PSEB Solutions for Class 10 Maths Chapter 6 Triangles

 PSEB Solutions for Class 10 Maths Chapter 6 Triangles

PSEB 10th Class Maths Solutions Chapter 6 Triangles

Ex 6.1

Question 1.
Fill in the blanks using the correct word given in brackets:
(i) All circles are ………………. (congruent, similar).
Solution:
All circles are similar.

(ii) All squares are ………………. (similar, congruent).
Solution:
MI squares are similar.

(iii) All ………………. triangles are similar. (isosceles, equilateral).
Solution:
All equilateral triangles are similar.

(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are __________ and
Solution:
equal

(b) their corresponding sides are ………………. (equal, proportional).
Solution:
proportional.

Question 2.
Give two different examples of pair
(i) similar figures
(ii) non-similar figures.
Solution:
(i) 1. Pair of equilateral triangle are similar figures.
2. Pair of squares are similar figures.

(ii) 1. A triangle and quadrilateral form a pair of non-similar figures.
2. A square and rhombus form pair of non – similar figures.

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

Solution:
The two quadrilaterals in the figure are not similar because their corresponding angles are not equal.

Ex 6.2

Question 1.
In fig. (i) and (ü), DE U BC. Find EC in (i) and AD in (ii).

 

Question 2.
E and F are points on the sides PQ and PR respectively of a APQR. 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 = 9cm.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
Solution:
In ∆PQR, E and F are two points on side PQ and PR respectively.

   

Hence proved.

Question 5.
In fig. DE || OQ and DF || OR. Show that EF || QR.

 

Question 6.
In flg., A, B and C points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show thatBC || QR.

 

Question 7.
Using Basic Proportionality theorem, 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 ¡t in class IX).
Solution:
Given: In ∆ABC, D is mid point of AB, i.e. AD = DB.
A line parallel to BC intersects AC at E as shown in figure. i.e., DE || BC.

Question 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 ¡tin Class IX).
Solution:
Given ∆ABC, D and E are mid points of AB and AC respectively such that AD = BD and AE = EC, D and Eare joined

    

∴ By using converse of basic
proportionlity Theorem,
EO || DC also EO || AB [Const]
⇒ AB || DC
∴ Quadrilateral ABCD is a trapezium with AB || CD.

Ex 6.3

Question 1.
State which pairs of triangles in Fig. 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) In ∆ABC and ∆PQR,
∠A = ∠P (each 60°)
∠B = ∠Q (each 80°)
∠C = ∠R (each 40°)
∴ ∆ABC ~ PQR [AAA Similarity criterion]

 

(vi) In ΔDEF, ∠D = 70°, ∠E = 80°
∠D + ∠E + ∠F = 180°
70° + 80° + ∠F = 180° [Angle Sum Propertyl
∠F= 180° – 70° – 80°
∠F = 30°
In ΔPQR,
∠Q = 80°, ∠R = 30°
∠P + ∠Q + ∠R = 180°
(Sum of angles of triangle)
∠P + 80° + 30° = 180°
∠P = 180° – 80° – 30°
∠P = 70°
In ΔDEF and ΔPQR,
∠D = ∠P (70° each)
∠E = ∠Q (80° each)
∠F = ∠R (30° each)
∴ ΔDEF ~ ΔPQR (AAA similarity criterion).

Question 2.
In Fig., ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. FInd ∠DOC, ∠DCO and ∠OAB.

Solution:
Given that: ∠BOC = 125°
∠CDO = 70°
DOB is a straight line
∴ ∠DOC + ∠COB = 180°
[Linear pair Axiom]
∠DOC + 125° = 180°
∠DOC = 180°- 125°
∠DOC = 55°
∠DOC = ∠AOB = 55°
[Vertically opposite angle]
But ΔODC ~ ΔOBA
∠D = ∠B = 70°
In ΔDOC, ∠D + ∠O + ∠C = 180°
70° + 55° + ∠C = 180°
∠C= 180° – 70° – 55°
∠C = 55°
∠C = ∠A = 55°
Hence ∠DOC = 55°
∠DCO = 55°
∴ ∠OAB = 55°.

Question 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:

Question 4.
In Fig.,  and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

 

Question 5.
S and T are points on skies PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:
S and T are the points on side PR and QR such that ∠P = ∠RTS.

To Prove. ∆RPQ ~ ∆RTS
Proof: In ∆RPQ and ∆RTS
∠RPQ = ∠RTS (given)
∠R = ∠R [common angle]
∴ RPQ ~ ARTS
[By AA similarity critierion which is the required result.]

Question 6.
In figure ∆ABE ≅ ∆ACD show that ∆ADE ~ ∆ABC.

Question 7.
In Fig., 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:
Given. ∆ABC, AD ⊥ BC CE⊥AB,

To Prove. (i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
Proof:
(i) In ∆AEP and ∆CDP,
∠E = ∠D (each 90°)
∠APE = ∠CPD (vertically opposite angle)
∴ ∆AEP ~ ∆CDP [By AA similarity criterion].

(ii) In ∆ABD and ∆CBE,
∠D = ∠E (each 90°)
∠B = ∠B (common)
∴ ∆ABD ~ ∆CBE [AA Similarity criterion]

(iii) In ∆AEP and ∆ADB.
∠E = ∠D (each 90°)
∠A = ∠A (common)
∴ ∆AEP ~ ∆ADB [AA similarity criterion].

(vi) In ∆PDC and ∆BEC,
∠C = ∠C
∠D = ∠E
∴ ∆SPDC ~ ∆BEC [AA similarity criterion].

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE Intersects CD at F. Show that AABE – &CFB.
Solution:
Given. Parallelogram ABCD. Side AD is produced to E, BE intersects DC at F.

To Prove. ∆ABE ~ ∆CFB
Proof. In ∆ABE and ∆CFB.
∠A = ∠C (opposite angle of || gm)
∠ABE = ∠CFB (alternate angle)
∴ ∆ABE ~ ∆CFB (AA similarity criterion)

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

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

 

(ii) In ∆DCB and ∆HGE,
∠B = ∠E [Proved above]
∠1 = ∠3 [Proved above]
∴ ∆DCB ~ ∆HGE [∵ AA similarity criterion]

(iii) In ∆DCA and ∆HGF
∠A = ∠F [Proved above]
∠2 = ∠4 [Proved above]
∴ ∆DCA ~ ∆HGF [∵ AA similarity criterion].

Question 11.
In Fig., E is a point on side CB produced of an Isosceles triangle ABC with AB = AC. IfAD ⊥BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.

Solution:
Given. ∆ABC, isosceles triangle with AB = AC AD ⊥ BC, side BC is produced to E. EF ⊥ AC
To Prove. ∆ABD ~ ∆ECF
Proof. ∆ABC is isosceles (given)
AB = AC
∴ ∠B = ∠C [Equal sides have equal angles opposite to it)
In ∆ABD and ∆ECF,
∠ABD = ∠ECF (Proved above)
∠ADB = ∠EFC (each 90°)
∴ ∠ABD – ∠ECF [AA similarity).

Question 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.). Show that ∆ABC ~ ∆PQR.

  

Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB. CD.
Solution:
Given. ∆ABC, D is a point on side BC such that ∠ADC = ∠BAC
To Prove. CA2 = BC × CD
Proof. In ABC and ADC,

Question 14.
Sides AB and AC and median AD of a triangle ABC are proportional to sides PQ and PR and median PM of another
triangle PQR. Prove that ∆ABC ~ ∆PQR.
Solution:

 

Question 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:

  

Ex 6.4

Question 1.
Let △ABC ~ △DEF and their areas be respectively 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
Solution:
△ABC ~ △DEF ;
area of △ABC = 64 cm2;
area of △DEF = 121 cm2;
EF= 15.4 cm

Question 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 traingles AOB and COD.
Solution:
ABCD is trapezium AB || DC. Diagonals AC and BD intersects each other at the point O. AB = 2 CD

∴ Required ratio of ar △AOB and △COD = 4 : 1.

Question 3.
In the fig., △ABC and △DBC are two triangles on the same base BC. If AD intersects BC at O show that

 

Hence proved.

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Given: Two ∆s ABC and DEF are similar and equal in area.
To Prove : ∆ABC ≅ ∆DEF

Question 5.
D, E and F are respectively the mid points of the sides BC, CA and AB of ∆ABC. Determine the ratio of the areas of triangles DEF and ABC.
Solution:

Given. D, E, F are the mid-point of the sides BC, CA and AB respectively of a tABC.
To find : ar (∆DEE) : ar (∆ABC)
Proof: In ∆ABC,
F is the mid-point of AB …(given)
E is the mid-point of AC …(given)
So, by the Mid-Foint Theorem
FE || BC and FE = 1/2 BC
⇒ FE || BD and FE = BD [∵ BD = 1/2 BC]
∴ BDEF is a || gm.
(∵ Opp. sides are || and equal)
In △s FBD and DEF,
FB = DE (opp. sides of || gm BDEF)
FD = FD .. .(common)
BD = FE
. ..(opp. sides of || gm BDEF)
∴ △FBD ≅ △DEF … (SSS Congruency Theorem)

Similary we can prove that:
△AFE ≅ △DEF
and △EDC ≅ △DEF
if △s are , then they are equal in area.
∴ ar (∆FBD) = ar. (∆DEF) ……………(1)
ar (∆AFE) = ar (∆DEF) ……………(2)
ar (∆EDC) = ar (∆DEF) ……………(3)
Now ar ∆ (ABC)
= ar (∆FBD) + ar (∆DEF) + ar (∆AFE) + ar (∆EDC)
= ar.(∆DEF) + ar (∆DEF) + ar (∆DEF) + ar. (∆DEF) [Using (1), (2) and (3)]
= 4 ar (∆DEF)
⇒ (∆DEF) = 1/4 ar(∆ABC)
⇒ ar.(ΔDEF)/ar.(ΔABC)=1/4
∴ ar (∆DEF) : ar (∆ABC) = 1 : 4.

Question 6.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
Given: ∆ABC ~ ∆DEF.
AX and DY are the medians to the side BC and EF respectively.

Question 7.
Prove that the areas of the equilateral triangle described on the side of a square Is half the area of the equilateral triangle described on its diagonal.
Solution:
Given: ABCD is a square. Equilateral ∆ABE is described on the side AB of the square and equilateral ∆ACF is desribed on the diagonal AC.

Question 8.
Tick the correct answer and justify: 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:

Question 9.
Tick the correct answer and justify: Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are ¡n the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81.
Solution:

Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm.
Solution:
(i) Let ∆ABC, with AB = 7 cm BC = 24 cm, AC = 25 cm
AB2 + BC2 = (7)2 + (24)2
= 49 + 576 = 625
AC2 = (25)2 = 625
Now AB2 + BC2 = AC2
∴ ∆ABC is right angled triangle. Hyp. AC = 25cm.

(ii) Let ∆PQR with PQ = 3 cm, QR = 8 cm PR = 6 cm
PQ2 + PR2 = (3)2 + (6)2
= 9 + 36 = 45
QR2 = (8)2 = 64.
Here PQ2 + PR2 ≠ QR2
∴ ∆PQR is not right angled triangle.

(iii) Let ∆MNP, with MN =50 cm, NP = 80 cm, MP = 100 cm
MN 2+ NP2 = (50)2 + (80)2
= 2500 + 6400 = 8900
MP2 = (100)2 = 10000
Here MP2 ≠ MN2 + NP2.
∴ ∆MNP is not right angled triangle.

(iv) Let ∆ABC, AB = 13 cm, BC = 12 cm, AC = 5 cm
BC2 + AC2 = (12)2 + (5)2
= 144 + 25 = 169
AB2 = (13)2 = 169
∴ AB2 = BC2 + AC2
∆ABC is right angled triangle.
Hyp. AB = 13 cm.

Question 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.
Solution:
Given: ∆PQR is right angled at P and M is a point on QR such that PM ⊥ QR.
To prove : PM2 = QM × MR

Proof: ∠P = 90° (Given)
∴ ∠1 + ∠2 = 90°
∠M = 900 (Given)
In ∆PMQ,
∠1 + ∠3 + ∠5 = 180°
=> ∠1 + ∠3 = 90° [Angle Sum Property] ………….(2) [∠5 = 90°]

Question 3.
In fig., ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB2 = BC.BD
(ii) AC2 = BC.DC
(üi) AD2 = BD.CD.
Solution:
Given. A right angled ∆ABD in which right angled at A and AC ⊥ BD.
To Prove:
(i) AB2 = BC.BD
(ii) AC2 = BC.DC .
(iii) AD2 = BD.ÇD .
Proof. In ∆DAB and ∆DCA,
∠D = ∠D (common)
∠A = ∠C (each 90°)
∴ ∆DAB ~ ∆DCA [AA similarity]
In ∆DAB and ∆ACB,
∠B = ∠B (common)
∠A = ∠C . (each 90°)
∴ ∆DAB ~ ∆ACB, .
From (1) and (2),
∆DAB ~ ∆ACB ~ ∆DCA.
(i) ∆ACB ~ ∆DAB (proved)

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution:
Given: ABC is an isosceles triangle right angled at C.
To prove : AB2 = 2AC2.
Proof: In ∆ACB, ∠C = 90° & AC = BC (given)

AB2 = AC2 + BC2
[By using Pythagoras Theorem]
=AC2 + AC2 [BC = AC]
AB2 = 2AC2
Hence proved.

Question 5.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is right triangle.
Solution:
Given: ∆ABC is an isosceles triangle AC = BC
To prove: ∆ABC is a right triangle.

Proof: AB2 = 2AC2 (given)
AB2 = AC2 + AC2
AB2 = AC2 + BC2 [AC = BC]
∴ By Converse of Pythagoras Theorem,
∆ABC is right angled triangle.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
∆ABC is equilateral triangle with each side 2a
AD ⊥ BC
AB = AC = BC = 2a
∆ADB ≅ ∆ADC [By RHS Cong.]
∴ BD = DC = a [c.p.c.t]

In right angled ∆ADB
AB2 = AD2 + BD2
(2a)2 = AD2 + (a)2
4a2 – a2 = AD2.
AD2 = 3a2
AD = √3a.

Question 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. [Pb. 2019]
Solution:
Given: Rhombus, ABCD diagonal AC and BD intersect each other at O.
To prove:
AB2 + BC2 + CD2 + AD2 = AC2 + BD2
Proof:The diagonals of a rhombus bisect each other at right angles.

∴ AO = CO, BO = DO
∴ ∠s at O are rt. ∠s
In ∆AOB, ∠AOB = 90°
∴ AB2 = AO2 + BO2 [By Pythagoras Theorem] …………..(1)
Similarly, BC2 = CO2 + BO2 ……………..(2)
CD2 = CO2 + DO2 ……………(3)
and DA2 = DO2 + AO2 ……………….(4)
Adding. (1), (2), (3) and (4), we get
AB2 + BC2 + CD2 + DA2 = 2AO2 + 2CO2 + 2BO2 + 2DO2
= 4AO2 + 4BO2
[∵ AO = CO and BO = DO]
= (2AO)2 + (2BO)2 = AC2 + BD2.

Question 8.
In fig., 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.

Solution:
Given: A ∆ABC in which OD ⊥ BC, 0E ⊥ AC and OF ⊥ AB.

To prove:
(i) AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Construction: Join OB, OC and OA.
Proof: (i) In rt. ∠d ∆AFO, we have
OA2 = OF2 + AF2 [By Pythagoras Theorem]
or AF2 = OA2 – OF2 …………..(1)

In rt. ∠d ∆BDO, we have:
OB2 = BD2+ OD2 [By Pythagoras Theorem]
⇒ BD2 = OB2 – OD2 …………..(2)

In rt. ∠d ∆CEO, we have:
OC2 = CE2 + OE2 [By Pythagoras Theorem]

⇒ CE2 = OC2 – OE2 ……………(3)

∴ AF2 + BD2 + CE2 = OA2 – OF2 + OB2 – OD2 + OC2 – 0E2
[On adding (1), (2) and (3)]
= OA2 + OB2 + OC2 – OD2 – OE2 – OF2
which proves part (1).
Again, AF2 + BD2 + CE2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)
= AE2 + CD2 + BF2
[∵AE2 = AO2 – OE2
CD2 = OC2 – OD2
BF2 = OB2 – OF2].

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Height of window from ground (AB) = 8m.

Length of ladder (AC) = 10 m
Distance between foot of ladder and foot of wall (BC) = ?
In ∆ABC,
AB2 + BC2 = AC2 [By Pythagoras Theorem]
(8)2 + (BC)2 = (10)2
64 + BC2 = 100
BC2 = 100 – 64
BC = √36
BC = 6 cm.
∴ Distance between fóot of ladder and foot of wall = 6 cm.

Question 10.
A guy wire attached to a vertical pole of height 18 m Is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Let AB is height of pole (AB) = 18 m
AC is length of wire = 24 m

Question 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 1½ hours?
Solution:
Speed of first aeroplane = 1000km/hr.

Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the
distance between their tops.
Solution:
Height of pole AB = 11 m
Height of pole (CD) = 6 m

Question 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:
Given: In right angled ∆ABC, ∠C = 90° ;
D and E are points on sides CA & CB respectively.
To prove: AE2 + BD2 = AB2 + DE2
Proof: In rt. ∠d ∆BCA,
AB2 = BC2 + CA2 …………..(1) [By Pythagoras Theorem]

In rt. ∠d ∆ECD,
DE2 = EC2 + DC2 ……………….(2) [By Pythagoras Theorem]
In right angled triangle ∆ACE,
AE2 = AC2 + CE2 ……………….(3)
In right angled triangle ∆BCD
BD2 = BC2 + CD2 ……………….(4)
Adding (3) and (4),
AE2 + BD2 = AC2 + CE2 + BC2 + CD2
= [AC2 + CB2] + [CE2 + DC2]
= AB2 + DE2
[From (1) and (2)]
Hence 2 + BD2 = AB2 + DE2.
Which is the required result.

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

Solution:
Given: ∆ABC, AD ⊥ BC
BD = 3CD.
To prove: 2AB2 = 2AC2 + BC2.

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9 AD2 = 7 AB2.
Solution:
Given: Equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC.
To prove: 9AD2 = 7 AB2.
Construction: AB ⊥ BC.
Proof: ∆AMB ≅ ∆AMC [By R.HS. Rule since AM = AM and AB = AC]

Question 16.
In an equilateral triangle, prove that three times the square of one side Ls equal to four times the square of one of its
altitudes.
Solution:
Given:
ABC is equilateral ∆ in which AB = BC = AC

To prove: 3 AB2 = 4 AD2
Proof: In right angled ∆ABD,
AB2 = AD2 + BD2 (Py. theorem)
AB2 = A BD2 (Py. theorem)

Question 17.
Tick the correct answer and justify: In ∆ABC, AB = 6 cm, AC = 12 cm and BC = 6√3 cm. [The angles of B are respectively
(A) 120°
(B) 64°
(C) 90°
(D) 45°
Solution.
AC = 12 cm
AB = 6√3 cm
BC = 6 cm
AC2 = (12)2 = 144 cm
AB2 + BC2 = (6√3)2 + (6)2
= 108 + 36
AB√3 + BC√3 = 144
∴ AB√3 + BC√3 = AC√3
Hence by converse of pythagoras theorem ∆ABC is right angred triangle right angle at B
∴ ∠B = 90°
∴ correct option is (C).

Ex 6.6

Question 1.
In figure, PS is bisector of ∠QPR of ∆PQR. Prove that = QS/SR=PQ/PR.

 

Question 2.
In the given fig., D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC, DN ⊥ AB, prove that:
(i) DM2 = DN.MC
(ii) DN2 = OMAN.

Solution:
Given: ∆ABC, DM ⊥ BC, DN ⊥ AB
To prove: DM2 = DN . AC
DN2 = DM . AN.
Proof: BD ⊥ AC (Given)
⇒ ∠BDC = 90°
⇒ ∠BDM + ∠MDC = 90°
In ∠DMC, ∠DMC = 90°
[∵ DM ⊥ BC (Given)]
⇒ ∠C + ∠MDC = 90°
From (1) and (2),
∠BDM + ∠MDC = ∠C + ∠MDC
∠BDM =∠C
[Cancelling ∠MDC from both sides]
Now in ∆BMD and ∆MDC,
∠BDM = ∠C [Proved)
∠BMD = ∠DMC [Each 90°]
∆BMD ~ ∆MDC [By AA criterion of similarity]
⇒ DM/BM=MC/DM
[∵ Corresponding sides of similar triangles are proportional]
⇒ DM2 = BM × MC
⇒ DM2 = DN × MC [∵ BM = DN]
Similarly, ∆NDA ~ ∆NBD
⇒ DN/BN=AN/DN
[∵ Corresponding sides of similar triangles are próportional]
⇒ DN2 = BN × AN
⇒ DN2 = DM × AN .
Hence proved.

Question 3.
In fig., ABC is triangle in which ∠ABC > 90° and AD ⊥ BC produced, prove that AC2 = AB2 + BC2 + 2BC.BD.

Solution:
Given: ∠ABC, AD ⊥ BC when produced, ∠ABC > 90°.
To prove : AC2 = AB2 + BC2 + 2BC. BD.
Proof: Let BC = a,
CA = b,
AB = c,
AD = h
and BD = x.
In right-angled ∆ADB,
Using Pythagoras Theorem.
AB2 = BD2 + AD2
i.e., c2 = x2 + h2
Again, in right-angled AADC,
AC2 = CD2 + AD2
i.e.. b2 = (a + x)2 + h2
= a2 + 2ax + x2 + h2
= a2 + 2ax + c2; [Using (1)]
b2 = a2 + b2 + 2w.
Hence, AB2 = BC2 + AC2 + 2BC × CD.

Question 4.
In fig., ABC is a triangle in which ∠ABC < 90°, AD ⊥ BC, prove that AC2 = AB2 + BC2 – 2BC.BD.

Solution:
Given: ∆ ABC, ∠ABC < 90°, AD ⊥ BC.
To prove : AC2 = AB2 + BC2 – 2BC BD.
Proof: ADC is right-angled z at D
AC2 = CD2 + DA2 (Pythagora’s Theorem) ……………..(1)
Also, ADB is right angled ∆ at D
AB2 = AD2 + DB2 ……………….(2)
From (1), we get:
AC2 = AD2 + (CB – BD)2
= AD2 + CB2 + BD2 – 2CB × BD
or AC2 = (BD2 + AD2) + CB2 – 2CB × BD
AC2 = AB2 + BC2 – 2BC × BD. [Using (2)]

Question 5.
In fig., AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

 

Question 6.
Prove that sum of squares of the diagonals of a parallelogram is equal to sum of squares of its sides.
Solution:

Hence, sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Question 7.
In fig., two chords AB and CD intersect each other at the point P prove that:
(i) ∆APC ~ ∆DPB
(ii) AP.PB = CP.DP.

To prove:
(i) ∆APC ~ ∆DPB
(ii) AP.PB = CP.DP.
Proof:
(i) In ∆APC and ∆DPB,
∠1 = ∠2 (Vertically opposite angle)
∠3 = ∠4 (angle on same segment)
∴ ∆APC ~ ∆DPB [AA similarity criterion]

(ii) ∆APC ~ ∆DPB (Proved above)
AP/DP=PC/PB
(If two triangles are sitnilar corresponding sides are proportional)
AP.PB = PC.DP
Which is the required result.

Question 8.
In fig., two chords AB and CD of a circle intersect each other at point P (when produced) outside the circle prove:
(i) ∆PAC ~ ∆PDB
(ii) PA.PB = PC.PD.

Solution:
Given: AB and CD are two chord of circle intersects each other at P (when produced)
To prove:
(i) ∆PAC ~ ∆PDB
(ii) PA.PB = PC.PD.
Proof:
(i) In ∆PAC and ∆PDB,
∠P = ∠P (Common)
∠PAC = ∠PDB.
(Exterior angle of cyclic quadrilqteral is equal to interior opposite angle)
∴ ∆PAC ~ ∆PDB [AA similarity criterion]

(ii) ∆PAB ~ ∆WDB
∴ PA/PD=PC/PB
[If two triangles are similar corresponding sides are proportional]
PA × PB = PC × PD.

 

In ∆ACE, we have:
AE = AC
⇒ ∠3 = ∠4 ………. (∠s opp. to equal sides)
Since CE || DA and AC cuts them, then:
∠2 = ∠4 ……….(alt ∠s)
Also CE || DA and BAE cuts them, then:
∠1 = ∠3 …………(Corr. ∠s)
Thus we have:
∠3 = ∠4
⇒ ∠3 = ∠1
But ∠4 = ∠2
⇒ ∠1 = ∠2.
HenCe AD bisects ∠BAC.

Question 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 the horizontal distance of the fly from her after 12 seconds?

Solution:
A right angled triangle, ABC, in which,
AB = 1.8 cm,
BC = 2.4 cm.
∠B = 90°
By Pythagoras Theorem,
AC2 = AB2 + BC2
AC2 = (1.8)2 + (2.4)2
AC2 = 3.24 + 5.76 =9
AC2 = (3)2
AC = 3 cm
Now, when Nazima pulls in the string at the rate of 5 cm/sec ; then the length of the string decrease = 5 × 12 = 60 cm
= 0.6 m in 12 seconds.
Let after 12 seconds, position of the fly will be at D.
∴ AD = AC – distance covered in 12 seconds
AD = (3 – 0.6) m
AD = 2.4 m
Also, in right angled ∆ABD,
Using Pythagoras Theorem,
AD2 = AB2 + BD2
(2.4)2 = (1.8)2 + BD2
BD2 = 5.76 – 3.24
BD2 = 2.52 m
BD = 1.587 m.
∴ Horizontal distance of the fly from Nazima = BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m
Hence, original length of string and horizontal distance of the fly from Nazima is 3 m and 2.79 m.