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Question-1

Give two different examples of pair of (i) similar figures, (ii) non-similar figures.

Solution:
Examples of similar figures:
i) Two square of sides 4 cm and 8 cm each.
ii) Two circles with the same centre and diameters 7.5 cm and 10 cm each.


Examples of non similar figures:
i) A square and a rhombus.
ii) A circle and a triangle.

Question-2

State whether the following quadrilaterals are similar or not
   

Solution:
The figures are not similar as they are not equiangular.

Question-3

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

Solution:
(i)

 

In ΔABC
DE is parallel to BC
By Basic Proportionality Theorem
------------------- (1)
Given: AD = 1.5 cm, DB = 3 cm, AE = 1 cm
Let EC = ‘x’ cm
Applying in (1)

1.5x = 3
x =
x = 2 cm
EC = 2cm


(ii)Given: DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm

Since DE || BC, using BPT

…………………………. (1) 

 Let AD be = x

sub. in (1)

x =
=
AD = 2.4 cm


Question-4

E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

 

 


Solution:

 
(i) PE = 3.9 cm, EQ = 3 cm
PF = 3.6 cm, FR = 2.4 cm
=
=


EF is not parallel to QR by convene of BDT
 

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


EF || QR by convene of BPT.

(iii) PQ = 1.28 cm ,  PR = 2.56 cm,  PE = 0.18 cm,  PF = 0.36 cm
EQ = PQ – PE
     = 1.28 – 0.18
     = 1.10
FR = PR – PF
     = 2.56 – 0.36
     = 2.20



EF is  parallel to QR

Question-5

In the figure, if LM || CB and LN || CD, prove that


Solution:

Given: LM || CB and LN || CD
To prove:
Proof:
In Δ ABC
LM || CB using basic proportionality Theorem
………………………….. (1)
Also in Δ ADC
LN || CD  
using basic proportionality 
  Theorem …………….. (2)

from (1) and (2)

Hence proved .

Question-6

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

Solution:

Given: ABC is a triangle and DE || AC and DF is parallel to AE
To prove:
Proof
In Δ ABC,
DE || AC (given)
(By BPT) …………………….. (1)

In Δ AEB,
DF || AE
(By BPT) ………………………. (2)
Comparing equation (1) and equation (2)

Hence proved.

Question-7

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

Solution:
    
Given: DE || OQ and DF || OR
To prove: EF || QR
Proof: 
In Δ POQ
DE || OQ (given)
By using BPT
………………………… (1)
In Δ POR
………………………….(2)
By comparing equations (1) and (2)

By using inverse of BPT
EF || QR
Hence proved.

Question-8

A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.
Show that BC || QR
.

Solution:


Given: A, B and C are points on OP, OQ and OR respectively such that AB || PQ, AC || PR
Proof: In Δ OPQ
AB || PQ (Given)
(By using BPT) ……………………………. (1)
In Δ OPR
Since AC || PR
(By using BPT) ………………….. (2)

By comparing (1) and (2)

 BC || QRBy using converse of BPT, 
Hence proved.

Question-9

Using Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Solution:


Given: ABCD is a trapezium and the diagonals AC and BD intersect at O.

To prove: The ratio
Construction : Draw OM || AB meeting BC at ‘M’
Proof: In Δ ACB, OM || AB
By using BPT ………………………………………. (1)
||ly In Δ BDC
OM || CD [ (OM || AB AND AB || CD OM || CD)]
using BPT
Taking the reciprocal
…………………………….(2)
from (1) and (2)

(or)
Hence proved.

Question-10

The diagonals of a quadrilateral ABCD intersect each other at the point O such that . Show that ABCD is a trapezium.

Solution:


Given: ABCD is a quadrilateral and the diagonals AC and BD intersect of ‘O’ such that

To prove: ABCD is a trapezium
Construction: Draw OM || AB
Proof:
In Δ ACB

OM || AB (By construction)
………………………. (1)
Given:
(or)
sub = in equation (1)
=
or =
In Δ BDC by converse of BPT
OM || DC. But OM || AB
AB || CD
ABCD is a trapezium.

Question-11

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) Similar
AAA Δ ABC Δ PQR
(ii) Similar
SSS, Δ ABC Δ QRP
(iii) Not similar
(iv) Not smilar
(v) Similar, 
SAS Δ ABC  Δ FGE
(vi)similar, AAA Δ DEF  Δ PQR

Question-12

ΔODC ~ Δ OBA, BOC = 125° and CDO = 70°. Find DOC, DCO and OAB.

Solution:

Given: Δ ODC Δ OBA
BOC= 125°
CDO = 70°
To find DOC, DCO and OAB
DOC= 180 – 125 = 55
DCO = 125 – 70 = 55
OAB  = 55 ( Q Δ ODC Δ OBA).

Question-13

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 
Solution:


Given: ABCD is a trapezium with AB || CD and the diagonals AC and BD intersect at ‘O’.
To prove: =
Proof:  In the figure consider the triangle OAB and OCD  
DOC  = AOB        (Vertically opposite angles are equal)
DCO  = OAB           (Alternate angles are equal)

By AA corollary of similar triangles.
Δ OAB Δ OCD When the two triangle are similar, the side are proportionally.
=
Hence proved.

Question-14

In the figure, = and 1 = 2. Show that Δ PQS ~ Δ TQR.
 

Solution:
Given: =
and
1 = 2
To prove: Δ PQS Δ TQR
Proof
In Δ PQR

1 = 2
PQR  = PRQ          
 
=>
PQ = PR
=
=
= ( QP = PR proved above )

1 is a common angle
By SAS similarity

D PQS ~ D TQR
Hence proved.

Question-15

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

Solution:

Given: S and T are the points on sides. PR and QR of a. Δ PQR such that P = RTS
To prove: Δ RPQ Δ RTS
Proof:

In Δ PQR and Δ TRS
QRP = TRS (common angle)
STR = QPR (given)
By AA similarity
Δ RPQ Δ RTS
Hence proved.

Question-16

In the figure, if ΔABE Δ ACD, show that Δ ADE ~ Δ ABC.

Solution:
Given: Δ ABE Δ ACD
To prove
: Δ ADE ~ Δ ABC
Prove: Δ ABE Δ ACD
AB = AC …………………… (1) (Since corresponding parts of triangles are equal)
and AD = AE …………………… (2)

and

A is a common angle
DAE = BAC
By SAS similarity condition
Δ ADE Δ ABC.

Question-17

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

Solution:



Given: Δ ABC in which the altitudes AD and CE intersect at P.
To prove:
(i) Δ AEP Δ CDP (ii) Δ ABD Δ CBE
(iii) Δ AEP Δ ADB (iv) Δ PDC Δ BEC
Proof:
(i) Δ AEP and Δ CDP
APE = CPD    (Vertically opposite angle and equal)
 
AEP = CDP   (Both are equal to 90° since CE and AD are altitudes)
BY AA similarity
Δ AEP Δ CDP
Hence proved.

(ii) To prove: Δ ABD Δ CBE

Proof: In Δ ABD and Δ CBE
CBE = ABD            (Common angle)
ADB = CEB            (Both are equal to 90°)
By AA similarity
Δ ABD  
D CBE

(iii) To prove: Δ AEP ADB
Proof: In Δ AEP and Δ ADB
PAE = DAB               (Common angle)
 
 AEP =  ADB              (Both are equal to 90°
By AA similarity
Δ AEP Δ ADB

(iv) To prove: Δ PDC Δ BEC
Proof: In Δ PDC and Δ BEC
 
PCD = ECB              (Common angle)
PDC = BEC               (Both are equal to 90°
Δ PDC  D BEC (AA similarity).

Question-18

In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) Δ ABC ~ Δ AMP
(ii)



Solution:

(i) To prove: Δ ABC Δ AMP
In Δ ABC and Δ AMP

  ABC =  AMP (Both are equal to 90°)
  BAC =  PAM(common angle)
BY AA corollary                   
Δ ABC AMP,  

(ii) To prove: 
Proof:
As proved above
Δ ABC Δ AMP
Since the corresponding sides of the similar triangles are proportional

 
Hence proved.

Question-19

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

(ii) Δ DCB ~ Δ HGE
(iii) Δ DCA ~ Δ HGF


Solution:

 
Given: Δ ABC Δ FEG and CD and GH are bisectors of the ACB and EGF of the Δ ACB and Δ EGF respectively.
To prove:
(i)
(ii) Δ DCB Δ HGE
(iii) Δ DCA Δ HGF
Proof: (i) Consider Δ ADC and Δ FGH
BAC =  EFG (Q D ABC ~ D FEG) is given)
(or)

DAC = HFG
Also
 
BCA = EGF (Q D ABC ~ D FEG is given)
× by
BCA = EGF
 =>
DCA = HGF
By AA corollary
Δ DAC Δ HFG



(ii) To prove: Δ DCB Δ HGE
In Δ DBC and Δ HEG
ABC FEG [QΔ ABC Δ FEG]
(or)
DBC = HEG [ Q D and H lie on the line AB and EF respectively]
BCA = EGF …………………..(1) [Q Δ ABC Δ FEG]
multiplying by
BCA EGF
(or)
DCB =  HGE …………………………..(2) [Q CD and GH are the bisectors of the ACB and EGF of the D ACB and D EGF respectively]
 
from (1) and (2)
Δ DCB Δ HGE [By AA corollary]

(iii) Already proved in (i)
In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC and EF perpendicular to AC prove that D ABD ~ D ECF.

Given: D ABC is isosceles with AB = AC. E is a point as CB such that EF perpendicular to AC. AD is perpendicular to BC.
To prove: D ABD ~ D ECF
Proof:

D ABC is isosceles with AB = AC (Given)
\
ABC = ACB ………………. (1) Property of an isosceles triangle
and
ADB = EFC ……………………(2) Both are equal to 90°
Now, in D ABD and D ECF
ABD = ECF From equation (1)
and

ADB = EFC From equation (2)
\ D ABD ~ D ECF.



 

Question-20

In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC and EF perpendicular to AC. Prove that  Δ ABD ~ Δ ECF.

Solution:

Given: ABC is an isosceles triangle, 
⇒ AB = AC
⇒  B =  C
⇒  ABD =  ECF
In Δ ABD and Δ ECF
  ADB =  EFC (Each 90°)
 ABD =  ECF  (Proved above)
 BAD =  CEF
∴  Δ ABD ~ Δ ECF. (By using AA similarity criterion)

Question-21

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:

  
In Δ ABC and PQR
ACB = PRQ      ( The shadow is cast of the since from the angle made by the sun’s ray is same )
 ABC = PQR     ( ∵ both are equal to 90°)
Δ ABC Δ PQR (By AA corollary)
=

or x =  = 42
Height of the tower = 42 m.

Question-22

If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that

Solution:

Given: Δ ABC Δ PQR
To prove:
Proof:
In Δ ABC Δ PQR
= ………………………………(1)
and
B = Q ……………………………..(2)
replacing BC = 2BD and QR = 2QM in equation (1)
……………………………..(3)
from (2) and (3)
Δ ABD Δ PQM By SAS similarly

Hence proved.

Question-23

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

Solution:
Since Δ ABC Δ DEF
=  (By Theorem 6.6)

(or) =
=
BC = × 15.4
BC = 11.2 cm.

Question-24

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:

In Δ AOB and COD
COD = AOB  (vertically opposite angles)
CDO = OBA (Alternate interior angles)
By AA similarity
Δ COD Δ AOB

 ==
=

Question-25

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

Solution:


Construction: From A and D draw perpendiculars to BC meeting BC at E and F respectively.
Proof:
In Δ AEO and Δ DFO
AEO = DFO = 90°
AOE = DOF   (vertically opposite angle)
Δ AOE Δ DOF (By AA similarity)
= = ………………………. (1)
=
Hence proved
.

Question-26

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

Solution:
Let ABC and DEF be two similar triangles where areas are equal.
(ie) area(Δ ABC) = area(Δ DEF)

= = =

since ar. Δ ABC = ar.Δ DEF

= = 1

= 1 or AB2 = DE2

AB = DF

||ly BC = EF and AC = DF

By SSS congruency

Δ ABC Δ DEF
Here the areas of two similar triangles are equal, those two triangles are congruent.

 

Question-27

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

Solution:

Since D and F are the midpoints of the AB and AC, DE || AF or DF || BE
Similarly EF || AB or EF || DB
AFED is a parallelogram as both pairs of opposite sides are parallel.
By the property of parallelogram

DBE = DFE = (In an parallelogram opposite angle are equal)
or DFE = ∠ABC   -------------------(1)
Similarly FEB = ∠ACB   …………………………..(2)
In Δ DEF and Δ ABC, from
Equation (1) and (2)
Δ DEF Δ CAB
= =
ar Δ DEF: ar Δ CAB = 1 : 4 .

Question-28

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

Solution:

Given: AM and DN are medians of the Δ ABC and Δ DEF and Δ ABC Δ DEF
To prove:

=
Proof: since Δ ABC Δ DEF

 
or
and

( QΔ ABC Δ DEF)

By SAS similarity
Δ ABM Δ DEN
or ………………………... (1)

 =
              = (From (1)
Hence proved.

Question-29

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

Solution:


Given: ABCD is a square and two equilateral triangle are drawn with BC and DB as are units side respectively.
To Prove: ar Δ PBC =
(ar ΔBDQ)
Proof:
Consider ΔPBC and Δ BQD
Since both these triangles are equilateral triangles, by the property of equilateral triangles
,
both the triangles are equiangular with each angle equal to 60°.
⇒  D PBC  D QBD
∴  = ………………………… (1)
Consider D BCP
Since it is a right angled triangle
BC2 + DC2 = BD2 ( By using Pythagoras theorem)
But BC = DC
2BC2 = BD2
Substituting in (1)
=
 ar D PBC = ar D BDQ
Hence the results.

Question-30

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) 72 = 49, 242 = 576, 252 = 625
576 + 49 = 625
Since the sum of the squares of the two sides of the
triangle is equal to the square of the third side, it is a right triangle.

(ii) 32 = 9, 82 = 64, 62 = 36
from the above 36 + 9 64
It is not right triangle.

(iii) 502 = 2500, 802 = 6400, 1002 = 10000
since 2500 + 6400 = 8900
8900 10000
It is not a right triangle. 

(iv) 132 = 169, 122 = 144, 52 = 25, 144 + 25 = 169
It is a right triangle.

Question-31

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

Solution:
 

To prove: PM2 = QM × MR

In Δ PMR

PR2 = PM2 + MR2 ………………….(1)
Also In Δ PMQ
PQ2 = PM2 + QM2 …………….. (2)
(1) + (2)
PR2 + PQ2 = 2PM2 + MR2 + QM2
QR2 = 2PM2 + MR2 + QM2
QR2 – MR2 = QM2 + 2PM2
(QR + MR) (QR - MR) – QM2 = 2PM2
(QR + MR) QM – QM2 =
2PM2  
QM( QR + MR – QM) = 2PM2
QM (MR +MR) = 2PM2
QM × 2MR = 2PM2
PM2 = QM × MR


Question-32

In the figure, ABD is a triangle right angled at A and AC perpendicular to BD. Show that
(i) AB
2 = BC . BD
(ii) AC
2 = BC . DC
(iii) AD
2 = BD . CD


Solution:
(i) To prove: AB2 = BC. BD
In Δ ABC and
Δ ABD
ABC = DBA  (common angle)
ACB = BAD  (equal to 90°
Δ ABC Δ DBA (By AA similarity)
 
⇒ AB2 = BC × BD

(ii) In
 Δ ADC and Δ ABC
 AD2 = AC2 + CD2 ..........(1)
AB2 = BC2 + AC2 ............(2)
(1) + (2) 
AD2 + AB2 = 2AC2 +CD2 + BC2
BD2 = 2AC2 + CD2 +BC2
BD2 - CD2 - BC2 = 2AC2
(BD + CD)(BD – CD) – BC2 = 2AC2

(BD + CD) BC – BC2 = 2AC2
(BC)(BD + CD – BC) = 2AC2
(BC)(CD + CD) = 2AC2    (
 BD – BC = CD)
 AC2 = BC×CD

(iii) In Δ ACD and D ABD
ADB = ADC (common angle)
BAD = DAC  (equal to 90°)
Δ ABD Δ ACD


 
⇒ AD2 = BD × DC

Question-33

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

Solution:


In Δ ABC since ACB= 90°
AC2 + BC2 = AB2 (By using Pythagoras theorem)
AC2 + AC2 = AB2 (
Q AC = BC )
AB2 = 2AC2 (proved)


Question-34

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

Solution:

Given: AB2 = 2AC2 and AC = BC
To prove: ABC is a right triangle
Proof:
Given: AB2 = 2AC2
(or) AB2 = AC2 +AC2 ………………(1)
since AC = BC given, equation (1) becomes
AB2 = AC2 + BC2
By the converse of Pythagoras theorem Δ ABC is a right triangle with
C = 90°.

Question-35

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

Solution:

Given: ABC is an equilateral triangle with each side equal to 2a
Construction: Draw AD r to BC
To find: Length of AD
In Δ ADB and
D ADC
ADB = ADC = 90°
AC = AB       (Q ABC is isosceles)
AD = AD       (common side)
Δ ADB Δ ADC (By RHS property)
BD = DC (CPCT)
BC = BD + DC
     = BD + BD
2a = 2BD
BD = a
 In Δ ABD, AB2 = AD2 + BD2
(2a)2 = AD2 + a2
4a2 = AD2 + a2
a2 – a2 = AB2
3a2 = AD2
AD = a
Similarly the other altitudes will also be
a only.

Question-36

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:

Given: ABCD is a rhombus and the diagonals AC and BD intersect at O.
To prove: AB2 + BC2 + DC2 + AD2 = AC2 + BD2
Proof: The diagonals of rhombus intersect at right angle.
∠AOB = 90°
In Δ AOB, AO2 + OB2 = AB2
+ = AB2
Hence AC2 + BD2 = 4AB2
Since 4AB2 can be taken as sum of the square of the 4 sides of the rhombus as each side of the rhombus equal, 

The sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Question-37

In the figure O is a point in the interior of a triangle ABC, OD perpendicular to BC, OE perpendicular to AC and OF perpendicular to AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2,

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

Solution:


(i) Given : In Δ ABC OD BC, OE AC and OF AB
Construction: Join OC, OA and OB.
In Δ OBD, OB2 = OD2 + BD2
In Δ OFA, OA2 = OF2 + AF2
In Δ OCE, OC2 = OE2 + CE2
OB2 + OA2 + OC2 = OD2 + OF2 + OE2 + BD2 + AF2 + CE2
(or)

OB2 + OA2 + OC2 – OD2 – OF2 – OE2 = BD2 + AF2 + CF2

(ii) From the result above
 AF2 + BD2 + CF2 = OB2 + OA2
OC2 OD2 – OF2 – OE2
                        = (OA2 – OE2) + (OC2 – OD2) +(OB2 – OF2)
                        = AE2 + CD2 + BF2

Question-38

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:

In the figure AC = ladder = 10 m
AB = Height of the window = 8 m
BC = Distance from the foot of the wall to the foot of the ladder = x m
ABC is a right angle triangle
AB2 + BC2 = AC2
(or) 82 + x2 = 102
x2 + 102 - 82
x2 = (10 + 8) (10 – 8)
    = (18)(2)
    = 36 x = = 6 m
The distance of the foot of the ladder from the wall = 6 m

Question-39

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:

In Δ ACB, AC = length of the wire = 24m and the largest of the tower = 18 m
182 + x2 = 242 (By using Pythagoras theorem)
x2 + 242 = 182
x2 = 576 – 324
    = 252
x =
   = 6m
Therefore, the stake should be driven to a distance of 6m so that the wire will be taut.

Question-40

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 1hours?

Solution:

Distance traveled by the aeroplane in the North directions is
AB = 1000 ´  
  = 1500 km. (Distance = speed × time)
And the Distance traveled by the aeroplane in the west direction is AC = 1200 × = 1800 km
BC is the distance between the two
 aeroplanes  
 AB2 + AC2 = BC2
(1500)2 + (1800)2 = BC2
2250000 + 3240000 = BC2
5490000 = BC2
BC = 100km.

Question-41

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:

Let AD be a pole of height 11m and BC be a pole of height 6m
Construction: Join CD and draw r from C to AD meeting AD at E.
Proof: ABCD is rectangle
AB = CE = 12 m
and BC = AE = 6 m
DE = AD – AE = 11 – 6 = 5 m
In Δ DEC
DE2 + EC2 = DC2
or 52 + 122 = DC2
25 + 144 = DC2
DC = = 13 m

 The distance between the two poles = 13 m.

Question-42

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:

To prove: AE2 + BD2 = AB2 + DE2
Proof:
In Δ AEC
AE2 = EC2 + AC2 ………………………….. (1)
In Δ BDC
BD2 = BC2 + CD2 ………………………….. (2)
(1) + (2)
AE2 + BD2 = EC2 + BC2 + AC2 + CD2
               = (EC2 + CD2) + (AC2 + BC2)
               = DE2 + AB2
Hence proved.

Question-43

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

Solution:


From ΔABD,
AD2 = AB2 
 BD2  ...................(1)

From
ΔADC,
AD2 = AC2  DC2  ......................(2)

From (1) and (2), AB2  BD2 = AC2  DC2 .................(3)

BD = 3 DC (given)
BC = BD + DC = 4 DC
BC2 = 16 DC2......................(4)
Now AB2 = AC2 + BD2 
 DC2
AB2 = AC2 + (3DC)2  DC2
AB2 = AC2 + 9DC2  DC2
       = AC2 + 8DC2
AB2 = AC2 +  
      = AC2 + BC2/2  [ from (4) ] 
2AB2 = 2AC2 + BC2

 

Question-44

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

Solution:

Given: Δ ABC is an equilateral
triangle. ‘D’ is a point on BC such that BD = BC.
To prove: 9AD2 = 7AB2
Construction: Draw AE r to BC meeting BC at E
Proof:

In an equilateral triangle the altitude and median are same line regret. Therefore AE is the median of the Δ ABC. E is the midpoint of BC
⇒ BE + EC = BC
    2EC = BC
BE = EC = BC ………………………….. (1)
BD = BC (Given)……………………(2)
DC = BC – BD
     = BC - BC =  BC ………………… (3)
DE = BE – BD
     = BC -
     = BC
     = = BC
In Δ ADE
Δ ADE
AD2 = AE2 + ED2 …………………(1)
But in Δ AEC
AC2 = AE2 + EC2
(or) AE2 = AC2 - EC2
substituting in (1)
AD2 = AC2 – EC2 + ED2
      = AC2 – DE2
– CE2
      = AC2 + (DC + CE)(DE – CE)
      = AC2 + (DC)(DE – CE)
      = AC2 + DC.DE – DC.CE
      = AC2 +
      = AC2 +
      = AC2 +
      = AC2 +
      = AC2 
 
      =
AD2 = AB2


∴ 9 AD2 = 7 AB2.

Question-45

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:

Let the side of the equilateral Δ ABC be ‘a’
Let AE be the altitude from A to BC meeting
BC at E.
E is the midpoint of C
Therefore BE = BC = a
In Δ ABE, AB2 = AE2 + BE2
AE2 = AB2 – BE2
      = a2 -
      = a2 -
      = =
AE = ………………………….. (1)
Now,
Three times, square of one of its side = 3a2
Four times the square of y
Its one of the altitude = 4 = 3a2
Therefore it is proved that in an equilateral triangle three times the square of one side is equal to four times the square of its altitude.

Question-46

Tick the correct answer and justify: In Δ ABC, AB = 63 cm, AC = 12 cm and BC = 6 cm. The angle B is:
(A) 120° (B) 60°
(C) 90° (D) 45°

Solution:
In Δ ABC,  AB = 6cm,  AC = 12 cm, BC = 6 cm

AB2 = (6)2 = 36 × 3 = 108
BC2 = 62 = 36
AC2 = 122 = 144
AB2 + BC2 = 108 + 36 = 144 = AC2 
∴ AB2 + BC2 = AC2  
 By converse of pythagoras theorem,  ∠ B = 90°    
Answer = (c) 90°.




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