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Question

Consider two triangular parks ABC and EFG such that a path from one corner of the park bisects the other side of the park as shown in the figure. If the dimensions of the park ABC are AB = 4 m, BC = 6 m and AC = 3 m and that of park EFG are EF = 2 m, EG = 3 m  and FG = 1.5 m.

Which of the following statements are true?


  1. ΔADCΔFHG
     

  2. CDGH=ABFE

  3. ΔCDBGHE

  4. CD = GH


Solution

The correct options are
A

ΔADCΔFHG
 


B

CDGH=ABFE


C

ΔCDBGHE


In ΔABC and ΔEFG
AB = 4 m
BC = 6 m
AC = 3 m
EF = 2 m
FG = 1.5 m
and EG = 3 m

CD and GH are medians to sides AB and EF, respectively.
Now in ΔABC and ΔEFG,
ABBC=EFEG
and BCAC=EGFG
ΔABCΔFEG(i)      [by SSS similarity]

A=F,B=E & C=G(ii)

Now, ABFE=ACFG=BCEG2AD2FH=ACFG=BCEGADFH=ACFG=BCEG(iii)   

(a) In ΔADC and ΔFHG, we have
ADFH=ACFG
and A=F  (from Eq. (iii))
ΔADCΔFHG      [by SAS similarity]

(b) We have,ΔADCΔFHGDCHG=ADFHCDGH=2AD2FHCDGH=ABFE[AB=2AD,FE=2FH]

(c) Now, ABFE=ACFG=BCEG(iv) [from Eq. (i)]
Also, CDGH=ABFE
CDGH=BCEG   [from Eq. (iv)]
ΔCDBΔGHE        [by SSS similarity]

 

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