $$\triangle$$ ACD.
AB = AC [By CPCT] ……………………………….(i)
And, AD = AE [By CPCT] ……………………………(ii)
In $$\triangle$$ ADE and $$\triangle$$ ABC, dividing eq.(ii) by eq(i),
$$\angle$$ A = $$\angle$$ A [Common angle]
$$\triangle$$ ADE ~ $$\triangle$$ ABC [SAS similarity criterion]

7. In the figure, altitudes AD and CE of $$\triangle$$ ABC intersect each other at the point P. Show that:
(i) $$\triangle$$ AEP ~ $$\triangle$$ CDP
(ii) $$\triangle$$ ABD ~ $$\triangle$$ CBE
(iii) $$\triangle$$ AEP ~ $$\triangle$$ ADB
(iv) $$\triangle$$ PDC ~ $$\triangle$$ BEC.

Given, altitudes AD and CE of $$\triangle$$ ABC intersect each other at the point P.
(i) In $$\triangle$$ AEP and $$\triangle$$ CDP,
$$\angle$$ AEP = $$\angle$$ CDP (90° each)
$$\angle$$ APE = $$\angle$$ CPD (Vertically opposite angles)
Hence, by AA similarity criterion,
$$\triangle$$ AEP ~ $$\triangle$$ CDP

(ii) In $$\triangle$$ ABD and $$\triangle$$ CBE,
$$\angle$$ ADB = $$\angle$$ CEB ( 90° each)
$$\angle$$ ABD = $$\angle$$ CBE (Common Angles)
Hence, by AA similarity criterion,
$$\triangle$$ ABD ~ $$\triangle$$ CBE

(iii) In $$\triangle$$ AEP and $$\triangle$$ ADB,
$$\angle$$ AEP = $$\angle$$ ADB (90° each)
$$\angle$$ PAE = $$\angle$$ DAB (Common Angles)
Hence, by AA similarity criterion,
$$\triangle$$ AEP ~ $$\triangle$$ ADB

(iv) In $$\triangle$$ PDC and $$\triangle$$ BEC,
$$\angle$$ PDC = $$\angle$$ BEC (90° each)
$$\angle$$ PCD = $$\angle$$ BCE (Common angles)
Hence, by AA similarity criterion,
$$\triangle$$ PDC ~ $$\triangle$$ BEC
Hence, by AA similarity criterion,
$$\triangle$$ PDC ~ $$\triangle$$ BEC

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

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 $$\triangle$$ ABE and $$\triangle$$ CFB,
$$\angle$$ A = $$\angle$$ C (Opposite angles of a parallelogram)
$$\angle$$ AEB = $$\angle$$ CBF (Alternate interior angles as AE || BC)
$$\triangle$$ ABE ~ $$\triangle$$ 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) $$\triangle$$ ABC ~ $$\triangle$$ AMP
(ii) CA/PA = BC/MP

Given, ABC and AMP are two right triangles, right angled at B and M respectively.
(i) In $$\triangle$$ ABC and $$\triangle$$ AMP, we have,
$$\angle$$ CAB = $$\angle$$ MAP (common angles)
$$\angle$$ ABC = $$\angle$$ AMP = 90° (each 90°)
$$\triangle$$ ABC ~ $$\triangle$$ AMP (AA similarity criterion)

(ii) As, $$\triangle$$ ABC ~ $$\triangle$$ 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 $$\angle$$ ACB and $$\angle$$ EGF such that D and H lie on sides AB and FE of $$\triangle$$ ABC and $$\triangle$$ EFG respectively. If $$\triangle$$ ABC ~ $$\triangle$$ FEG, Show that: (i) CD/GH = AC/FG (ii) $$\triangle$$ DCB ~ $$\triangle$$ HGE (iii) $$\triangle$$ DCA ~ $$\triangle$$ HGF

Given, CD and GH are respectively the bisectors of $$\angle$$ ACB and $$\angle$$ EGF such that D and H lie on sides AB and FE of $$\triangle$$ ABC and $$\triangle$$ EFG respectively.
(i) From the given condition,
$$\triangle$$ ABC ~ $$\triangle$$ FEG.
$$\angle$$ A = $$\angle$$ F, $$\angle$$ B = $$\angle$$ E, and $$\angle$$ ACB = $$\angle$$ FGE
Since, $$\angle$$ ACB = $$\angle$$ FGE
$$\angle$$ ACD = $$\angle$$ FGH (Angle bisector)
And, $$\angle$$ DCB = $$\angle$$ HGE (Angle bisector)
In $$\triangle$$ ACD and $$\triangle$$ FGH,
$$\angle$$ A = $$\angle$$ F
$$\angle$$ ACD = $$\angle$$ FGH
$$\triangle$$ ACD ~ $$\triangle$$ FGH (AA similarity criterion)
CD/GH = AC/FG

(ii) In $$\triangle$$ DCB and $$\triangle$$ HGE,
$$\angle$$ DCB = $$\angle$$ HGE (Already proved)
$$\angle$$ B = $$\angle$$ E (Already proved)
$$\triangle$$ DCB ~ $$\triangle$$ HGE (AA similarity criterion)

(iii) In $$\triangle$$ DCA and $$\triangle$$ HGF,
$$\angle$$ ACD = $$\angle$$ FGH (Already proved)
$$\angle$$ A = $$\angle$$ F (Already proved)
$$\triangle$$ DCA ~ $$\triangle$$ 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 $$\perp$$ BC and EF $$\perp$$ AC, prove that $$\triangle$$ ABD ~ $$\triangle$$ ECF.

Given, ABC is an isosceles triangle.
AB = AC
$$\angle$$ ABD = $$\angle$$ ECF
In $$\triangle$$ ABD and $$\triangle$$ ECF,
$$\angle$$ ADB = $$\angle$$ EFC (Each 90°)
$$\angle$$ BAD = $$\angle$$ CEF (Already proved)
$$\triangle$$ ABD ~ $$\triangle$$ ECF (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 $$\triangle$$ PQR (see Fig 6.41). Show that $$\triangle$$ ABC ~ $$\triangle$$ PQR.

Given, $$\triangle$$ ABC and $$\triangle$$ PQR, AB, BC and median AD of $$\triangle$$ ABC are proportional to sides PQ, QR and median PM of $$\triangle$$ PQR
i.e. AB/PQ = BC/QR = AD/PM
We have to prove: $$\triangle$$ ABC ~ $$\triangle$$ PQR
As we know here,
$${{AB} \over {PQ}} = {{{{1} \over {2}}BC} \over {{{1} \over {2}}QR}} = {{AD} \over {PM}}$$
AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)
$$\triangle$$ ABD ~ $$\triangle$$ PQM [SSS similarity criterion]
$$\angle$$ ABD = $$\angle$$ PQM [Corresponding angles of two similar triangles are equal]
$$\angle$$ ABC = $$\angle$$ PQR
In $$\triangle$$ ABC and $$\triangle$$ PQR
AB/PQ = BC/QR ………………………….(i)
$$\angle$$ ABC = $$\angle$$ PQR ……………………………(ii)
From equation (i) and (ii), we get,
$$\triangle$$ ABC ~ $$\triangle$$ PQR [SAS similarity criterion]

13. D is a point on the side BC of a triangle ABC such that $$\angle$$ ADC = $$\angle$$ BAC. Show that $$CA^2 = CB.CD$$

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

In $$\triangle$$ ADC and $$\triangle$$ BAC,
$$\angle$$ ADC = $$\angle$$ BAC (Already given)
$$\angle$$ ACD = $$\angle$$ BCA (Common angles)
$$\triangle$$ ADC ~ $$\triangle$$ BAC (AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
CA/CB = CD/CA
$$CA^2 = CB.CD$$
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 $$\triangle$$ ABC ~ $$\triangle$$ PQR.

Given: Two triangles $$\triangle$$ ABC and $$\triangle$$ PQR in which AD and PM are medians such that;
We have to prove, $$\triangle$$ ABC ~ $$\triangle$$ 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 $$\triangle$$ ABD and $$\triangle$$ CDE, we have
BD = DC [Since, AP is the median]
and, $$\angle$$ ADB = $$\angle$$ CDE [Vertically opposite angles]
$$\triangle$$ ABD $$\cong$$ $$\triangle$$ CDE [SAS criterion of congruence]
AB = CE [By CPCT] …………………………..(i)
Also, in $$\triangle$$ PQM and $$\triangle$$ MNR,
PM = MN [By Construction.]
QM = MR [Since, PM is the median]
and, $$\angle$$ PMQ = $$\angle$$ NMR [Vertically opposite angles]
$$\triangle$$ PQM = $$\triangle$$ MNR [SAS criterion of congruence]
PQ = RN [CPCT] ………………………………(ii)
Now, AB/PQ = AC/PR = AD/PM
From equation (i) and (ii),
CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]
$$\triangle$$ ACE ~ $$\triangle$$ PRN [SSS similarity criterion]
Therefore, $$\angle$$ 2 = $$\angle$$ 4
Similarly, $$\angle$$ 1 = $$\angle$$ 3
$$\angle$$ 1 + $$\angle$$ 2 = $$\angle$$ 3 + $$\angle$$ 4
$$\angle$$ A = $$\angle$$ P …………………………………………….(iii)
Now, In $$\triangle$$ ABC and $$\triangle$$ PQR, we have
From equation (iii),
$$\angle$$ A = $$\angle$$ P
$$\triangle$$ ABC ~ $$\triangle$$ PQR [ SAS similarity criterion]

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

Given, Length of the vertical pole = 6 m
Shadow of the pole = 4 m
Let Height of tower = h m
Length of shadow of the tower = 28 m

In $$\triangle$$ ABC and $$\triangle$$ DEF,
$$\angle$$ C = $$\angle$$ E (angular elevation of sum)
$$\angle$$ B = $$\angle$$ F = 90°
$$\triangle$$ ABC ~ $$\triangle$$ DEF (AA similarity criterion)
AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)
6/h = 4/28
h =( 6×28)/4
h = 6 × 7
h = 42 m
Hence, the height of the tower is 42 m.

16. If AD and PM are medians of triangles ABC and PQR, respectively where $$\triangle$$ ABC ~ $$\triangle$$ PQR prove that AB/PQ = AD/PM.

Given, $$\triangle$$ ABC ~ $$\triangle$$ PQR

We know that the corresponding sides of similar triangles are in proportion.
AB/PQ = AC/PR = BC/QR……………………………(i)
Also, $$\angle$$ A = $$\angle$$ P, $$\angle$$ B = $$\angle$$ Q, $$\angle$$ C = $$\angle$$ R ………….…..(ii)
Since AD and PM are medians, they will divide their opposite sides.
BD = $$\frac{BC}{2}$$ and QM = $$\frac{QR}{2}$$ ……………..………….(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM ……………………….(iv)
In $$\triangle$$ ABD and $$\triangle$$ PQM,
From equation (ii), we have
$$\angle$$ B = $$\angle$$ Q
From equation (iv), we have,
AB/PQ = BD/QM
$$\triangle$$ ABD ~ $$\triangle$$ PQM (SAS similarity criterion)

#### Solution for Exercise 6.4

1. Let $$\triangle$$ ABC ~ $$\triangle$$ DEF and their areas be, respectively, 64 $$cm^2$$and 121 $$cm^2$$. If EF = 15.4 cm, find BC.

Given, $$\triangle$$ ABC ~ $$\triangle$$ DEF,
Area of $$\triangle$$ ABC = 64 $$cm^2$$
Area of $$\triangle$$ DEF = 121 $$cm^2$$
EF = 15.4 cm
$${{Area of \triangle ABC} \over {Area of \triangle DEF}} = {{AB^2} \over {DE^2}}$$
As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,
= $$\frac{AC^2}{DF^2} = \frac{BC^2}{EF^2}$$
$$\frac{64}{121} = \frac{BC^2}{EF^2}$$
$$(\frac{8}{11})^2 = (\frac{BC}{15.4})^2$$
$$\frac{8}{11} = \frac{BC}{15.4}$$
BC = $$8×\frac{15.4}{11}$$
BC = 8 × 1.4
BC = 11.2 cm

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

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

In $$\triangle$$ AOB and $$\triangle$$ COD, we have
$$\angle$$ 1 = $$\angle$$ 2 (Alternate angles)
$$\angle$$ 3 = $$\angle$$ 4 (Alternate angles)
$$\angle$$ 5 = $$\angle$$ 6 (Vertically opposite angle)
$$\triangle$$ AOB ~ $$\triangle$$ 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 ($$\triangle$$ AOB)/Area of ($$\triangle$$ COD) = $$AB^2/CD^2$$
= $$(2CD)^2/CD^2$$ [ AB = CD]
Area of ($$\triangle$$ AOB)/Area of ($$\triangle$$ COD)
= $$4CD^2/CD$$ = 4/1
Hence, the required ratio of the area of $$\triangle$$ AOB and $$\triangle$$ 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 ($$\triangle$$ ABC)/area ($$\triangle$$ DBC) = AO/DO.

Given, ABC and DBC are two triangles on the same base BC. ADintersects BC at O.
We have to prove: Area ($$\triangle$$ ABC)/Area ($$\triangle$$ DBC) = AO/DO
Let us draw two perpendiculars AP and DM on line BC.

We know that area of a triangle = 1/2 × Base × Height
$${{ar(\triangle ABC) } \over {ar(\triangle DEF}} = {{{{1} \over {2}}BC(AP)}\over {{{1} \over {2}}BC(DM)}} ={{AP} \over {DM}}$$
In $$\triangle$$ APO and $$\triangle$$ DMO,
$$\angle$$ APO = $$\angle$$ DMO (Each 90°)
$$\angle$$ AOP = $$\angle$$ DOM (Vertically opposite angles)
$$\triangle$$ APO ~ $$\triangle$$ DMO (AA similarity criterion)
AP/DM = AO/DO
Area ($$\triangle$$ ABC)/Area ($$\triangle$$ DBC) = AO/DO.

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

Say $$\triangle$$ ABC and $$\triangle$$ PQR are two similar triangles and equal in area

Now let us prove $$\triangle$$ ABC $${\displaystyle \cong }$$ $$\triangle$$ PQR.
Since, $$\triangle$$ ABC ~ $$\triangle$$ PQR
Area of ($$\triangle$$ ABC)/Area of ($$\triangle$$ PQR) = $$BC^2/QR^2$$
$$BC^2/QR^2$$ = 1 [Since, Area($$\triangle$$ ABC) = ($$\triangle$$ PQR)
$$BC^2/QR^2$$
BC = QR
Similarly, we can prove that
AB = PQ and AC = PR

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

Given, D, E and F are respectively the mid-points of sides AB, BC and CA of $$\triangle$$ ABC.

In $$\triangle$$ ABC,
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
FE || BC and FE || BD [BD = 1/2BC]
Since, opposite sides of parallelogram are equal and parallel
BDEF is parallelogram.
Similarly in $$\triangle$$ FBD and $$\triangle$$ DEF, we have
FB = DE (Opposite sides of parallelogram BDEF)
FD = FD (Common sides)
BD = FE (Opposite sides of parallelogram BDEF)
$$\triangle$$ FBD $${\displaystyle \cong }$$ $$\triangle$$ DEF
Similarly, we can prove that
$$\triangle$$ AFE $${\displaystyle \cong }$$ $$\triangle$$ DEF
$$\triangle$$ EDC $${\displaystyle \cong }$$ $$\triangle$$ DEF
As we know, if triangles are congruent, then they are equal in area.
So,
Area($$\triangle$$ FBD) = Area($$\triangle$$ DEF) ……………………………(i)
Area($$\triangle$$ AFE) = Area($$\triangle$$ DEF) ……………………………….(ii)
and,
Area($$\triangle$$ EDC) = Area($$\triangle$$ DEF) ………………………….(iii)
Now,
Area($$\triangle$$ ABC) = Area($$\triangle$$ FBD) + Area($$\triangle$$ DEF) + Area($$\triangle$$ AFE) + Area($$\triangle$$ EDC) ………(iv)
Area($$\triangle$$ ABC) = Area($$\triangle$$ DEF) + Area($$\triangle$$ DEF) + Area($$\triangle$$ DEF) + Area($$\triangle$$ DEF)
From equation (i), (ii) and (iii),
Area($$\triangle$$ DEF) = (1/4)Area($$\triangle$$ ABC)
Area($$\triangle$$ DEF)/Area($$\triangle$$ ABC) = 1/4
Hence, Area($$\triangle$$ DEF):Area($$\triangle$$ ABC) = 1:4

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

Given: AM and DN are the medians of triangles ABC and DEF respectively and $$\triangle$$ ABC ~ $$\triangle$$ DEF.

We have to prove: Area($$\triangle$$ ABC)/Area($$\triangle$$ DEF) = AM2/DN2
Since, $$\triangle$$ ABC ~ $$\triangle$$ DEF (Given)
Area($$\triangle$$ ABC)/Area($$\triangle$$ DEF) = $$(AB^2/DE^2)$$ ……………………………(i)
and, AB/DE = BC/EF = CA/FD ………………………………………(ii)
$${{AB} \over {DE}} ={{{{1} \over {2}}BC} \over {{{1} \over {2}}EF}} = {{CD} \over {FD}}$$
In $$\triangle$$ ABM and $$\triangle$$ DEN,
Since $$\triangle$$ ABC ~ $$\triangle$$ DEF
$$\angle$$ B = $$\angle$$ E
AB/DE = BM/EN [Already Proved in equation (i)]
$$\triangle$$ ABC ~ $$\triangle$$ DEF [SAS similarity criterion]
AB/DE = AM/DN …………………………………………………..(iii)
$$\triangle$$ ABM ~ $$\triangle$$ DEN
As the areas of two similar triangles are proportional to the squares of the corresponding sides.
area($$\triangle$$ ABC)/area($$\triangle$$ DEF) = $$AB^2/DE^2 = AM^2/DN^2$$
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.

Given, ABCD is a square whose one diagonal is AC. $$\triangle$$ APC and $$\triangle$$ BQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

Area($$\triangle$$ BQC) = $${{1} \over {2}}$$ Area($$\triangle$$ APC)
Since, $$\triangle$$ APC and $$\triangle$$ BQC are both equilateral triangles, as per given,
$$\triangle$$ APC ~ $$\triangle$$ BQC [AAA similarity criterion]
area($$\triangle$$ APC)/area($$\triangle$$ BQC) = (AC2/BC2) = AC2/BC2
Since, Diagonal = $$\sqrt{2}$$ side = $$\sqrt{2}$$ BC
$$({{\sqrt{2} BC } \over {BC}})^2 = 2$$
area($$\triangle$$ APC) = 2 × area($$\triangle$$ BQC)
area($$\triangle$$ BQC) = $${{1} \over {2}}$$area($$\triangle$$ APC)
Hence, proved.

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

Given, $$\triangle$$ ABC and $$\triangle$$ BDE are two equilateral triangle. D is the midpoint of BC.
BD = DC = 1/2BC
Let each side of triangle is 2a.
As, $$\triangle$$ ABC ~ $$\triangle$$ BDE
Area($$\triangle$$ ABC)/Area($$\triangle$$ BDE) = $$AB^/BD^2 = (2a)^2/(a)^2 = 4a^2/a^2 = 4/1 = 4:1$$
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

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

Let ABC and DEF are two similar triangles, such that,
$$\triangle$$ ABC ~ $$\triangle$$ 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,
Area($$\triangle$$ ABC)/Area($$\triangle$$ DEF) = $$AB^2/DE^2$$
Area($$\triangle$$ ABC)/Area($$\triangle$$ DEF) = $$(4/9)^2$$ = 16/81 = 16:81
Hence, the correct answer is (D).

#### Solution for Exercise 6.5

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

(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 $$\ne$$ 64
Or, $$3^2 + 6^2 \ne 8^2$$
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 triangle’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 $$\ne$$ 10000
Or, $$50^2 + 80^2 \ne 100^2$$
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 $$\perp$$ QR. Show that $$PM^2 = QM × MR$$.

Given, $$\triangle$$ PQR is right angled at P is a point on QR such that PM ?QR

We have to prove, $$PM^2 = QM × MR$$
In $$\triangle$$ PQM, by Pythagoras theorem
$$PQ^2 = PM^2 + QM^2$$
Or, $$PM^2 = PQ^2 – QM^2$$ ……………………………..(i)
In $$\triangle$$ PMR, by Pythagoras theorem
$$PR^2 = PM^2 + MR^2$$
Or, $$PM^2 = PR^2 – MR^2$$ ………………………………………..(ii)
Adding equation, (i) and (ii), we get,
$$2PM^2 = (PQ^2 + PM^2) – (QM^2 + MR^2)$$
= $$QR^2 – QM^2 – MR^2$$ [Thus $$QR^2 = PQ^2 + PR^2$$]
= $$(QM + MR)^2 – QM^2 – MR^2$$
= 2QM × MR
$$PM^2 = QM × MR$$

3. In Figure, ABD is a triangle right angled at A and AC $$\perp$$ BD. Show that
(i) $$AB^2 = BC × BD$$
(ii) $$AC^2 = BC × DC$$
(iii) $$AD^2 = BD × CD$$

(i) In $$\triangle$$ ADB and $$\triangle$$ CAB,
$$\angle$$ DAB = $$\angle$$ ACB (Each 90°)
$$\angle$$ ABD = $$\angle$$ CBA (Common angles)
$$\triangle$$ ADB ~ $$\triangle$$ CAB [AA similarity criterion]
AB/CB = BD/AB
$$AB^2 = CB × BD$$

(ii) Let $$\angle$$ CAB = x
In $$\triangle$$ CBA,
$$\angle$$ CBA = 180° – 90° – x
$$\angle$$ CBA = 90° – x
Similarly, in $$\triangle$$ CAD
$$\angle$$ CAD = 90° – $$\angle$$ CBA
= 90° – x
$$\angle$$ CDA = 180° – 90° – (90° – x)
$$\angle$$ CDA = x
In $$\triangle$$ CBA and $$\triangle$$ CAD, we have
$$\angle$$ CBA = $$\angle$$ CAD
$$\angle$$ CAB = $$\angle$$ CDA
$$\angle$$ ACB = $$\angle$$ DCA (Each 90°)
$$\triangle$$ CBA ~ $$\triangle$$ CAD [AAA similarity criterion]
AC/DC = BC/AC
$$AC^2 = DC × BC$$

(iii) In $$\triangle$$ DCA and $$\triangle$$ DAB,
$$\angle$$ DCA = $$\angle$$ DAB (Each 90°)
$$\angle$$ CDA = $$\angle$$ ADB (common angles)
$$\triangle$$ DCA ~ $$\triangle$$ DAB [AA similarity criterion]
DC/DA = DA/DA
$$AD^2 = BD × CD$$

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

Given, $$\triangle$$ ABC is an isosceles triangle right angled at C.

In $$\triangle$$ ACB, Angle C = 90°
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.

Given, $$\triangle$$ABC is an isosceles triangle having AC = BC and $$AB^2 = 2AC^2$$

In $$\triangle$$ ACB,
AC = BC
$$AB^2 = 2AC^2$$
$$AB^2 = AC^2 + AC^2$$
= $$AC^2 + BC^2$$ [Since, AC = BC]
Hence, by Pythagoras theorem $$\triangle$$ ABC is right angle triangle.

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

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

Draw, AD $$\perp$$ BC
In $$\triangle$$ ADB and $$\triangle$$ ADC,
AB = AC
$$\angle$$ ADB = $$\angle$$ ADC [Both are 90°]
Therefore, $$\triangle$$ ADB $${\displaystyle \cong }$$ $$\triangle$$ ADC by RHS congruence.
Hence, BD = DC [by CPCT]
In right angled $$\triangle$$ ADB,
$$AB^2 = AD^2 + BD^2$$
$$(2a)^2 = AD^2 + a^2$$
$$AD^2 = 4a^2 – a^2$$
$$AD^2 = 3a^2$$
$$AD = \sqrt{3} a$$

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

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 $$\triangle$$ AOB,
$$\angle$$ AOB = 90°
$$AB^2 = AO^2 + BO^2$$ …………………….. (i) [By Pythagoras theorem]
Similarly,
$$AD^2 = AO^2 + DO^2$$ …………………….. (ii)
$$DC^2 = DO^2 + CO^2$$ …………………….. (iii)
$$BC^2 = CO^2 + BO^2$$ …………………….. (iv)
Adding equations (i) + (ii) + (iii) + (iv), we get,
$$AB^2 + AD^2 + DC^2 + BC^2 = 2(AO^2 + BO^2 + DO^2 + CO^2 )$$
= $$4AO^2 + 4BO^2$$ [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 $$\perp$$ BC, OE $$\perp$$ AC and OF $$\perp$$ AB. Show that:
(i) $$OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2 = AF^2 + BD^2 + CE^2$$ ,
(ii) $$AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2$$.

Given, in $$\triangle$$ ABC, O is a point in the interior of a triangle.
And OD $$\perp$$ BC, OE $$\perp$$ AC and OF $$\perp$$ AB
Join OA, OB and OC

(i) By Pythagoras theorem in $$\triangle$$ AOF, we have
$$OA^2 = OF^2 + AF^2$$
Similarly, in $$\triangle$$ BOD
$$OB^2 = OD^2 + BD^2$$
Similarly, in $$\triangle$$ COE
$$OC^2 = OE^2 + EC^2$$
$$OA^2 + OB^2 + OC^2 = OF^2 + AF^2 + OD^2 + BD^2 + OE^2 + EC^2$$
$$OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2 = AF^2 + BD^2 + CE^2.$$

(ii) $$AF^2 + BD^2 + EC^2 = (OA^2 – OE^2) + (OC^2 – OD^2) + (OB^2 – OF^2)$$
$$AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2$$.

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

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,
$$AC^2 = AB^2 + BC^2$$
$$10^2 = 8^2 + BC^2$$
$$BC^2 = 100 – 64$$
$$BC^2 = 36$$
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?

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,
$$AC^2 = AB^2 + BC^2$$
$$24^2 = 18^2 + BC^2$$
$$BC^2 = 576 – 324$$
$$BC^2 = 252$$
$$BC = 6 \sqrt{7}$$m
Therefore, the distance from the base is $$6 \sqrt{7}$$ m

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 $${1} {\dfrac{1}{2}}$$ hours?

Given,
Speed of first aeroplane = 1000 km/hr
Distance covered by first aeroplane flying due north in
$${1} {\dfrac{1}{2}}$$ hours = 100 × 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr
Distance covered by second aeroplane flying due west in
$${1} {\dfrac{1}{2}}$$ hours (OB) = 1200 × 3/2 km = 1800 km

In right angle $$\triangle$$ AOB, by Pythagoras Theorem,
$$AB^2 = AO^2 + OB^2$$
$$AB^2 = (1500)^2 + (1800)^2$$
$$AB = \sqrt {(2250000 + 3240000)}$$
= $$\sqrt{5490000}$$
AB = $$300 \sqrt{61}$$ 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.

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 6 m and 11 m.
Therefore, CP = 11 – 6 = 5m
From the figure, it can be observed that AP = 12 m
By Pythagoras theorem for $$\triangle$$ APC, 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 $$AE^2 + BD^2 = AB^2 + DE^2$$.

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 $$\triangle$$ ACE, we get
$$AC^2 + CE^2 = AE^2$$ ………………………………………….(i)
In $$\triangle$$ BCD, by Pythagoras theorem, we get
$$BC^2 + CD^2 = BD^2$$ ………………………………..(ii)
From equations (i) and (ii), we get,
$$AC^2 + CE^2 + BC^2 + CD^2 = AE^2 + BD^2$$ …………..(iii)
In $$\triangle$$ CDE, by Pythagoras theorem, we get
$$DE^2 = CD^2 + CE^2$$
In $$\triangle$$ ABC, 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 $$\triangle$$ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that $$2AB^2 = 2AC^2 + BC^2$$.

Given, the perpendicular from A on side BC of a $$\triangle$$ ABC intersects BC at D such that;
DB = 3CD.
In $$\triangle$$ ABC,
AD $$\perp$$ BC and BD = 3CD
$$AB^2 = AD^2 + BD^2$$ ……………………….(i)
$$AC^2 = AD^2 + DC^2$$ ……………………………..(ii)
Subtracting equation (ii) from equation (i), we get
$$AB^2 – AC^2 = BD^2 – DC^2$$
= $$9CD^2 – CD^2$$ [Since, BD = 3CD]
= $$9CD^2 = 8(BC/4)^2 [Since, BC = DB + CD = 3CD + CD = 4CD]$$
Therefore, $$AB^2 – AC^2 = {{BC^2} \over {2}}$$
$$2(AB^2 – AC^2) = BC^2$$
$$2AB^2 – 2AC^2 = BC^2$$
$$2AB^2 = 2AC^2 + BC^2$$.

15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that $$9AD^2 = 7AB^2$$.

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 $$\triangle$$ ABC.
BE = EC = BC/2 = a/2
And, $$AE = {{a \sqrt{3}} \over {2}}$$
Given, BD = 1/3BC ? BD = a/3 DE = BE – BD = a/2 – a/3 = a/6 In $$\triangle$$ ADE, by Pythagoras theorem, $$AD^2 = AE^2 + DE^2$$ $$AD^ = ({{a\ sqrt{3}} \over {2}})^2 + ({{a} \over {6}})^2$$
$$= ({{3a^2} \over {4}}) + ({{a^2} \over {36}})$$
$$={{28a^2} \over {36}}$$
$$={{7} \over {9}} AB^2$$
$$9 AD^2 = 7 AB^2$$

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

Given, an equilateral triangle say ABC,

Let the sides of the equilateral triangle be of length a, and AE be the altitude of ?ABC.
BE = EC = BC/2 = a/2
In $$\triangle$$ ABE, by Pythagoras Theorem, we get
$$AB^2 = AE^2 + BE^2$$
$$a^2 = AE^2 + ({{a} \over {2}})^2$$
$$AE^2 = a^2 - {{a^2} \over {4}}$$
$$AE^2 = {{3a^2} \over {4}}$$
$$4AE^2 = 3a^2$$
4 × (Square of altitude) = 3 × (Square of one side)
Hence, proved.

17. Tick the correct answer and justify: In $$\triangle$$ ABC, AB = $$6 \sqrt {3}$$ cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°
(B) 60°
(C) 90°
(D) 45°

Given, in $$\triangle$$ ABC, AB = $$6 \sqrt {3}$$ 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, $$\triangle$$ ABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
$$\angle$$ B = 90°
Hence, the correct answer is (C).

#### Solution for Exercise 6.6

1. In Figure, PS is the bisector of $$\angle$$ QPR of $$\triangle$$ PQR. Prove that QS/PQ = SR/PR

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 $$\angle$$ QPR. Therefore,
$$\angle$$ QPS = $$\angle$$ SPR………………………………..(i)
As per the constructed figure,
$$\angle$$ SPR=$$\angle$$ PRT(Since, PS||TR)……………(ii)
$$\angle$$ QPS = $$\angle$$ QRT(Since, PS||TR) …………..(iii)
From the above equations, we get,
$$\angle$$ PRT=$$\angle$$ QTR
Therefore,
PT=PR
In $$\triangle$$ QTR, 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 $$\triangle$$ ABC, such that BD $$\perp$$ AC, DM $$\perp$$ BC and DN $$\perp$$ AB. Prove that:
(i) $$DM^2 = DN . MC$$
(ii) $$DN^2 = DM . AN.$$

Let us join Point D and B.
(i) Given,
BD $$\perp$$ AC, DM $$\perp$$ BC and DN $$\perp$$ AB
Now from the figure we have,
DN || CB, DM || AB and $$\angle$$ B = 90 °
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.
$$\angle$$ CDB = 90° => $$\angle$$ 2 + $$\angle$$ 3 = 90° ……………………. (i)
In $$\triangle$$ CDM, $$\angle$$ 1 + $$\angle$$ 2 + $$\angle$$ DMC = 180°
$$\angle$$ 1 + $$\angle$$ 2 = 90° …………………………………….. (ii)
In $$\triangle$$ DMB, $$\angle$$ 3 + $$\angle$$ DMB + $$\angle$$ 4 = 180°
$$\angle$$ 3 + $$\angle$$ 4 = 90° …………………………………….. (iii)
From equation (i) and (ii), we get
$$\angle$$ 1 = $$\angle$$ 3
From equation (i) and (iii), we get
$$\angle$$ 2 = $$\angle$$ 4
In $$\triangle$$ DCM and $$\triangle$$ BDM,
$$\angle$$ 1 = $$\angle$$ 3 (Already Proved)
$$\angle$$ 2 = $$\angle$$ 4 (Already Proved)
$$\triangle$$ DCM ~ $$\triangle$$ BDM (AA similarity criterion)
BM/DM = DM/MC
DN/DM = DM/MC (BM = DN)
$$DM^2 = DN × MC$$
Hence, proved.

(ii) In right triangle DBN,
$$\angle$$ 5 + $$\angle$$ 7 = 90° ……………….. (iv)
In right triangle DAN,
$$\angle$$ 6 + $$\angle$$ 8 = 90° ………………… (v)
D is the point in triangle, which is foot of the perpendicular drawn from B to AC.
$$\angle$$ ADB = 90° => $$\angle$$ 5 + $$\angle$$ 6 = 90° ………….. (vi)
From equation (iv) and (vi), we get,
$$\angle$$ 6 = $$\angle$$ 7
From equation (v) and (vi), we get,
$$\angle$$ 8 = $$\angle$$ 5
In $$\triangle$$ DNA and $$\triangle$$ BND,
$$\angle$$ 6 = $$\angle$$ 7 (Already proved)
$$\angle$$ 8 = $$\angle$$ 5 (Already proved)
$$\triangle$$ DNA ~ $$\triangle$$ BND (AA similarity criterion)
AN/DN = DN/NB
$$DN^2 = AN × NB$$
$$DN^2 = AN × DM$$ (Since, NB = DM)
Hence, proved.

3. In Figure, ABC is a triangle in which $$\angle$$ ABC > 90° and AD $$\perp$$ CB produced. Prove that $$AC^2= AB^2+ BC^2+ 2 BC.BD.$$

By applying Pythagoras Theorem in $$\triangle$$ ADB, we get,
$$AB^2 = AD^2 + DB^2$$ ……………………… (i)
Again, by applying Pythagoras Theorem in $$\triangle$$ ACD, we get,
$$AC^2 = AD^2 + DC^2$$
$$AC^2 = AD^2 + (DB + BC)^2$$
$$AC^2 = AD^2 + DB^2 + BC^2 + 2DB × BC$$
From equation (i), we can write,
$$AC^2 = AB^2 + BC^2 + 2DB × BC$$
Hence, proved.

4. In Figure, ABC is a triangle in which $$\angle$$ ABC < 90° and AD $$\perp$$ BC. Prove that $$AC^2= AB^2+ BC^2 – 2 BC.BD.$$

By applying Pythagoras Theorem in $$\triangle$$ ADB, we get,
$$AB^2 = AD^2 + DB^2$$
We can write it as;
$$AD^2 = AB^2 - DB^2$$……………….. (i)
By applying Pythagoras Theorem in $$\triangle$$ ADC, we get,
$$AD^2 + DC^2 = AC^2$$
From equation (i),
$$AB^2 - BD^2 + DC^2 = AC^2$$
$$AB^2 - BD^2 + (BC - BD)^2 = AC^2$$
$$AC^2 = AB^2 - BD^2 + BC^2 + BD^2 - 2BC × BD$$
$$AC^2= AB^2 + BC^2 - 2BC × BD$$
Hence, proved.

5. In Figure, AD is a median of a triangle ABC and AM $$\perp$$ BC. Prove that :
(i) $$AC^2 = AD^2 + BC.DM + 2 (BC/2)^2$$
(ii) $$AB^2 = AD^2 – BC.DM + 2 (BC/2)^2$$
(iii) $$AC^2 + AB^2 = 2 AD^2 + {{1} \over {2}} BC^2$$

(i) By applying Pythagoras Theorem in ?AMD, we get,
$$AM^2 + MD^2 = AD^2$$ ………………. (i)
Again, by applying Pythagoras Theorem in $$\triangle$$ AMC, 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,
$$AD^2 + (BC/2)^ 2 + 2MD.(BC/2)^ 2 = AC^2$$
$$AD^2 + (BC/2)^ 2 + 2MD × BC = AC^2$$
Hence, proved.

(ii) By applying Pythagoras Theorem in -ABM, we get;
$$AB^2 = AM^2 + MB^2$$
=> $$(AD^2 - DM^2 ) + MB^2$$
=> $$(AD^2 - DM^2 ) + (BD - MD)^2$$
=> $$AD^2 - DM^2 + BD^2 + MD^2 - 2BD × MD$$
=> $$AD^2 + BD^2 - 2BD × MD$$
=> $$AD^2 + (BC/2)^ 2 – 2(BC/2) MD$$
=> $$AD^2 + (BC/2)^2 – BC.MD$$
Hence, proved.

(iii) By applying Pythagoras Theorem in $$\triangle$$ ABM, we get,
$$AM^2 + MB^2 = AB^2$$ ………………….… (i)
By applying Pythagoras Theorem in $$\triangle$$ AMC, we get,
$$AM^2 + MC^2 = AC^2$$ …………………..… (ii)
Adding both the equations (i) and (ii), we get,
$$2AM^2 + MB^2 + MC^2 = AB^2 + AC^2$$
$$2AM^2 + (BD - DM)^2 + (MD + DC)^2 = AB^2 + AC^2$$
$$2AM^2+BD^2 + DM^2 - 2BD.DM + MD^2 + DC^2 + 2MD.DC = AB^2 + AC^2$$
$$2AM^2 + 2MD^2 + BD^2 + DC62 + 2MD (- BD + DC) = AB^2 + AC^2$$
$$2(AM^2+ MD^2) + (BC/2) ^2 + (BC/2)^ 2 + 2MD(-BC/2 + BC/2)^ 2 = AB^2 + AC^2$$
$$2AD^2 + BC^2/2 = AB^2 + AC^2$$

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

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 $$\triangle$$ DEA, we get,
$$DE^2 + EA^2 = DA^2$$ ……………….… (i)
By applying Pythagoras Theorem in $$\triangle$$ DEB, we get,
$$DE^2 + EB^2 = DB^2$$
$$DE^2 + (EA + AB)^2 = DB^2$$
$$(DE^2 + EA^2 ) + AB^2 + 2EA × AB = DB^2$$
$$DA^2 + AB^2 + 2EA × AB = DB^2$$ ……………. (ii)
By applying Pythagoras Theorem in $$\triangle$$ ADF, we get,
$$AD^2 = AF^2 + FD^2$$
Again, applying Pythagoras theorem in $$\triangle$$ AFC, we get,
$$AC^2 = AF^2 + FC^2 = AF^2 + (DC - FD)^2$$
= $$AF^2 + DC^2 + FD^2 - 2DC × FD$$
= $$(AF^2 + FD^2 ) + DC^2 - 2DC × FD (AC^2)$$
$$AC^2= AD^2 + DC^2 - 2DC × FD$$ ………………… (iii)
Since ABCD is a parallelogram,
AB = CD ………………….…(iv)
And BC = AD ………………. (v)
In $$\triangle$$ DEA and $$\triangle$$ ADF,
$$\angle$$ DEA = $$\angle$$ AFD (Each 90°)
$$\angle$$ EAD = $$\angle$$ ADF (EA || DF)
$$\triangle$$ EAD $${\displaystyle \cong }$$ $$\triangle$$ FDA (AAS congruence criterion)
EA = DF ……………… (vi)
Adding equations (i) and (iii), we get,
$$DA^2 + AB^2 + 2EA × AB + AD^2 + DC^2 - 2DC × FD = DB^2 + AC^2$$
$$DA^2 + AB^2 + AD^2 + DC^2 + 2EA × AB - 2DC × FD = DB^2 + AC^2$$
From equation (iv) and (vi),
$$BC^2 + AB^2 + AD^2 + DC^2 + 2EA × AB - 2AB × EA = DB^2 + AC^2$$
$$AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$$

7. In Figure, two chords AB and CD intersect each other at the point P. Prove that :
(i) $$\triangle$$ APC ~ $$\triangle$$ DPB
(ii) AP . PB = CP . DP

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

(i) In $$\triangle$$ APC and $$\triangle$$ DPB,
$$\angle$$ APC = $$\angle$$ DPB (Vertically opposite angles)
$$\angle$$ CAP = $$\angle$$ BDP (Angles in the same segment for chord CB)
Therefore,
$$\triangle$$ APC ~ $$\triangle$$ DPB (AA similarity criterion)

(ii) In the above, we have proved that $$\triangle$$ APC ~ $$\triangle$$ DPB
We know that the corresponding sides of similar triangles are proportional.
AP/DP = PC/PB = CA/BD
AP/DP = PC/PB
AP. PB = PC. DP
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) $$\triangle$$ PAC ~ $$\triangle$$ PDB
(ii) PA . PB = PC . PD.

(i) In $$\triangle$$ PAC and $$\triangle$$ PDB,
$$\angle$$ P = $$\angle$$ P (Common Angles)
As we know, exterior angle of a cyclic quadrilateral is $$\angle$$ PCA and $$\angle$$ PBD is opposite interior angle, which are both equal.
$$\angle$$ PAC = $$\angle$$ PDB
Thus, $$\triangle$$ PAC ~ $$\triangle$$ PDB(AA similarity criterion)

(ii) We have already proved above,
$$\triangle$$ APC ~ $$\triangle$$ DPB
We know that the corresponding sides of similar triangles are proportional.
Therefore,
AP/DP = PC/PB = CA/BD
AP/DP = PC/PB
AP. PB = PC. DP

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

In the given figure, let us extend BA to P such that;
AP = AC.
Now join PC.

Given, BD/CD = AB/AC
BD/CD = AP/AC
By using the converse of basic proportionality theorem, we get,
$$\angle$$ BAD = $$\angle$$ APC (Corresponding angles) ……………….. (i)
And, $$\angle$$ DAC = $$\angle$$ ACP (Alternate interior angles) …….… (ii)
By the new figure, we have;
AP = AC
$$\angle$$ APC = $$\angle$$ ACP ……………………. (iii)
On comparing equations (i), (ii), and (iii), we get,
$$\angle$$ BAD = $$\angle$$ 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? 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?

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 $$\triangle$$ ABC, is such way;
$$AC^2 = AB^2+ BC^2$$
$$AB^2 = (1.8 m)^2 + (2.4 m)^2$$
$$AB^2 = (3.24 + 5.76) m^2$$
$$AB^2 = 9.00 m^2$$
$$AB = \sqrt{9} m = 3 m$$
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 × 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 - String pulled by Nazima in 12 seconds
= (3.00 - 0.6) m
= 2.4 m
In $$\triangle$$ ADB, by Pythagoras Theorem,
$$AB^2 + BD^2 = AD^2$$
$$(1.8 m)^2 + BD2 = (2.4 m)^2$$
$$BD^2 = (5.76 - 3.24) m^2 = 2.52 m^2$$
BD = 1.587 m
Horizontal distance of fly = BD + 1.2 m
= (1.587 + 1.2) m = 2.787 m
= 2.79 m