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Triangles Exercise 6.3

Hi Friends and Champs

Before Proceeding to next exercise, pls read and understand the following theorems.


Theorem 6.3 : If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. 

This criteria is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.

( pls see proof from book )


Fig 6.3.1

In figure 6.3.1 , If 

<A = <D, <B = <E , <C = <F , then




Theorem 6.4 : If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.

This criteria is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.

( pls see proof from book )

Reverse of theorem is also true i.e. If :

 then  <A = <D, <B = <E , <C = <F  is also true

Theorem 6.5 : If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. 

 This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two triangles.

( pls see proof from book )


                                                                        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 

Fig 6.34

Solution:

(i) Similar  by  AAA

(ii) Similar , by SSS

(iii) Not similar

(iv) Similar , by SAS

(v) Not similar

(vi) Similar ,  by AA


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

Fig 6.35
Solution:

< BOC = 1250 ( Given )
< DOC = 1800 - < COB ( By linear pair angle  property )
=> < DOC = 1800 - 1250
                  = 550
and so , < DCO = 1800 - ( < CDO + < DOC )     ( By angle sum property of triangle )
                          = 1800 - ( 700 + 550 )
                          = 1800 - 1250
                          =550
Since, Δ ODC ~ Δ OBA 
           it's implies that:
                                    < DAB = < DCO = 550 ( By AAA criterion of similarity)

 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:
                                                                                                          
Let ABCD is a trapezium with AB || DC.

In Δ AOB and ΔDOC

< ABO = < CDO ( alternate angle )
< BAO = < DCO ( alternate angle )
and
< AOB = < DOC ( vertically opposite angle )

=> Δ AOB ~ ΔDOC ( AAA criterion similarity )

so, OA / OC = OB / OD ( hence proved )


4. In Fig. 6.36, QR/QS =  QT/PR and <1 = <2. Show and that Δ PQS ~ Δ TQR.
Fig 6.36

Solution:  

In  Δ PQR , <1 = (2 ( given )
=>  Δ PQR is an isosceles triangle,
       so PQ = PR ----------(1)

In  Δ PQS and in  Δ TQR
QR/QS = QT / PR                             ( given )
=> QR / QS = QT / PQ                     ( from eq (1) PQ = PR )
=> PS || TR                                         ( theorem 6.2 If a line divides any two sides of a triangle                                                                          in the same ratio, then the line is parallel to the third side )
=> < QPS = < QTR and
      < QSP = < QRT
so Δ PQS ~ Δ TQR                               ( from AAA criterion of similar traiangles )
Hence proved

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

Solution: 
                    < P = < RTS


In ΔRPQ and ΔRTS 

<P = < RTS ( given )
< R = < R ( common angle of both triangle )

then < S = < Q ( by angle sum property , if 2 corresponding angles of triangles  are equal then 3rd will                             also be  equal )  

Hence  Δ RPQ ~ Δ RTS   (  AAA criterion of similar triangles )



 6. In Fig. 6.37, if Δ ABE ~ ΔACD, show that Δ ADE ~ Δ ABC.
Fig 6.37
Solution :

                Since, Δ ABE ~ ΔACD  ( given )
                        => AD/AE = AB/AC
                        => DE || BC             ( theorem 6.2 If a line divides any two sides of a triangle                                                                          in the same ratio, then the line is parallel to the third side )
            Since DE || BC
                    => <D = <B and <E = <C
so Δ ABC ~ Δ ADE        ( AAA criterion of similar triangles )
Hence Proved
                
 
7. In Fig. 6.38, 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:
(i) Δ AEP ~ Δ CDP 
                                 < D = ,<E ( right angle )
                                 <CPD = < APE
                               => Δ AEP ~ Δ CDP  ( AA criterion of similar triangle )

(ii) Δ ABD ~ Δ CBE 
                                    < D = ,E ( right angle )
                                    <B = <B ( equal angle )
                                => Δ ABD ~ Δ CBE  ( AA criterion of similar triangle )

(iii) Δ AEP ~ Δ ADB 
                                    <A = <A ( equal angle )
                                    <E =  <D ( right angle ) 
                                => Δ AEP ~ Δ ADB  ( AA criterion of similar triangle )

(iv) Δ PDC ~ Δ BEC
                                    < D = ,E ( right angle )
                                    <DCP = < BCE ( equal angle )
                                  => Δ PDC ~ Δ BEC  ( AA criterion of similar triangle )

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



Solution: In Δ ABE and Δ CFB 
                            < A = <C           ( opposite angles are equal of a parallelogram) 
                            < E = < CBE     ( alternate angles  of parallel sides , since AE || BC)
                            <ABE = < CFB ( alternate angle )

                            => Δ ABE ~ Δ CFB   ( by AAA criterion of similar triangles )
                                   Hence Proved
                            



 9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: 
 (i)  Δ ABC ~ Δ AMP 
 (ii) CA / PA  = BC / MP


Solution:
(i)  Δ ABC ~ Δ AMP 
                                     < M = < B              (right angle )
                                     < A = <A                 ( equal angle )
                                     < P = <C                 ( equal from angle sum property )
                                    => Δ ABC ~ Δ AMP    ( by AAA criterion or similar angle )
 (ii) CA / PA  = BC / MP
                                    From (i) Δ ABC ~ Δ AMP
                                                => BC / MP = CA / PA 

 10. 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) CD/GH = AC /FG 
 (ii) Δ DCB ~ Δ HGE 
 (iii) Δ DCA ~ Δ HGF

Solution:

<ACD = < BCD    ( since CD is a bisector, given) and

<FGH = < EGH    ( since GH is a bisector, given)

(i) CD/GH = AC /FG 
                                    Since Δ ABC ~ Δ FEG ( given )
                                             <A = <F
                                             <C = <G and
                                               <B = <E
                                  since CD and GH are bisector of <C and <G respectively , so
                                            < ACD = <FGH
                                           => <ADC = < FHG ( from angle sum property )
                                           => Δ DCA ~ Δ HGF
                                           => CD/GH = AC /FG ( corresponding sides of similar triangle )

 (ii) Δ DCB ~ Δ HGE 
                                      Since Δ ABC ~ Δ FEG ( given )
                                             => <B = <E
                                             => <C = <G and
                                             =>  <B = <E
                                      Since CD and GH are bisector of <C and <G respectively , so
                                            < BCD = <EGH
                                           => <BDC = < EHG ( from angle sum property )
                                           => Δ DCB ~ Δ HGE
                                           => CD/GH = AC /FG ( corresponding sides of similar triangle )
 (iii) Δ DCA ~ Δ HGF
                                        Since Δ ABC ~ Δ FEG ( given )
                                             <A = <F
                                             <C = <G and
                                               <B = <E
                                        Since CD and GH are bisector of <C and <G respectively , so
                                            < ACD = <FGH
                                           => <ADC = < FHG ( from angle sum property )
                                           => Δ DCA ~ Δ HGF

                                

11. In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF.
Solution:
                In  Δ ABC AB = AC  ( given  isosceles triangle )
                    and AD is the bisector of < A
                       => <BAD = <CAD   ( since AD is a bisector )
                       and <BDA = <CDA ( right angle )
                       and < B = <C ( since ABC is a isosceles triangle )
                       Hence Δ ABD ~ Δ ADC  --------------------- (1)
                       
                In Δ ECFand Δ ADC 
                              < C = <C                 ( equal angle )
                              <EFC = < ADC        ( right angle )
                               <FEC = < CAD        ( angle sum property )
                    Hence  Δ ECF ~ Δ ADC -------------------------(2)

                From (1) and (2)
                                Δ ABD ~ Δ ADC ~ Δ ECF 
               Hence Δ ABD ~ Δ ECF     (proved)


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

Solution: 

AB/PQ = BC/QR = AD / PM    --------------------- (1)  ( given )

since AD and PM are the medians of side BC and QM respectively , therefore

BD = 1/2 BC  and QM = 1/2 QR

From eq (1)

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

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

=> Δ ABD ~ ΔPQM

=> < B = <Q     --------------------- (2)   ( since Δ ABD ~ ΔPQM already proved )

In Δ ABC and Δ PQR

<B = <Q     from eq (2) , already proved

and AB/PQ = BC/QR                   from eq (1) ( given )

so from the theorem if one angle of triangles are equal and corresponding sides are in same proportion then triangles will be similar, therefore:

Δ ABC ~ Δ PQR  ( hence proved )



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

Solution:


In Δ ABC and Δ ADC 
< BAC = < ADC     ( given )

and <C = <C    ( equal angle )

=> <CAD = <ABC (  angle sum property of triangle )

=> Δ ABC ~ Δ ADC   ( AAA criterion of triangle similarity )

lets draw above figure in parts to understand ( cut the part Δ ADC  and rotate , we can easily see the proportionate sides )



=> CA / CD = CB / CA  

=> CA2 = CB . CD
Hence proved

 14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that Δ ABC ~ Δ PQR.

Solution:



AB / PQ = AC / PR  = AD /PM ------------------------ (1)  (given)

Since AD and PM are median , therefore 

BD = DC  and

QM = MR 

now produce AD to E such that AD = DE and produce PM to N such that PM = MN and join CE and RN.




In Δ ABD and Δ CDE 

BD = DC  ( given, since AD is a median )
<D = <D   ( vertically opposit angle )
AD = DE  ( by construction )

=> Δ ABD ≅ Δ CDE   ( congruence by SAS ) 
=> AB = CE   ----------(2)  ( by CPCT )

Similary in In Δ PQM and Δ MRN 

We can prove that Δ PQM ≅ Δ MRN ( congruence by SAS ) and so,
 PQ = RN ---------------(3) 

From eq (1)
AB / PQ = AC / PR  = AD /PM

CE / RN = AC / PR = AD/PM

or  CE / RN = AC / PR = 2AD/2PM             ( multiply and divide by 2 )

or  CE / RN = AC / PR = AE/PN

=> Δ ACE ~ Δ PRN ( By correspondence sides criterion of triangle similarity),
=> < EAC = < RPN ------------(4)

similarly we can prove 
< BAD = <QPM ---------------(5)

from eq (4) and eq (5 ) 
<BAC = <QPR

In Δ ABC and Δ PQR
<BAC = <QPR    ( proved above )
and AB / PQ = AC / PR  ( given )

then Δ ABC ~ Δ PQR  (one equal angle and corresponding sides in proportion criterion)
Hence 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:


Since both shadows are at the same time and Pole and tower are vertical to the earth , the resultant triangle 3 -d shape will be similar ,
Let x meter is the height of tower then criterion of similar triangle,

AB / PQ  = BC / QR

=> 6 / x = 4 / 28
=> x = (6 x 28 ) / 4
=> x = 42 m 
 


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

Solution:

Δ ABC ~ Δ PQR ( given )

so AB / PQ = AC/PR = BC/QR    ----------------- (1)

BD = DC and QM = MR ----------------------(2) ( since AD and PM are medians )

from eq(1)
AB/PQ = BC/QR
=> AB/PQ = 2BD / 2QM     ( from eq (2) )
=> AB/PQ =BD/QM
and <B = <Q             ( since Δ ABC ~ Δ PQR given )
hence Δ ABD ~ Δ PQM  ( criterion of similarity due to one equal angle and same proportion
of corresponding sides )

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




Thanks for reading , kindly comment for your suggestions. 🙏🙏🙏

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Location: Faridabad, Haryana, India

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Statistics Exercise -13.2

Hi Friends and Champs! We will talk about the mode of some supplied grouped data today. Remember that you have already learnt the mode for ungrouped data from the last class. As you are previously aware, the mode for ungrouped data is the frequency that is repeated more frequently. As an illustration: Example - Run score by a batsman in 8 matches are 50,98,99,98,76,75,99,98.Find the mode ? Solution : 98 runs repeated 3 times , Hence mode is 98. It is evident from the example above that figure 98 is the mode because it is repeated three times.Thus, it is easy to find the mode for ungrouped data. However, it is a little more complicated with aggregated data. For ungrouped data, we select the class with the highest frequency and use the formula below to determine the mode. \[Mode=l + \frac{f1-f0} {2f1-f0-f2}×h \] where l = lower limit of the modal class, h = size of the class interval (assuming all class sizes to be equal), f1 = frequency of the modal class, f0 = frequency of t...
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