NCERT Solutions Class 10 Maths Chapter 6 Triangles

NCERT Solutions Class 10 Maths Chapter 6 Triangles

NCERT Solutions Class 10 Maths Chapter 6 Triangles: National Council of Educational Research and Training Class 10 Maths Chapter 6 Solutions – Triangles. NCERT Solutions Class 10 Maths Chapter 6 PDF Download.

 

NCERT Solutions Class 10 Maths Chapter 6: Overview

Board NCERT
Class 10
Subject Maths
Chapter 6
Chapter Name Triangles
Topic Exercise Solutions

 

NCERT Solutions Class 10 Maths

Chapter 6 – Triangles

 

Exercise 6.1

 

(1) Fill in the blanks using the correct word given in brackets:

(i) All circles are ——-. (Congruent, similar)

Ans: All circles are Similar.

 

(ii) All squares are ——-. (Similar, congruent)

Ans: All squares are similar.

 

(iii) All ——- triangles are similar. (Isosceles, equilateral)

Ans: All equilateral triangles are similar.

 

(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)

Ans: Two polygons of the same number of sides are similar, if (a) Their corresponding angles are equal and (b) Their corresponding sides are proportional.

 

(2) Give two different examples of pair of

(i) Similar Figures:

 

(ii) Non-similar figures:

 

(3) State whether the following quadrilaterals are similar or not:

Solution:

Two Polygons of same number of side are similar if;

(a) Their corresponding angles are equal and

(b) Their corresponding sides are in the same ratio.

In this case ∠SPQ ≠ 90° but ∠DAB = 90°

Hence, they are not similar.

 

Exercise – 6.2

 

(1) In Fig. 6.17, (i) and (ii), DE || BC.

Find EC in (i) and AD in (ii).

Solution:

(i)

Hence, DE || BC,

So, AD/BD = AE/EC

=> 1.5/3 = 1/EC

=> EC = 3/1.5 = 2

∴ Ans: EC = 2cm

 

(ii)

Here, DE || BC

Then, AD/DB = AE/EC

= AB/7.2 = 1.8/5.4

= AD = 7.2×1.8/5.4

= 2.4

Ans: AD = 2.4 cm

 

(2) E and F are points on the sides PQ and PR respectively of a ∆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

 Solution:

Here PE = 3.9 cm, EQ = 3cm, PE = 3.6 cm and FR = 2.4 cm

Now, PE/EQ ≠ PE/FR

= 3.9/3 ≠ 3.6/2.4

So, in this case EF || QR is not satisfied.

 

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

Solution:

PE = 4cm, QE = 4.5cm, PF = 8cm and RF = 9cm.

Now, PE/QE = 4/4.5 = 8/9 and PF/RF = 8/9

As, PE/QE = PF/RF, So in this case EF||QR is satisfied.

 

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

Solution:

PQ = 1.28cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Now, PF/RF = RF/PR-PF = 0.36/2.2 = 9/55

As, PE/QE = PF/RF, so in this case EF||QR is satisfied.

 

(3) In Fig. 6.18, if LM||CB and LN||CD, Prove that AM/AB = AN/AD

Solution:

Here, LM||CB and LN||CD

Now, for ∆ABC, LM||CB, then AM/BM = AL/CL —– (i) and, from ∆ADC we get, AN/DN = AL/CL —– (ii) from (i) and (ii), AM/BM = AN/BM = AN/DN = BM/AM = DN/AN

= BM/AM + 1 = DN/AN + 1 = AB/AM = AD/AN = AM/AB = AN/AD (Ans)

 

(4) In Fig. 6.19, DE||AC and DF||AE, Prove that BF/FE = BE/EC

Solution:

From, ∆ABC we have, DE||AC then, BE||EC = BD/DA, DF||AE.

So, BF/FE = BD/DA —— (ii)

Now, from (i) and (ii) we have, BF/FE = BE/EC (Proved)

 

 

(5) In Fig. 6.20, DE||OQ and DF||QR. show that EF||QR.

Solution:

Hence, DE||OQ and DF||OR;

From, ∆PRO we have DE||OQ,

Then, DE/QE = PD/DO —– (i)

And, ∆PRQ we have DF||OR, So, PF/FR = PD/DO —– (ii)

From (i) and (ii) we get,

PE/QE = PF/FR

Hence, this is shows, that EF||QR as E and F lies, in PQ and PR respectively. (Proved)

 

(6) In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB||PQ and AC||PR. Show that BC||QR.

Solution:

Here, ∆PQO and we have AB||PQ, So, PA/AO = QB/BO —— (i)

Similarly, ∆PRO and AC||PR, So, PA/AO = OC/CR —– (ii)

From (i) and (ii) we get,

QB/BO = OC/CR

Now, ∆QRO and QB/BO = OC/CR, So, according to the definition, we have BC||QR. (Proved)

 

(7) Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution:

Let us assume ∆ABC where DE is parallel to BC and D is the mid-point of AB [i.e. AD = DB]

Now, we have to prove that E is the mid-point AC.

Therefore,

∆ABC and DE||BC,

We know that if a line drawn parallel to one side of triangle, intersects the other two sides in distinct points, then it divides the other two side in same ratio. So,

AD/DB = AE/EC

=> DB/DB = AE/EC

=> 1 = AE/EC

=> AE = EC

Hence, it is proved that, E is the mid-point of AC, (Proved)

 

(8) Using Theorem 6.2, 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:

Let us assume ∆ABC is the given triangle and D and E are the mid points of AB and AC;

So, AD = DB and AE = EC

We have to prove that, DE||BC

Now, AD/DB = DB/DB = 1

And AE/EC = EC/E = 1 [As, AD = DB and AE = EC]

So, AD/DB = AE/EC

According to the theorem this shows that, DE||BC;

 

(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:

Let, ABCD be the given trapezium in which AB||DC and it’s diagonals AC and BD intersects D each other at the point O.

Now, we have to prove, AO/BO = CO/DO.

At first to prove this theorem we draw a straight line parallel to AB and DC.

Therefore,

∆ADC we have EO||DC then, AE/ED = AO/CD —– (i) And ∆ABD we have EO||AB so, AE/ED = BO/DO —- (ii)

From (i) and (ii) we get,

AO/CO = BO/DO = AO/BO = CO/DO (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:

It is given that, the diagonals of quadrilateral ABCD intersect each other at the point O such that,

AO/BO = CO/DO

We have to prove that, ABCD is a trapezium. i.e, AB||CD

To, prove this we draw a line EF parallel to AB which intersects the O point.

Now, AO/BO = CO/DO

=> AO/CO = BO/BO —– (i)

Then, in ∆ADB, EO||AB (As EF||AB)

So, AE/DE = BO/DO = AO/CO [From (i)]

Thus, in ∆ADC, the line EO divides the triangle in the same ratio.

∴ EO||DC

Also, EO||AB

= EO||AB||DC

= AB||DC

It is clear that, one pair of opposite sides of quadrilateral ABCD are parallel, therefore ABCD is a trapezium. (Proved)

 

EXERCISE 6.3

 

(1) State which pairs of triangles in Fig. 6.34 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)

According to the figure,

∠A = ∠P

∠B = ∠Q

∠C = ∠R

Then according to the A-A-A criterion we get, ∆ABC ~ ∆PQR (Ans)

 

(ii)

According to the figure,

AB/QR = 2/4 = 1/2

AC/PQ = 3/6 = 1/2

BC/PR = 2.5/5 = 1/2

Hence, AB/QR = AC/PQ = BC/PR

Then by the 3 – 5 – 5 criterion we get, ∆ABC ~ ∆PQR.

 

(iii)

According to the figure,

MP/DE = 2/4 = 1/2; LP/DF = 3/6 = 1/2 and ML/EF = 2.7/5

Now, it is clear that the ratio of the corresponding sides of the triangles are not same.

Hence, ∆LMP and ∆DEF are not similar.

 

(iv)

According to the figure,

LM = LQ

MN/PQ = 2.5/5 = 1/2

ML/QR = 5/10 = 1/2

Hence, by the S-A- S similarity criterion of two triangle,

∆MNL ~ ∆PQR;

 

(v)

According to the figure,

∠A = ∠F = 80°

Now, AB/FD = 2.5/5 = 1/2 and BC/DE = 3/DE

As, DE is not given then, AB/FD ≠ BC/DE

So, ∆ACB and ∆FDE are not similar.

 

(vi)

In ∆DEF we have ∠D = 70°, ∠E = 80°

∴ ∠F = 180° – (∠D + ∠E) = 30°

And ∆PQR we have ∠Q = 80°, ∠R = 30°

∴ ∠P = 180° – (∠Q + ∠R) = 70°

Therefore by A-A-A similarity criterion for two triangles,

∆DEF ~ ∆PQR;

 

(2) In Fig. 6.35, ∆ODC ~ ∆OBA, BOC = 125° and CDO = 70°. Find DOC, DCO and OAB

Solution:

According to the question,

∆ODC ~ ∆OBA and ∠BOC – 125°, ∠CDO = 70°

Here, DB is a straight line.

So, ∠BOC + ∠DOC = 180°

= ∠DOC = 180° – 125° – 55°

Again, ∆ODC ~ ∆OBA

Hence, ∠DCO = ∠OAB —– (i)

 

and from ∆DOC,

∠DOC + ∠OCD + ∠ODC = 180°

= 70° + 55° + ∠OCD = 180°

= ∠OCD = ∠DCO = 55°

∴ ∠DOC = 55°, ∠DCO = 55° and ∠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 OA/OC = OB/OD

Solution:

ABCD is a trapezium with AB||CD and BD intersecting O.

Now, in ∆AOB and ∆COD we have,

∠AOB = ∠DOC (Vertically opposite angles)

∠ABO = ∠CDO (Since AB||CD with BD as traversal, alternate angle are equal)

∠BAO = ∠OCD (Since AB||CD with AC as transversal, alternate angle are equal)

Hence, ∆AOB ~ ∆COD (By the A-A-A similarity criterion)

So, OA/OC = OB/OD (As if 2 triangles are similar their corresponding sides are proportion)

 

(4) In Fig. 6.36 QR/QS = QT/PR and 1 = 2. Show that ∆PQS ~ ∆TQR.

Solution:

In the figure, QR/QS = QT/PR and ∠1 = ∠2

Prove:

We have to prove,

∆PQS ~ ∆TQR

Prove:

As, ∠1 = ∠2 we have PR = PQ —– (i)

Now, QR/QS = QT/PR

= QR/QS = QT/PQ [As PR = PQ] —– (ii)

In ∆PQS and ∆TQR

∠PQS = ∠TQR (Both are common angles) and QR/QS = QT/PQ [From (ii)]

∴ ∆PQS ~ ∆TQR (By S-A-S criterion) Hence it is proved.

 

(5) S and T are points on sides PR and QR of ∆PQR such that P = RTS. Show that ∆RPQ ~ ∆RTS

Solution:

S and T are points on sides PR and QR of ∆PQR Such that ∠P = ∠RTS;

In ∆PQR and ∆RTS, ∠QPR = ∠RTS (Given) and ∠TRS = ∠QRP (Common)

Hence, ∆PQR ~ ∆RTS (By A-A-A criterion)

= ∆RPQ ~ ∆RTS (Proved)

 

(6) In Fig. 6.37, if ∆ABE ∆ACD, show that ∆ADE ~ ∆ABC

Solution:

Here, ∆ABE ≅ ∆ACD;

We have to prove,

∆ADE ~ ∆ABC

Now, ∆ABE ≅ ∆ACD

So, AB = AC —- (i)

AD = AE —– (ii)

(i)/(ii) we get, AB/AD = AC/AE — (iii)

Therefore, In ∆ADE and ∆ABC ∠A = ∠A (Common on angle)

And AB/AD = AC/AE

Hence, ∆ADE ~ ∆ABC (By S-A-S) (Proved)

 

(7) In Fig. 6.38, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that:

(i) ∆AEP ~ ∆CDP

(ii) ∆ABD ~ ∆CBE

(iii) ∆AEP ~ ∆ADB

(iv) ∆PDC ~ ∆BEC

Solution:

In this figure altitudes AD and CE of ∆ABC intersect each other at the point P.

(i) In ∆AEP and ∆CDP

∠AEP = ∠CDP

(As AD and CE are altitudes)

∠APE = ∠CPD (As vertically opposite angles)

Hence, ∆AEP ~ ∆CDP (AA similarity)

 

(ii) In ∆ABD ~ ∆CBE

∠ABD = ∠CBE (Common angle)

∠ADB = ∠CEB (As AD and CE are altitudes)

Hence, ∆ABD ~ ∆CBE (AA similarity)

 

(iii) In ∆AEP and ∆ADB

∠PAE = ∠DAB (Common angle)

∠PEA = ∠ADB (As AD and CE are altitudes)

Hence, ∆AEP ~ ∆CBE

 

(iv) In ∆PDC and ∆BEC.

∠PCD = ∠BCE (Common angle)

∠PDC = ∠BEC (As AD and CE are altitudes)

Hence, ∆PDC ~ ∆BEC (Proved)

 

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

Solution:

Here, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD — F.

We have to prove,

∆ABE ~ ∆CFB

Now in ∆ABE and ∆CFB

∠BAE = ∠BCF [The opposite angles of the parallelogram ABCD]

And, ∠AEB = ∠CBF [As AE||BC and BE is the transversal]

Hence, ∆ABE ~ ∆CFB (A.A similarity Proved)

 

(9) In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:-

(i) ∆ABC ~ ∆AMP

(ii) CA/PA = BC/MP

Solution:

Here, ∆ABC = ∆AMP are two right angled triangles, and ∠ABC = ∠AMP

[Both are 90°]

Now,

(i) ∆ABC and ∆AMP

∠ABC = ∠AMP (Both 90°]

∠BAC = ∠PAM (Common angle)

Hence, ∆ABC ~ ∆AMP (A-A Similarity)

 

(ii) We know that if two triangles are similar, the ratio of their corresponding sides is proportional.

So, CA/PA = BC/MP = AB/AM

Hence, CA/PA = BC/MP (Proved)

 

(10) CD and GH are respectively the bisectors of ACB and EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that:

(i) CD/GH = AC/FG

(ii) ∆DCB ~ ∆HGE

(iii) ∆DCA ~ ∆HGF

Solution:

Here, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. Also ∆ABC ~ ∆FEG

(i) Here, ∆ABC ~ ∆FEG So,

PHOTO

Now, 1/2 ∠ACB = ½ ∠EGF

= ∠ACD = ∠FGH [As CD and GH are respectively bisects ∠ACB and ∠EGF]

In ∆ACD and ∆FGH

∠DAC = ∠HFG [As ∠BAC = ∠EFG]

And, ∠ACD = ∠FGH [From (ii)]

Hence, ∆ACD ~ ∆FGH

So, AC/FG = AD/FH = CD/GH [As, if two triangles are similar the ratio of their corresponding sides is proportional]

∴ CD/GH = AC/FG (Proved)

 

(ii) In ∆DCB ~ ∆HGF

∠DBC = ∠HEG [From (i) ∠ABC = ∠FEG]

and, ∠BCD = ∠HGE [As CD and GH are bisectors of the angles ∠ACB and ∠EGF]

Hence, ∆DCB ~ ∆HGF (AA Similarity)

 

(iii) In ∆DCA and HGF

∠DAC = ∠HFG [From (i) ∠BAC = ∠EFG]

And,

∠DCA = ∠HGF [As CD and GH are bisectors of the angles ∠ACB and ∠EGF]

Hence, ∆DCA ~ ∆HGF (AA similarity) Hence Proved. (Proved)

 

(11) In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If ADBC and EFAC, prove that ∆ABD ~ ∆ECF.

Solution:

Here, AB = AC

AD⊥BC and EF⊥AC

Now, ∠ADB = ∠EFC = 90° —— (i)

(As AD⊥BC and EF⊥AC)

Also, AB = AC they

∠ABC = ∠ACB

=> ∠ABD = ∠FCE —- (ii)

In ∆ABD and ∆ECF

∠ADB = ∠EFC (From (i))

∠ABD = ∠FCE (From (ii))

Hence, ∆ABD ~ ∆ECF (AA Similarity)

 

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

Solution:

Here, AD and PM are the median of the triangles ∆ABC & ∆PQR respectively.

and

AB/PQ = BC/QR = AD/PM

Since, AD is median so,

BD = CD = 1/2 BC

Similarity QM = RM = 1/2 QR

Now, AB/PQ = BC/QR = AD/PM

= AB/PQ = 2BD/2QM = AD/PM

= AB/PQ = BD/QM = AD/PM

So, ∆ABD ~ ∆PQM (S-S-S similarity)

Hence, ∠ABD = ∠PQM (Corresponding angles of similar triangles are equal)

= ∠ABC = ∠PQR —– (i)

Then, in ∆ABC and ∆PQR

∠ABC = ∠PQR (From (i))

And AB/PQ = BC/QR

Hence by S-A-S Similarity, ∆ABC ~ ∆PQR (Proved)

 

(13) D is a point on the side BC of a triangle ABC such that ADC = BAC. Show that CA2 = CB.CD.

Solution:

Here, D is a point on BC.

∠ADC = ∠BAC

Now, In ∆ABC and ∆ADC

∠BCA = ∠DCA (common angle)

∠ADC = ∠BAC (Given)

Hence, ∆ABC ~ ∆ADC (AA similarity)

Now, AB/CD = CB/CA = CA/CD [As if the two triangles are similar their corresponding sides are proportional]

Hence, CB/CA = CA/CD

= CA2 = CB.C (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 ∆ABC ~ ∆ PQR.

Solution:

Here, AD and PM are median of ∆ABC and ∆PQR respectively.

So,

BD = CD = 1/2 BC and QM = RM = 1/2 QR

It is also given, AB/PQ = BC/QR = AD/PM

= AB/PQ = 2BD/2QM = AD/PM

= AB/PQ = BD/QM = AD/PM

So, ∆ABD ~ ∆PQM (S-S-S similarity)

Hence, ∠ABD = ∠PQM (Corresponding angles)

= ∠ABC = ∠PQR —– (i)

Then, In ∆ABC and ∆PQR

∠ABC = ∠PQR (From (i))

And AB/PQ = BC/QR

Hence, by S-A-S Similarity, ∆ABC ~ ∆PQR (Proved)

 

(15) A vertical pole of 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.

Solution:

Here, height of pole, AB = 6m

Length of pole of shadow, BC = 4m

Length of shadow of tower, EF = 28m

We have to find the height of the tower, So ED =?

Now, in ∆ABC and ∆DEF

∠B = ∠E = 90°

∠C = ∠F (Some equation in both the cases as both shadows are cast at the same time)

∴ ∆ABC ~ ∆DEF (AA similarity)

We knows that if two triangles are similar ratio of their sides are in proportion so,

AB/DE = BC/EF = 6/DE = 4/28 = 1/7

= DE = 6×7 = 42

Hence, the height of the tower is 42 meters. (Ans)

 

(16) If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that AB/PQ = AD/PM

Solution:

Here, AD is the median of ∆ABC and PM is the median of ∆PQR

So, BD = DC = ½ BC

Similarly, QM = MR = 1/2 QR

Also, ∆ABC ~ ∆PQR,

So AB/PQ = BC/QR = AC/PR

= AB/PQ = 2BD/2QM = AC/PR

= AB/PQ = BD/QM = AC/PR —- (i)

In ∆ABD and ∆PQM

∠ABD = ∠PQM (∆ABC ~ ∆PQR)

And, AB/PQ = BD/QM [From (i)]

Hence, ∆ABD ~ ∆PQM, so

AB/PQ = BD/QM = AD/PM [As if 2 triangles are similar then their side are in proportional]

Hence, AB/PQ = AD/PM (Proved)

 

Exercise – 6.4

 

(1) Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Solution:

Here, ∆ABC ~ ∆DEF and area of ∆ABC = 64 cm2

Area of ∆DEF = 121 cm2

And, EF = 15.4 cm, we have to find to BC =?

We know that if two triangle are similar, ratio of areas is equal to square of ratio of its corresponding sides.

Hence,

area of ∆ABC/Area of ∆DEF = (BC/EF)2

=> (BC/15.4)2 = 64/121

=> BC/15.4 = 8/11

=> BC = 8×15.4/11

=> 8×1.4

= 11.2

Hence, BC = 11.2 cm (Ans)

 

(2) Diagonals of a trapezium ABCD with AB||DC intersect each other at the point O.

If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Solution:

Here, ABCD is a trapezium with AB||DC it’s diagonals AC and BD intersect each other at the point O.

Also, AB = 2CD

In, ∆AOB and ∆COD

∠COD = ∠AOB (Vertically opposite angle)

∠ABO = ∠CDO (As AB||CD with AC as traversal)

Hence, ∆AOB ~ ∆COD (A-A similarity)

Now if two triangles are similar then their ratio of areas is equal to the square of the ratio of it’s corresponding sides.

Hence, Area of ∆AOB/Area of ∆COD = (AB/CD)2 = (2CO/CD)2 = 4/1 (As AB = 2CD given)

Hence, the required ratio = 4:1 (Ans)

 

(3) In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

ar (ABC)/ar (DBC) = AO/DO

Solution:

Here, ∆ABC and ∆DBC are two triangles with common base BC.

To, prove this we draw two altitude AE and DE such that, AE⊥BC and DE⊥BC;

We know that the area of a triangle = ½ × Base × altitude.

So, area of ∆ABC = 1/2 × AE × BC and,

Area of ∆DBC = ½ × DF × BC

Now, ar(ABC)/ar(DBC) = (1/2) × AE × BC/(1/2) × DF × BC = AE/DF —– (i)

In ∆AOE and ∆DOF

∠AEO = ∠DFO (both are 90°)

∠AOE = ∠DOF (Vertically opposite angle)

Hence, ∆AOE ~ ∆DOF

So, AO/DO = AE/DF = EO/FO

Now, from (i),

ar(ABC)/ar(DBC) = AO/DO (As AO/DO = AE/DF) (Proved)

 

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

Solution:

Let, ∆ABC and ∆DEF are the two similar triangles whose areas are equal.

i.e, ∆ABC ~ ∆DEF

and area of ∆ABC = area of ∆DEF

We know that, if two triangles are similar, then ration of areas is equal to square of ratio of its corresponding sides.

So, ar(∆ABC)/ar(∆DEF) = (AB/DE)2 = (BC/EF)2 = (AC/DF)2

= (AB/DE)2 = (BC/EF)2 = (AC/DF)2 [As area of ∆ABC = area of ∆DEF]

Now,

(AB/DE)2 = 1 = AB/DE = 1 = AB = DE

Similarly, BC = EF and AC = DF

In ∆ABC and ∆DEF,

EF = BC

AB = DE

AC = DF

Hence, by S-S-S congruency ∆ABC

∆ABC ≅ ∆DEF (Proved)

 

(5) D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.

Solution:

Here, D, E and F be the mid points of AB. BC and AC respectively.

So, AD = DB = 1/2 AB

BE = EC = 1/2 BC

AF = FC = 1/2 AC

Now we join the points D, E, F.

We know that,

Line joining mid-points of two sides of a triangle is parallel to the 3rd side.

So, DF||BC = DF||BE (As E is the mid-point of BC)

Similarly, DE||EC, DB||EF and DE||EF

Hence, DBEF is a parallelogram (as DF||BE and DB||EF)

So, DECF is also a parallelogram they ∠DFE = ∠EBD and

∠EDF = ∠ECF

[As the opposite angles of a parallelogram is always equal]

Now, in ∆DEF and ∆ABC

∠DFE = ∠ABC [As ∠DFE = ∠EBD]

∠EDF = ∠BCA [As ∠EDF = ∠ECF]

Hence, ∆DEF ~ ∆ABC

Now, area of ∆DEF/area of ∆ABC = (DE/AC)2

= (FC/AC)2 [As DECF is a parallelogram]

PHOTO

∴ The require ratio = 1:4; (Ans)

 

(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:

Here, ∆ABC and ∆PQR are two similar triangles. I.e. ∆ABC ~ ∆PQR; and AD and PS  is the medians of ∆ABC and ∆PQR respectively.

I.e. BD = DC = 1/2 BC and QS = SR = ½ QR

As, ∆ABC ~ ∆PQR,

So,

ar(∆ABC)2/ar(∆PQR) = (AB/PQ)2 = (BC/QR)2 = (AC/PR)2 — (i)

Again, AB/PQ = BC/QR = 2BD/2QS = BD/QS

In ∆ABD and ∆PQS

∠ABD = ∠PQS [As, ∆ABC ~ ∆PQR]

and AB/PQ = BD/QS

So, ∆ABD ~ ∆PQS [S-A-S Similarity]

Hence, AB/PQ = BD/QS = AD/PS [As ∆ABD ~ ∆PQS]

Now, from (i) and we get,

ar(∆ABC)/ar(∆PQR) = (AB/PQ)2  (AD/PS)2 [As, AB/PQ = AD/PS]

Hence, it is proved (Prove)

 

(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:

Here, ABCD is a square and ∆DBF, ∆BCE are two equilateral triangle.

We have to prove, ar(∆BCE)/ar(∆FDB) = 1/2

As, ∆DBF and ∆BCE are equilateral So,

DF/CE = FB/EB = DB/CB

Hence, S-S-S similarity,

∆DBF ~ ∆BCE

Then, ar(∆DBF)/ar(∆BCE) = (DB/BC)2 —- (i)

Again, ABCD is a square and DB is a diagonal

So, DB = √2 BC

From (i), ar(∆DBF)/ar(∆BCE) = (√2 BC/BC)2 = 2

= ar (∆BCE) = 1/2 × ar (∆DBF)

Hence, it is proved.

 

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

(A) 2:1

(B) 1:2

(C) 4:1

(D) 1:4

Solution:

∆ABC and ∆BDE are two equilateral. And D is the mid-point of BC.

i.e, BD = ½ BC

Then,

∆ABC ~ ∆BDE

Now, AB/BE = AC/ED = BC/BD

Therefore,

ar(∆ABC)/ar(∆BDE) = (BC/BD)2 = (2BD/BD)2 = 4/1

Hence, ans: (c) 4:1

 

(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:

It is given that the ratio of sides of two similar triangles = 4:9

Now,

Area of triangle 1/Area of triangle 2 = (Side of triangle 1/Side of triangle 2)2

= (4/9)2 = 16/81

Hence, ans -> (D) 16:81

 

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

Solution:

Side of the given triangle 7cm, 24cm, 25cm

Here, (24cm)2 + (7cm)2 = (25cm)2

Hence, it is a right triangle and its length of hypotenuse = 25 cm;

 

(ii) 3 cm, 8 cm, 6 cm

Solution:

Sides are 3 cm, 8 cm, 6 cm

Here, (3cm)2 + (6cm)2 ≠ (8cm)2

So, it is not a right triangle.

 

(iii) 50 cm, 80 cm, 100 cm

Solution:

Sides are 50 cm, 80 cm, 100 cm,

Here, (50cm)2 + (80cm)2 ≠ (100cm)2

So, it is not a right triangle.

 

(iv) 13 cm, 12 cm, 5 cm

Solution:

Sides are 13cm, 12cm, 5cm

Here, (12cm)2 + (5cm)2 = (13cm)2

So, it is a right triangle and it’s length of hypotenuse = 13cm;

 

(2) PQR is a triangle right angled at P and M is a point on QR such that PMQR. Show that PM2 = QM. MR.

Solution:

Here ∆PQR is a right triangle.

∠P = 90° and

PM⊥QR i.e, ∠M = 90°

Now, it is clear that, ∆PQR, ∆PMR and ∆PQM are right angled triangle.

Now, according to the Pythagoras theorem we have,

QR2 = PR2 + PQ2 (As in ∆PQR and ∠P = 90°)

PR2 = MR2 + PM2 (As in ∆PMR, ∠M = 90°)

PQ2 = PM2 + QM2 (As in ∆PQM, ∠M = 90°)

Also, QR = MR + QM

Now, QR2 = PR2 + PQ2

=> (MR + QM)2 = MR2 + PM2 + PM2 + QM2

=> 2PM2 = 2QM.MR

=> PM2 = QM.MR

Hence, it is proved

 

(3) In Fig. 6.53, ABD is a triangle right angled at A and ACBD. Show that –

(i) AB2 = BC.BD

Solution:

Here ∠A = 90° and AC⊥BD

In ∆ABD and ∆ABC

∠BAD = ∠ACD [Both 90°]

∠ABD = ∠ABC [Common angle]

Hence, ∆ABD ~ ∆ABC

Now, we known that, if two triangle are similar then their corresponding sides are in equal ratio.

So, AB/CB = AD/AC = BD/AB

∴ AB/BC = BD/AB = AB2 = BC.BD (proved)

 

(ii) AC2 = BC.DC

Solution:

In ∆ABC and ∆ACD

∠ACB = ∠ACD (Both 90°)

∠BAC = ∠CAD

∴ ∆ABC ~ ∆ACD

So, AC/DC = BC/AC = AC2 = BC.DC (Proved)

 

 

(iii) AD2 = BD.CD

Solution:

In ∆ABD and ∆ACD

∠BAD = ∠ACD (Both 90°)

∠BDA = ∠ADC (Common angle)

∴ ∆ABD ~ ∆ACD

So, AD/CD = BD/AD = AD2 = BD.CD (Proved)

 

(4) ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2

Solution:

Here, ∆ABC is an isosceles triangle and ∠C = 90°;

Now, AB = Hypotenuse and AC = BC [As ∆ABC is an isosceles]

According to the Pythagoras theorem in ∆ABC we have,

AB2 = AC2 + BC2

=> AB2 = AC2 + AC2 [As AC = BC]

=> AB2 = AC2 + AC2

=> AB2 = 2AC2 (Proved)

 

(5) ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.

Solution:

Here, ∆BC is an isosceles triangle.

AC = BC and AB2 = 2AC2

= AB2 = AC2 + BC2

= AB2 = AC2 + BC2 [As, AC = BC]

From the Pythagoras theorem, we have,

(Hypotenuse)2 = (Height)2 + (Base)2 in the ∆ABC, the largest side is AB. i.e. AB = Hypotenuse,

Hence, AB2 = AC2 + BC2

So, ∆ABC is a right angled triangle as it follows the Pythagoras theorem. (Proved)

 

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

Solution:

Here, ∆ABC is an equilateral triangle.

AB = BC = CA = 2a (Given) and we draw the altitudes AD, BE, CF with respectively AB, AC and BC;

In ∆AFC and ∆BCF,

AC = BC (Given)

CF = CF (Common sides)

∠AFC = ∠BFC (Both 90°)

∴ ∆AFC ≅ ∆BCF

So, AF = BF = ½ AB = ½ × 2a = a

Again, ∆AFC is a right triangle as ∠AFC = 90°;

Now, AC2 = AF2 + CF2

=> (2a)2 = (a)2 + (CF)2 [as, AC = 2a, AF = a]

=> (CF)2 = 4a2 – a2 = 3a2

=> CF = √3a2 = √3a

Similarly, length of altitudes, AD = √3a and length of altitude BE = √3a

 

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

Solution:

Here, ABCD is a rhombus and AC and BD its diagonals, intersect at the point O. we have to prove,

Sum of square of all sides = Sum of the squares of its diagonals

i.e, AB2 + BC2 + CD2 + AD2 = AC2 + BD2

We know that,

Diagonals of a rhombus bisect each other at right angles.

Therefore,

∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

Also, AO = CO = 1/2 AC and BO = DO = 1/2 BD

Now, ∆AOB is a right angled triangle. Using Pythagoras theorem,

(Hypotenuse)2 = (Height)2 + (Base)2

=> (AB)2 = (OA)2 + (OB)2

=> (AB)2 = (1/2 AC)2 +  (1/2 BD)2

=> 4AB2 = AC2 + BD2

=> AB2 + AB2 + AB2 + AB2 = AC2 + BD2

=> AB2 + BC2 + CD2 + AD2 = AC2 + BD2

(As, we know that rhombus 4 sides are equal)

Hence, it is proved.

 

(8) In Fig. 6.54, O is a point in the interior of a triangle ABC, ODBC, OEAC and OFAB. Show that

(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2,

(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Solution:

Here, ∆ABC is a triangle. and,

OD⊥BC

OE⊥AC

OF⊥AB

(i) We have to prove,

OA2 + OB2 + OC2 – OD2 – OE2 – OF2

= AF2 + BD2 + CE2

At first we join the point O from A, B and C;

Now, we use the Pythagoras theorem,

In ∆OAF,

(OA)2 = (AF)2 + (OF)2 —– (i)

In ∆ODB,

(OB)2 = (OD)2 + (BD)2 —— (ii)

In ∆OEC,

(OC)2 = (OE)2 + (EC)2 —– (iii)

Now, adding (i), (ii) and (iii) we get,

(OA)2 + (OB)2 + (OC)2 = (AF)2 + (OF)2 + (OD)2 + (BD)2 + (OE)2 + (EC)2

= OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 (Proved)

 

(ii) We have to prove,

AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Using Pythagoras theorem,

In ∆ODB, OB2 = OD2 + BD2 —– (1)

In ∆OFB, OB2 = OF2 + FB2 —– (2)

In ∆OFA, OA2 = OF2 + AF2 —- (3)

In ∆OEA, OA2 = OE2 + AE2 —– (4)

In ∆OEC, OC2 = OE2 + CE2 —– (5)

In ∆ODC, OC2 = OD2 + CD2 —- (6)

Now, L.H.S.

AF2 + BD2 + CE2

= (OA2 – OF2) + (OB2 – OD2) + (OC2 – OE2)

= (OA2 – OE2) + (OB2 – OF2) + (OC2 – OD2)

= AE2 + BF2 + CD2

= AE2 + CD2 + BF2 = R.H.S

Hence proved.

 

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

Solution:

Here, AC = Height of window = 8m

AB = Length of ladder = 10m

We have to find the distance of foot of ladder from wall i.e, BC

Since, ∠BCA = 90°

∴ ∆ABC is a right triangle now using Pythagoras theorem,

(AB)2 = (AC)2 + (BC)2

= (10)2 = (8)2 + (BC)2

= BC = √102 – 82 = √100 – 64 = 6

Hence, the required distance is 6m; (Ans)

 

(10) A guy wire attached to a vertical pole of height 18m is 24m 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:

Here, AB = Height of vertical Pole = 18m

AC = Length of wire = 24

We have to find,

Distance from the base of the pole to the another end of the wire i.e. BC;

Here, ∠ABC = 90° So, ∆ABC is a right triangle. [As, the pole will be perpendicular by using Pythagoras theorem]

(AC)2 = (AB)2 + (BC)2

=> (24)2 = (18)2 + (BC)2

=> BC = √242 – 182

= √252

= 6√7

Hence the required distance is = 6√7 m; (Ans)

 

(11) An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1 1/2 hours?

Solution:

According to the question, the speed of the north aeroplane is = 1000 km/hr and the west aeroplane is 1200 km/hr.

Now, we have to find the distance between the two planes after 1.5 hours i.e. BC

Therefore, Speed = Distance/Time

=> Distance = Speed × Time

Let, AB = Distance covered by north aeroplane

And AC = Distance covered by west aeroplane

So, AB = (1000 × 1.5) km = 1500 km

AC = (1200 × 1.5) km = 1800 km

North and west are perpendicular. So, ∠BAC = 90°

So, ∆ABC is a right triangle.

Now, by using Pythagoras theorem

(BC)2 = (AB)2 + (AC)2

=> (BC)2 = (1500)2 + (1800)2

=> BC = √225000 + 3240000

=> √5490000

=> √3×3×61×100×100
=> 3×100×√61

=> 300 √61

Hence, the distance between the two aeroplane is 300√61 km (Ans)

 

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

Solution:

Here, AD and BE are the height of the two poles.

AD = 6m

BE = 11m

And the distance between the feet of the poles, AB = 12m

Now, CD = AB = 12m and BE = BC + CE

=> CE = BE – BC

= (11 – 6) m

= 5m

And ∠ECO = 90° So, ∆CDE right triangle we have to find the distance between the tops of the pole, i.e. E by Pythagoras theorem, in ∆CDE.

(ED)2 = (CD)2 + (EC)2

=> ED = √52 + 122

= √169

= 13

Hence, the required distance is 13m; (Ans)

 

(13) D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C Prove that AE2 + BD2 = AB2 + DE2.

Solution:

Here, ∠ACB = 90° and D, E are points on the sides CA and CB respectively. Now, we joint the points A and E & band D

Then we have 4 right triangle, using Pythagoras theorem,

In ∆ACE, AE2 = AC2 + CE2 —— (1)

In ∆BDC, BD2 = DC2 + BC2 ——- (2)

In ∆CDE, ED2 = CD2 + EC2 —— (3)

In ∆ABC, AB2 = AC2 + BC2 ——- (4)

Now,

L.H.S –

AE2 + BD2

= AC2 + CE2 + DC2 + BC2 [From (1) and (2)]

= (AC2 + BC2) + (CD2 + EC2)

= AB2 + ED2 [From (3) and (4)]

= AB2 + DE2 (R.H.S)

Hence it is proved.

 

(14) The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3 CD (See fig. 6.55). Prove that 2AB2 = 2AC2 + BC2.

Solution:

Here, AD⊥BC and DB = 3CD

Let, CD = x

=> DB = 3x

Now, BC = CD + DB

= 4x

Hence, ∆ADC and ∆BDA both are right triangle [As AD⊥BC] Using Pythagoras theorem,

AC2 = CD2 + AD2 and

AB2 = DB2 + AD2

Now, AB2 – AC2 = DB2 + AD2 – CD2 – AD2

=> AB2 – AC2 = (3x)2 – (x)2

=> AB2 – AC2 = 8x2

=> 2AB – 2AC2 = 16x2 = (BC)2

=> 2AB2 = BC2 + 2AC2 (Proved)

 

(15) 5. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD2 = 7AB2.

Solution:

Here, ∆ABC is an equilateral triangle.

BD = 1/3 BC;

We have to prove,

9AD2 = 7AB2  and to prove this we construct AE⊥BC;

In ∆ABE and ∆AEC

AB = AC (Given)

AE = AE (common side)

∠AEB = ∠AEC (Both 90°, AE⊥BC)

∴ ∆AEB ≅ ∆AEC

∴ BE = EC = 1/2 BC

Let, AB = AC = EC = x/2

By Pythagoras theorem,

In ∆ABE, AB2 = BC2 + AE2 —– (i)

In ∆ADE, AD2 = AE2 + DE2 —– (ii)

Now, (ii) —- (i) We get,

AD2 – AB2 = DE2 – BE2

=> AD2 = AB2 + DE2 – BE2

= x2 + x2/36 – x2/4

= 28x2/36

Here,

DE = (BE – BD)

= (x/2 – x/3)

= x/6

As, BD = 1/3 BC

 

Now,

9AD2 = 9 × 28x2/36 = 7x2 = 7AB2

∴ 9AD2 = 7AB2 (Proved)

 

(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:

Here, ABC is an equilateral triangle.

I.e. AB = BC = CD

We have to prove,

3 × square of one Side of the triangle = 4 × square of one of its altitude.

i.e, 3AC2 = 4 CD2

First we constract CD⊥AB,

Here CD is the altitude

In ∆ADC and ∆BCD,

AC = BC (Given)

CD = CD (Common side)

∠ADC = ∠BDC (Both 90°)

∴ ∆ADC ≅ ∆BCD

So, AD = BD = 1/2 AB

Let, AB = BC = AC = a

∴ AD = BD = a/2

Using Pythagoras theorem

In ∆ADC, AC2 = AD2 + CD2

= a2 = (a/2)2 + CD2

= CD2 = a2 – a2/4

= CD = 3a2/4

= 4CD2 = 3a2 = 3AB2

= 3AC2

= 3BC2

[As, AB = BC = AC.a]

Hence it is proved.

 

(17) Tick the correct answer and justify: In ∆ABC, AB = 6√3 cm, AC = 12 cm and BC = 6cm. The angle B is:

(A) 120°

(B) 60°

(C) 90°

(D) 45° 

Solution:

In ∆ABC, AB = 6√3 cm

AC = 12cm

BC = 6cm

Now, (AB)2 + (BC)2 = (AC)2

So, AC = Hypotenuse

∴ The angle B is: (c) 90° (Ans).

 

Exercise – 6.6

 

(1) In Fig. 6.56, PS is the bisector of QPR of ∆PQR. Prove that QS/SR = PQ/PR

Solution:

Here, PS is the bisector of ∠QPR

i.e. ∠QPS = ∠SPR

Here we constract,

RT such that,

PS||RT;

Therefore applying

Basic proportionality theorem in ∆QRT

We get, QS/SR = PQ/PT —– (i)

Now, we need to prove, PT = PR

So, ∠SPR = ∠PRT (Alternate interior angles)

and ∠QPS = ∠PTR (Corresponding angles)

Againg ∠QPS = ∠SPR [Given]

So, ∠PRT = ∠PTR Then in ∆PTR we get,

PT = PR

Now, from (i),

QS/SR = PQ/PR [As, PT = PR] (Proved)

 

(2) In Fig 6.57, D is a point on hypotenuse AC of ∆ABC, such that BDAC, DMBC and DNAB, Prove that:

(i) DM2 = DN.MC

(ii) DN2 = DM.AN

Solution:

Here, DM⊥BC and DN⊥AB

(i)  We have to prove, DM2 = DN.MC

We know that, It a perpendicular is drawn the vertex of the right angle of a right triangle to the hypotenuse, the triangle on both sides of the perpendicular are similar to the whole triangle and to each other.

Now, we apply, the theorem,

In ∆BDC we get, ∆BMD ~ ∆DMC (As DM⊥BC, given)

And in ∆BDA

∆AND ~ ∆DND (As DN⊥AD, given)

Using, ∆BMD ~ ∆DMC

BM/DM = MD/MC

=> (DM)2 = BM.MC —– (i)

We need to prove, BM = DN

It is given, AB⊥BC and DM⊥BC

=> AB||DM

=> NB||DM

Also, CB⊥AB and DN⊥AB

=> CB||DN

=> MB||DN

So, DNBM is a parallelogram, as ND||DM and MB||DN

Since, DNBM is a parallelogram, its opposite sides are always equal.

So, DN = MB and DM = NB —- (ii)

Now, (DM)2 = DN.MC (Proved)

 

(ii) Again, ∆AND ~ ∆DNB

So, AN/DN = DN/BN

= (DN)2 = AN.BN

= (DN)2 = AN.BM

(From (ii), NB = DM)

 

 

(3) In Fig. 6.58, ABC is a triangle in which ABC > 90° and ADCB produced. Prove that AC2 + BC2 + 2BC.BD.

 Solution:

It is given that ABC is a triangle in which ∠ABC > 90° and AD⊥CB extended D.

and, ∠ADB = ∠ADC = 90°

So, Applying Pythagoras theorem,

In ∆ADB, AB2 = AD2 + BD2 —— (i)

In ∆ABC, AC2 = AD2 + CD2 ——– (ii)

= AD2 + (DB + BC)2

= (AD2 + BD2) + 2BD.BC + BC2

= AB2 + BC2 + 2BD.BC

(As, AB2 = AD2 + BD2)

Hence, It is Proved.

 

(4) In Fig. 6.59, ABC is a triangle in which ABC < 90° and ADBC. Prove that AC2 = AB2 + BC2 – 2BC.BD.

Solution:

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

Here, ∠ADB = ∠ADC = 90°

So, applying Pythagoras theorem,

In ∆ADB, AB2 = AD2 + BD2 —- (i)

In ∆ADC, AC2 = AD2 + CD2 —– (ii)

Now, AC2 = AD2 + CD2

= AD2 + (BC – BD)2

= AD2 + BC2 – 2BC.BD+BD2

= (AD2 + BD2) + BC2 – 2BC.BD

= AB2 + BC2 – 2BC.BD

Hence, AC2 = AB2 + BC2 – 2BC.BD (Proved)

 

(5) In Fig. 6.60, AD is a median of a triangle ABC and AMBC. Prove that:

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

(ii) AB2 = AD2 – BC.DM + (BC/2)2

(iii) AC2 + AB2 = 2AD2 + 1/2 BC2 

Solution:

Here AD is a median and AM⊥BC.

∴ BD = CD = 1/2 BC —– (i)

(i) Using Pythagoras theorem,

In ∆AMD

AD2 = AM2 + MD2 —– (2)

In ∆AMC,

AC2 = AM2 + CM2 —- (3)

Subtracting (3) —- (2)

AC2 – AD2 = CM2 – MD2

= AC2 = AD2 + (MD + CD)2 – DM2 [As, CM = MD + CD]

=> AC2 = AD2 + DM2 + CD2 + 2CD.DM – DM2]

= AD2 + CD2 + 2CD.DM

= AD2 + (BC/2)2 + BC.DM [As, CD = BC/2]

Hence Proved,

(ii) In ∆AMB,

AB2 = AM2 + BM2 — (4)

Now, subtracting,

(4) —— (2)

AB2 – AD2 = BM2 – MD2

=> AB2 = AD2 + (BD – MD)2 – MD2

= AD2 + BD – 2BD.MD + MD2 – MD2

= AD2 – BC.DM + (BC/2)2 [BD = ½ BC]

Hence Proved

(iii) From the result (i) and (ii)

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

AB2 = AD  – BC.DM + (BC/2)2
_________________________
Adding, AC2 + AB2 = 2nd + 2 (BC/2)2

=> AC2 + AB2 = —- AD2 + 1/2 BC2

Hence, Proved;

 

(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:

Here, ABCD is the parallelogram; we know that, the diagonals of a parallelogram bisect each other.

i.e, BO = OD = 1/2 BD

OA = OC = 1/2 AC

Now, by Apollonius theorem we get,

In ∆CBD, CB2 + CD2 = 2CO2 + 2BO2 —– (1)

In ∆ABD, AB2 + AD2 = 2AO2 + 2BO2 —– (2)

Adding (1) and (2),

AB2 + AD2 + CB2 + CD2

= 2AO2 + 2BO2 + 2CO2 + 2BO2

= 2AO2 + 2BO2 + 2AO2 + 2BO2 [∵ OA = OC]

= 4AO2 + 4BO2

= (2AO)2 + (2BO)2

= AC2 + BD2 [∵ OA = 1/2 AC, OB = 1/2 BD]

Hence it is proved

 

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

(i) ∆APC ~ ∆DPB

(ii) AP.PB = CP.DP

Solution:

Here, AB and CD are two chords of a circle intersecting at point P.

(i) In ∆ADC and ∆DPB

∠APC = ∠BPD

(Vertically opposite angle]

∠CAP = ∠BDP (Angles in the same segment of a circle)

∴ ∆APC ~ ∆DPB (AA similarity)

Hence proved

(ii) In the first result,

∆APC ~ ∆DPB,

Thus, AP/DP = CP/BP

[If two triangles are similar, then the ration of the corresponding sides are equal]

= AP.BP = CP.DP (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) ∆PAC ~ ∆PDB

(ii) PA.PB = PC.PD

Solution:

(i) Here, the chords AB and CD intersect each other at the point P, outside of the circle.

Now, ABDC is a cyclic quadrilateral,

So, the sum of its opposite angle is 180°;

Thus, ∠BAC + ∠PDB = 180° —– (1)

Also, ∠BAC + ∠PAC = 180° (By linear pair)

Now, ∠BAC = 180° – ∠PAC

From (1) we get, ∠PAC + ∠PDB = 180°

= ∠PDB = ∠PAC —– (2)

In ∆PAC and ∆PDB

∠APC – ∠BPD [Common angle]

∠PDB = ∠PAC [From (2)]

∴ ∆PAC ~ ∆PDB [A-A Similarly]

Hence Proved

(ii) Now, ∆PAC ~ ∆PDB

Thus, PA/PD = PC/PB [If two triangles are similar then the ratio of their corresponding sides is equal]

= PA.PB = PC.PD

Hence proved.

 

(9) In Fig. 6.63, D is a point on side BC of ∆ABC Such that BD/CD = AB/AC. Prove that AD is the bisector of BAC.

Solution:

Here, BD/CD = AB/AC

We have to bisector of ∠BAC.

i.e. ∠BAD = ∠CAD

To prove this, first construct a line AE by produce BA such that AE = AC;

In ∆AEC,

AE = AC [By construction]

∠AEC = ∠ACE [Angles opposite to the equal sides are equal]

Also, BD/CD = AB/AC

= BD/CD = AB/AE [As AE = AC]

Then in ∆BEC we have a line AD, which divides the two sides BE and be of ∆BEC in same ratio.

Now, by theorem,

OA||CE [If a line divides any two sides of a triangle, in the same ratio then the line is parallel to the 3rd side]

Now,

DA||CE and CA is transversal so,

∠BAD = ∠AEC (Corresponding angle)

∠DAC = ∠ACE (Alternate angles)

And we proved,

∠AEC = ∠ACE

Thus, ∠BAD = ∠DAC

Hence it is 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 Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Solution:

Let, A be the position of the tip of the fishing rod above the surface of the water.

Thus, AB = 1.8m and according to the question,

CD = 3.6m

CB = 2.4m

∴ BD = CD – CB = 1.2m

Since AB is a straight line,

∠ABC = 90°, Do, AC2 = AB2 + BC2

= AC = √(1.8)2 + (2.4)2

= √9 = 3

Now, Nazima pulls the string at the rate of 5cm per second.

So, length of string pulled in 12 seconds, is = (5×12) cm = 60cm = 0.6m

So, our new figure will be,

AP = AC – 0.6

= (3 – 0.6) m = 2.4m

Now we need to find the horizontal distance of the fly form Nazima.

So, we find the distance BP.

Hence, AP2 = AB2 + BP2

= BP = √AP2 – AB2

= √(3.4)2 – (1.8)2

= √2.52

= ± 1.58

∴ BP = 1.58m (as distance cannot be negative)

∴ The total horizontal distance of the fly form Nazima,

DP = BP + DP

= (1.58 + 1.2) m

= 2.78m (Ans)

 

 

Updated: December 4, 2021 — 1:28 pm

Leave a Reply

Your email address will not be published.