Triangles

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NCERT Solutions for Chapter 6 Triangles Class 10 Maths

Book Solutions

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)

Answer

(i) Similar
(ii) Similar
(iii) Equilateral 
(iv) (a) Equal, (b) Proportional
Exercise 6.1 Page Number 122

2

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

Answer

(i) Two twenty-rupee notes, Two two rupees coins.
(ii) One rupee coin and five rupees coin, One rupee not and ten rupees note.
Exercise 6.1 Page Number 122

3

State whether the following quadrilaterals are similar or not:

Answer

The given two figures are not similar because their corresponding angles are not equal.
Exercise 6.1 Page Number 122

1

In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Answer

(i) In △ ABC, DE || BC (Given)
Exercise 6.2 Page Number 128

2

E and F are points on the sides PQ and PR respectively of a ΔPQR. 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

Answer

In ΔPQR, E and F are two points on side PQ and PR respectively.
(i) PE = 3.9 cm, EQ = 3 cm(Given)
PF = 3.6 cm, FR = 2,4 cm (Given)

 

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

 

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

Exercise 6.2 Page Number 128

3

In the fig 6.18, if LM || CB and LN || CD, prove thatAM/MB = AN/AD

Answer

In the given figure, LM || CB
By using basic proportionality theorem, we get,

Exercise 6.2 Page Number 128

4

In the fig 6.19, DE||AC and DF||AE. Prove that

BF/FE = BE/EC

Answer

In ΔABC, DE || AC (Given)
Exercise 6.2 Page Number 128

5

In the fig 6.20, DE||OQ and DF||OR, show that EF||QR.

Answer

In ΔPQO, DE || OQ (Given)
Exercise 6.2 Page Number 129

6

In the 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.

Answer

Exercise 6.2 Page Number 129

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).

Answer

Given:

ΔABC in which D is the mid point of AB such that AD=DB.
A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.
To Prove: E is the mid point of AC.
Proof: D is the mid-point of AB.
∴ AD=DB

∴ AE =EC
Hence, E is the mid point of AC.

Exercise 6.2 Page Number 129

8

Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX)

Answer

Given:

ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.
To Prove: DE || BC
Proof: D is the mid point of AB (Given)
∴ AD=DB

Exercise 6.2 Page Number 129

9

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

Answer

Given:

ABCD is a trapezium in which AB || DC in which diagonals AC and BD intersect each other at O.

Exercise 6.2 Page Number 129

10

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

Answer

Exercise 6.2 Page Number 129

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:

Answer

(i) In  ΔABC and ΔPQR, we have
∠A = ∠P = 60° (Given)
∠B = ∠Q = 80°(Given)
∠C = ∠R = 40°(Given)
∴ ΔABC ~ ΔPQR (AAA similarity criterion)

 

(ii) In  ΔABC and ΔPQR, we have

∴  ΔABC ~ ΔQRP (SSS similarity criterion)

 

(iii) In ΔLMP and ΔDEF, we have
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6

 

(iv) In ΔMNL and ΔQPR, we have

∠M = ∠Q = 70°
∴ ΔMNL ~ ΔQPR (SAS similarity criterion)

 

(v) In ΔABC and ΔDEF, we have
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°

 

(vi) In ΔDEF,we have
∠D + ∠E + ∠F = 180° (sum of angles of a triangle)
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° - 70° - 80°
⇒ ∠F = 30°
In PQR, we have
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° - 80° -30°
⇒ ∠P = 70°
In ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Hence, ΔDEF ~ ΔPQR (AAA similarity criterion)

Exercise 6.3 Page Number 138

2

In the fig 6.35, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB. 

Answer

DOB is a straight line.
Therefore, ∠DOC + ∠ COB = 180°
⇒ ∠DOC = 180° - 125°
= 55°
In ΔDOC,
∠DCO + ∠ CDO + ∠ DOC = 180°
(Sum of the measures of the angles of a triangle is 180º.)
⇒ ∠DCO + 70º + 55º = 180°
⇒ ∠DCO = 55°
It is given that ΔODC ~ ΔOBA.
∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠ OAB = 55°
∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠OAB = 55°
Exercise 6.3 Page Number 139

3

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

Answer

In ΔDOC and ΔBOA,
∠CDO = ∠ABO [Alternate interior angles as AB || CD]
∠DCO = ∠BAO [Alternate interior angles as AB || CD]
∠DOC = ∠BOA [Vertically opposite angles]
∴ ΔDOC ~ ΔBOA [AAA similarity criterion]
Exercise 6.3 Page Number 139

4

In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Answer

In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR ...(i)
Exercise 6.3 Page Number 140

5

S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that 
ΔRPQ ~ ΔRTS.

Answer

In ΔRPQ and ΔRST,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)
∴ ΔRPQ ~ ΔRTS (By AA similarity criterion)
Exercise 6.3 Page Number 140

6

In the fig 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

Answer

It is given that ΔABE ≅ ΔACD.
∴ AB = AC [By cpct] ...(i)
And, AD = AE [By cpct] ...(ii)
In ΔADE and ΔABC,

∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [By SAS similarity criterion]
Exercise 6.3 Page Number 140

7

In the fig 6.38, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC

Answer

(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP(Each 90°)
∠APE = ∠CPD(Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔCDP

 

(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB (Each 90°)
∠ABD = ∠CBE (Common)
Hence, by using AA similarity criterion,
ΔABD ~ ΔCBE

 

(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB (Each 90°)
∠PAE = ∠DAB (Common)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔADB

 

(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
ΔPDC ~ ΔBEC

Exercise 6.3 Page Number 140

8

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Answer

In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (By AA similarity criterion)
Exercise 6.3 Page Number 140

9

In the fig 6.39, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

(i) ΔABC ~ ΔAMP

(ii) CA/PA = BC/MP

Answer

(i) In ΔABC and ΔAMP, we have
∠A = ∠A (common angle)
∠ABC = ∠AMP = 90° (each 90°)
∴  ΔABC ~ ΔAMP (By AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (By AA similarity criterion)
If two triangles are similar then the corresponding sides are equal,

Exercise 6.3 Page Number 140

10

CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:

(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF

Answer

(i) It is given that ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,
∠A = ∠F (Proved above)
∠ACD = ∠FGH (Proved above)
∴ ΔACD ~ ΔFGH (By AA similarity criterion)

 

(ii) In ΔDCB and ΔHGE,
∠DCB = ∠HGE (Proved above)
∠B = ∠E (Proved above)
∴ ΔDCB ~ ΔHGE (By AA similarity criterion)

 

(iii) In ΔDCA and ΔHGF,
∠ACD = ∠FGH (Proved above)
∠A = ∠F (Proved above)
∴ ΔDCA ~ ΔHGF (By AA similarity criterion)

Exercise 6.3 Page Number 140

11

In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

Answer

It is given that ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∠ADB = ∠EFC (Each 90°)
∠BAD = ∠CEF (Proved above)
∴ ΔABD ~ ΔECF (By using AA similarity criterion)
Exercise 6.3 Page Number 141

12

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

Answer

Given:

ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR

Exercise 6.3 Page Number 141

13

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

Answer


In ΔADC and ΔBAC,
∠ADC = ∠BAC(Given)
∠ACD = ∠BCA(Common angle)
∴ ΔADC ~ ΔBAC (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
Exercise 6.3 Page Number 141

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 ΔABC ~ ΔPQR.

Answer



To Prove: ΔABC ~ ΔPQR
Construction: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.
Proof: In ΔABD and ΔCDE, we have
AD = DE  [By Construction]
BD = DC [∴ AP is the median]
and, ∠ADB = ∠CDE [Vertically opp. angles]
∴ ΔABD ≅ ΔCDE [By SAS criterion of congruence]
⇒ AB = CE [CPCT] ...(i)
Also, in ΔPQM and ΔMNR, we have
PM = MN [By Construction]
QM = MR [∴ PM is the median]
and, ∠PMQ = ∠NMR [Vertically opposite angles]
∴ ΔPQM = ΔMNR [By SAS criterion of congruence]
⇒ PQ = RN [CPCT] ...(ii)

∴ ΔACE ~ ΔPRN [By SSS similarity criterion]
Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P ...(iii)
Now, In ΔABC and ΔPQR, we have
Exercise 6.3 Page Number 141

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.

Answer

Length of the vertical pole = 6m (Given)
Shadow of the pole = 4 m (Given)
Let Height of tower = h m
Length of shadow of the tower = 28 m (Given)
In ΔABC and ΔDEF,
∠C = ∠E (angular elevation of sum)
∠B = ∠F = 90°
∴ ΔABC ~ ΔDEF (By AA similarity criterion)
Exercise 6.3 Page Number 141

16

If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR Prove that AB/PQ = AD/PM.

Answer


Exercise 6.3 Page Number 141

1

Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Answer

It is given that,
Area of ΔABC = 64 cm2
Area of ΔDEF = 121 cm2
EF = 15.4 cm
and, ΔABC ~ ΔDEF
Exercise 6.4 Page Number 143

2

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

Answer


ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.
In ΔAOB and ΔCOD, we have
∠1 = ∠2(Alternate angles)
∠3 = ∠4(Alternate angles)
∠5 = ∠6(Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [By AAA similarity criterion]
Now, Area of (ΔAOB)/Area of (ΔCOD)
Exercise 6.4 Page Number 143

3

In the fig 6.53, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that  area of ∆ ABC/ area of  ∆ DBC = AO/ DO .

Answer


Exercise 6.4 Page Number 144

4

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

Answer


Given: ΔABC and ΔPQR are similar and equal in area.
To Prove: ΔABC ≅ ΔPQR
Proof: Since, ΔABC ~ ΔPQR

Similarly, we can prove that
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [BY SSS criterion of congruence]
Exercise 6.4 Page Number 144

5

D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

Answer


Given: D, E and F are the mid-points of the sides AB, BC and CA respectively of the ΔABC.
To Find: area(ΔDEF) and area(ΔABC)
Solution: In ΔABC, we have
F is the mid point of AB (Given)
E is the mid-point of AC (Given)
So, by the mid-point theorem, we have

∴ BDEF is parallelogram [Opposite sides of parallelogram are equal and parallel]
Similarly in ΔFBD and ΔDEF, we have
FB = DE (Opposite sides of parallelogram BDEF)
FD = FD (Common)
BD = FE (Opposite sides of parallelogram BDEF)
∴ ΔFBD ≅ ΔDEF
Similarly, we can prove that
ΔAFE ≅ ΔDEF
ΔEDC ≅ ΔDEF
If triangles are congruent,then they are equal in area.
So, area(ΔFBD) = area(ΔDEF) ...(i)
area(ΔAFE) = area(ΔDEF) ...(ii)
and, area(ΔEDC) = area(ΔDEF) ...(iii)
Now, area(ΔABC) = area(ΔFBD) + area(ΔDEF) + area(ΔAFE) + area(ΔEDC) ...(iv)
area(ΔABC) = area(ΔDEF) + area(ΔDEF) + area(ΔDEF) + area(ΔDEF)
Exercise 6.4 Page Number 144

6

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

Answer



Exercise 6.4 Page Number 144

7

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Answer

Given: ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.
Exercise 6.4 Page Number 144

8

Tick the correct answer and justify:
Q8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4

Answer

Exercise 6.4 Page Number 144

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

Answer


Exercise 6.4 Page Number 144

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

Answer

(i) Given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of these sides, we will get 49, 576, and 625.
49 + 576 = 625
(7)2 + (24)2 = (25)2
The sides of the given triangle are satisfying Pythagoras theorem.Hence, it is right angled triangle.
Length of Hypotenuse = 25 cm

 

(ii) Given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.
However, 9 + 36 ≠ 64
Or, 32 + 62 ≠ 82
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.

 

(iii) Given that sides are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 502 + 802 ≠ 1002
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.

 

(iv) Given that sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.
Clearly, 144 +25 = 169
Or, 122 + 52 = 132
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
Length of the hypotenuse of this triangle is 13 cm.

Exercise 6.5 Page Number 150

2

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.

Answer


Given: ΔPQR is right angled at P is a point on QR such that PM ⊥QR.
To prove: PM2 = QM × MR
Proof: In ΔPQM, we have
PQ2 = PM2 + QM2 [By Pythagoras theorem]
Or, PM2 = PQ2 − QM2 ...(i)
In ΔPMR, we have
PR2 = PM2 + MR2 [By Pythagoras theorem]
Or, PM2 = PR2 − MR2 ...(ii)
Adding (i) and (ii), we get
2PM2 = (PQ2 + PM2) - (QM2 + MR2
= QR2 − QM2 − MR2        [∴ QR2 = PQ2 + PR2]
= (QM + MR)2 − QM2 − MR2
 = 2QM × MR
∴ PM2 = QM × MR
Exercise 6.5 Page Number 150

3

In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB2 = BC × BD
(ii) AC2 = BC × DC
(iii) AD2 = BD × CD

Answer

(i) In ΔADB and ΔCAB, we have
∠DAB = ∠ACB (Each equals to 90°)
∠ABD = ∠CBA (Common angle)
∴ ΔADB ~ ΔCAB [AA similarity criterion]

 

(ii) Let ∠CAB = x
In ΔCBA,
∠CBA = 180° - 90° - x
∠CBA = 90° - x
Similarly, in ΔCAD
∠CAD = 90° - ∠CBA
= 90° - x
∠CDA = 180° - 90° - (90° - x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA (Each equals to 90°)
∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]

 

(iii) In ΔDCA and ΔDAB, we have
∠DCA = ∠DAB (Each equals to 90°)
∠CDA = ∠ADB (common angle)
∴ ΔDCA ~ ΔDAB [By AA similarity criterion]

Exercise 6.5 Page Number 150

4

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

Answer


Given that ΔABC is an isosceles triangle right angled at C.
In ΔACB, ∠C = 90°
AC = BC (Given)
AB2 = AC2 + BC2 [By using Pythagoras theorem]
= AC2 + AC2 [Since, AC = BC]
AB2 = 2AC2

Exercise 6.5 Page Number 150

5

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

Answer

 

Given that ΔABC is an isosceles triangle having AC = BC and AB2 = 2AC2
In ΔACB,
AC = BC (Given)
AB2 = 2AC2 (Given)
AB2 = AC2 + AC2
= AC2 + BC2 [Since, AC = BC]
Hence, By Pythagoras theorem ΔABC is right angle triangle.

Exercise 6.5 Page Number 150

6

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

Answer

ABC is an equilateral triangle of side 2a. 
Draw, AD ⊥ BC
In ΔADB and ΔADC, we have
AB = AC[Given]
AD = AD[Given]
∠ADB = ∠ADC[equal to 90°]
Therefore, ΔADB ≅ ΔADC by RHS congruence.
Hence, BD = DC [by CPCT]
In right angled ΔADB, 
AB2 = AD2 + BD2
(2a)2 = AD2 + a2
⇒ AD2  = 4a2 - a2
⇒ AD2  = 3a2
⇒ AD  = √3a 

Exercise 6.5 Page Number 150

7

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

Answer


ABCD is a rhombus whose diagonals AC and BD intersect at O. [Given]
We have to prove that, 
AB2 + BC2 + CD2 + AD2 = AC2 + BD2
Since, the diagonals of a rhombus bisect each other at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB,
∠AOB = 90°
AB2 = AO2 + BO2 ... (i) [By Pythagoras]
Similarly, 
AD2 = AO2 + DO2 ... (ii)
DC2 = DO2 + CO2 ... (iii)
BC2 = CO2 + BO2 ... (iv)
Adding equations (i) + (ii) + (iii) + (iv)  we get,
AB2 + AD2 + DC2 + BC2  =  2(AO2 + BO2 + DO2 + CO2 )
= 4AO2 + 4BO2 [Since, AO = CO and BO =DO]
= (2AO)2 + (2BO)2 = AC2 + BD2

Exercise 6.5 Page Number 150

8

In Fig. 6.54, O is a point in the interior of a triangle
ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

Answer


(i) Applying Pythagoras theorem in ΔAOF, we have
OA2 = OF2 + AF2
Similarly, in ΔBOD
OB2 = OD2 + BD2
Similarly, in ΔCOE
OC2 = OE2 + EC2
Adding these equations,
OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2.
(ii) AF2 + BD2 + EC2 = (OA2 - OE2) + (OC2 - OD2) + (OB2 - OF2)
∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Exercise 6.5 Page Number 151

9

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

Answer


Let BA be the wall and Ac be the ladder,
Therefore, by Pythagoras theorem,we have
AC2 = AB2 + BC2
102 = 82 + BC2
BC2 = 100 – 64
BC2 = 36
BC = 6m
Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.
Exercise 6.5 Page Number 151

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 ?

Answer


Let AB be the pole and AC be the wire.
By Pythagoras theorem,
AC2 = AB2 + BC2
242 = 182 + BC2
BC2 = 576 - 324
BC2 = 252
BC = 6√7m
Therefore, the distance from the base is 6√7m.

Exercise 6.5 Page Number 151

11

An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after   hours ?

Answer


Speed of first aeroplane = 1000 km/hr
Distance covered by first aeroplane due north in 1 hours (OA) = 1000 × 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr 
Distance covered by second aeroplane due west in 1  hours (OB) = 1200 × 3/2km = 1800 km
In right angle ΔAOB, we have
AB2 = AO2 + OB2
⇒ AB2 = (1500)2 + (1800)2
⇒ AB = √2250000 + 3240000
= √5490000
⇒ AB = 300√61 km
Hence, the distance between two aeroplanes will be 300√61 km.

Exercise 6.5 Page Number 151

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.

Answer


Let CD and AB be the poles of height 11 m and 6 m.
Therefore, CP = 11 - 6 = 5 m
From the figure, it can be observed that AP = 12m
Applying Pythagoras theorem for ΔAPC, we get
AP2 = PC2 + AC2
(12m)2 + (5m)2 = (AC)2
AC2 = (144+25)m2 = 169 m2
AC = 13m
Therefore, the distance between their tops is 13.
Exercise 6.5 Page Number 151

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. 

Answer


Applying Pythagoras theorem in ΔACE, we get
AC2 + CE2 = AE2....(i)
Applying Pythagoras theorem in ΔBCD, we get
BC2 + CD2 = BD2....(ii)
Using equations (i) and (ii), we get
AC2 + CE2 + BC2 + CD2 = AE2 + BD2...(iii)
Applying Pythagoras theorem in ΔCDE, we get
DE2 = CD2 + CE2
Applying Pythagoras theorem in ΔABC, we get
AB2 = AC2 + CB2
Putting these values in equation (iii), we get
DE2 + AB2 = AE2 + BD2.
Exercise 6.5 Page Number 151

14

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

Answer

Given that in ΔABC, we have
AD ⊥BC and BD = 3CD
In right angle triangles ADB and ADC, we have
AB2 = AD2 + BD2...(i)
AC2 = AD2 + DC2...(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB2 - AC2 = BD2 - DC  
= 9CD2 - CD2[∴ BD = 3CD]
= 9CD2 = 8[Since, BC = DB + CD = 3CD + CD = 4CD
Therefore, AB2 - AC2 =  
⇒ 2(AB2 - AC2) = BC2
⇒ 2AB2 - 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.
Exercise 6.5 Page Number 151

15

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

Answer


Exercise 6.5 Page Number 151

16

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

Answer


Exercise 6.5 Page Number 151

17

Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°
(B) 60°
(C) 90° 
(D) 45°

Answer


Given: AB = 6√3 cm, AC = 12 cm, and BC = 6 cm
We can observe that
AB2 = 108
AC2 = 144
And, BC2 = 36
AB2 + BC2 = AC2
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct option is (C).
Exercise 6.5 Page Number 151

1

In figure, PS is the bisector of ∠QPR of ∆PQR, prove that  QS/SR = PQ/PR.

Answer

Given: In figure, PS is the bisector of ∠QPR of ∆PQR. Now, draw RT||SP  to meet QP produced in T.
∵ RT||SP and transversal PR intersects them
∴ ∠ 1 = ∠2 (Alternate interior angle)…(i)
∵ RT ||SP  and transversal QT intersects them

∴ ∠3 = ∠4 (Corresponding angle) …(ii)
But ∠1 = ∠3 (Given)
∴ ∠2 = ∠4 [From Eq. (i) and (ii)]
∴ PT = PR ….(iii) (∵Sides opposite to equal angles of a triangle are equal)
Now, in ∆QRT,
PS|| RT  (By construction)
Exercise 6.6 Page Number 152

2

In figure, D is a point on hypotenuse AC of ∆ABC, such that  BD⊥ AC  , DM ⊥BC  and DN⊥ AB  . Prove that.
(i) DM2 = DN. MC
(ii) DN2 = DM. AN

Answer

Given: D is a point on hypotenuse AC of ∆ABC, DM⊥BC and DN⊥ AB .

Exercise 6.6 Page Number 152

3

In figure 6.58, ABC is a triangle in which ∠ABC >90° and AD⊥ CB produced.
Prove that  AC2 =  AB2 + BC2=2BC .BD.

Answer

Given: In figure, ABC is a triangle in which ∠ABC> 90°and AD ⊥CB produced.
Proof : In right ∆ABC,
∵ ∠D= 900
∴ AC2 = AD2 +DC2 (By Pythagoras theorem)
⇒ AC2 = AD2 +(DB2 + BC2) [∵ DC= DB+BC]
⇒ AC2 = (AD2+DB2 )+BC2 + 2DB.BC [∵ (a+b)2 = a2 +b2+2ab]
⇒ AC2 = AB2+BC2 +2DB . BC
∵In right ∆ADB with ∠D = 900 , AB2 = AD2+DB2 [By Pythagoras theorem)
Exercise 6.6 Page Number 152

4

In figure 6.59, ABC is a triangle in which ∠ ABC <900 and AD⊥ BC .
Prove that AC2 = AB2 + BC2=2BC .BD.

Answer

Given: In figure, ABC is a triangle in which ∠ABC< 90°and AD ⊥CB.
Proof: In right ∆ ADC
∠D = 900
∴ AC2 = AD2+DC2  (By Pythagoras theorem)
⇒ AC2= AD2+(BC – BD)2  (∵ BC = BD+DC)
⇒ AC2= AD2 + BC2+BD2 –2BC . BD  [∵ (a – b)2 = a2+b2 – 2ab ]
⇒ AC2= (AD2 +BD2)+BC2 –2BC . BD
⇒ AC2= AB2 +BC2 –2BC . BD
∵In right ∆ADB with ∠D = 900, AB2 = AD2+BD2  [By Pythagoras theorem)
Exercise 6.6 Page Number 152

5

In figure 6.60, AD is a median of triangle ABC and AM ⊥ BC. Prove that

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

(ii) AB2 = AD2 – BC.DM + (BC/2)2

(iii) AC2 + AB2 = 2AD2 + ½ BC2

Answer

Given: In figure, AD is a median of a ∆ABC and AM⊥ BC.
Proof :

(i) In right ∆AMC,
∵  ∠M =90°
∴ AC2 = AM2+MC2 (By Pythagoras theorem)
∵ MC = MD+DC
⇒ AC2 = AM2+(MD+DC)2
⇒ AC2 = AM2 + MD2 +DC2 +2MD. DC [∵ (a+ b)2  = a2+b2+2ab ]
⇒ AC2 =  (AM2 + MD2 )+ DC2 +2MD. DC
⇒ AC2 = AD2+ DC2 +2DC. MD
∵In right ∆AMD with ∠M = 90°,

AD2 = AM2+MD2 [By Pythagoras theorem)

 

(ii) In right ∆AMB,
∵  ∠M = 90°
∴ AB2 = AM2+BM2     (By Pythagoras theorem)
∵ BD  = BM +MD
⇒ AB2 = AM2 + (BD – MD)2
⇒ AB2 = AM2 + BD2 +MD2 –2BD. MD [∵ (a – b)2 = a2 +b2 –2ab ]
⇒ AB2 = (AM2 +MD2)+BD2 – 2BD. MD
∵In right ∆AMD with ∠M = 90°,

AD2 = AM2+MD2  [By Pythagoras theorem)
⇒ AB2 = AD2 + BD2 – 2BD. MD
⇒ AB2 = AD2 + (1/2BC)2 – BC. MD [∵2BD = BC, (AD is a median of ∆ABC)]
∴  AB2 = AD2 + (1/2BC)2 – BC. MD  … (ii)

(iii) On adding Eq. (i) and (ii), we get

Exercise 6.6 Page Number 152

6

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

Answer

Given: ABCD is a parallelogram whose diagonals are AC and BD.

Now, draw AM⊥DC and BN ⊥ D (produced). Proof In right ∆AMD and ∆BNC,
AD= BC    (Opposite sides of a parallelogram)
AM= BN   (Both are altitudes of the same parallelogram to the same base)
∴ ∆ AMD≅ ∆ BNC   (RHS congruence criterion)
∴ MD=NC     (CPCT)…(i)
In right ∆BND,
∵ ∠N = 90°
∴  BD2 = BN2 +DN2  (By Pythagoras theorem )
⇒BD2 = BN2 +(DC+CN)2  (∵DN = DC+CN)
⇒BD2 = BN2 + DC2 +CN2  +2DC . CN  [(a+ b)2 = a2+b2 +2ab]
⇒BD2 = (BN2+ CN2 )+ DC2 +2DC . CN
∵In right ∆BNC with ∠N = 900, BC2 = BN2+NC2  [By Pythagoras theorem)
⇒BD2 = BC2+ DC2 +2DC . CN  …. (ii)
In right ∆AMC, ∠M= 90°
∴ AC2 = AM2 +MC2  (∵ MC = DC – DM)
⇒ AC2 = AM2 + (DC – DM)2        [(a+ b)2 = a2+b2 +2ab]
⇒ AC2 = AM2 + DC2 +DM2 –2DC .DM
⇒ AC2 = (AM2 + DM2) + DC2 –2DC .DM
∵In right ∆AMD with ∠M = 900, AD2 = AM2+MD2  [By Pythagoras theorem)
⇒ AC2 = AD2 + DC2 –2DC .DM
⇒ AC2 = AD2 + AB2 −2DC .CN  … (iii)

(∵ DC = AB , opposite sides of parallelogram and BM= CN from Eq. (i)]
Now, on adding Eq. (iii) and (ii), we get
AC2 = (D2 + AB2 − 2DC .CN
BD2 = BC2+ DC2 +2DC . CN
⇒ AC2 + BD2 = (AD2 + AB2)+ (BC2+ DC2)
⇒ AC2 + BD2 = AD2 + AB2+BC2+ DC2

Exercise 6.6 Page Number 153

7

In figure 6.61, two chords AB and CD intersect each other at the point P. Prove that
(i) ∆APC ~∆DPB
(ii) AP. PB =CP .DP

Answer

(i) ∆APC and ∆DPB
∠APC= ∠DPB  (Vertically opposite angles)
∠CAP= ∠BDP (Angles in the same segment)
∴ ∆APC ~ ∆DPB  (AA similarity criterion)

 

(ii) ∆APC ~ ∆DPB  [Proved in (i)]

⇒ AP. BP = CP . DP
⇒ AP. PB = CP. DP (Proved)

Exercise 6.6 Page Number 153

8

In figure 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

Answer

(i) We know that, in a cyclic quadrilaterals, the exterior angle is equal to the interior opposite angle.
Therefore,  ∠PAC = ∠PDB  …(i)
and   ∠PCA = ∠PBD  …(ii)
In view of Eq. (i) and (ii), we get
(∵ AA similarity criterion)
∆PAC ~ ∆PDB  (Similar)

 

(ii) ∆PAC ~∆PDB  [Proved in (i)]

Exercise 6.6 Page Number 153

9

In figure 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 .

Answer


∴ In ∆BCE,
AD||CE (By converse of basic proportionality theorem)
∴ ∠BAD = ∠AEC (Corresponding angle)  …(i)
and ∠CAD= ∠ACE (Alternate interior angle) …(ii)
∵  AC = AE (By construction)
∴ ∠AEC = ∠ACE  …(iii)

(Angles opposite equal sides of a triangle are equal)
Using Eq. (i), (ii) and (iii), we get
∠BAD= ∠CAD
i.e., AD is the bisector of ∠BAC.

Exercise 6.6 Page Number 153

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 cms−1, what will be the horizontal distance of the fly from her after 12 s?

Answer


∴ Length of remaining string left out =3.0 −0.6 = 2.4m
BD2 = AC2 – AB2  (By Pythagoras theorem)
⇒  BD2= (2.4)2 – (1.8)2  = 5.76 – 3.24 = 2.52
Exercise 6.6 Page Number 153

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