+0

# Triangle

0
386
6
+2493

excuse me for writings

Solveit  Jan 13, 2016

#5
+82801
+15

Note that we have similar triangles, such that :

BDA ≈ BAC   ....so....

BA/DA  = BC/AC

And BC  = sqrt(10^2 + 20^2)  = sqrt(500)  = 10sqrt(5)    .......so....

10/DA  = 10sqrt(5) /20

10/DA  = sqrt(5) / 2    cross-multiply

20 = sqrt(5) * DA

DA = 20/sqrt(5)  = 20sqrt(5)  / 5   = 4sqrt(5)  →  (C)

CPhill  Jan 13, 2016
Sort:

#1
+8613
0

Hayley1  Jan 13, 2016
#2
+2493
0

Solveit  Jan 13, 2016
#3
+8613
0

Ughh. Blockeed. Can you messege me the picture? :)

Hayley1  Jan 13, 2016
#4
+2493
0

enter from the link below may be it will show

Solveit  Jan 13, 2016
#5
+82801
+15

Note that we have similar triangles, such that :

BDA ≈ BAC   ....so....

BA/DA  = BC/AC

And BC  = sqrt(10^2 + 20^2)  = sqrt(500)  = 10sqrt(5)    .......so....

10/DA  = 10sqrt(5) /20

10/DA  = sqrt(5) / 2    cross-multiply

20 = sqrt(5) * DA

DA = 20/sqrt(5)  = 20sqrt(5)  / 5   = 4sqrt(5)  →  (C)

CPhill  Jan 13, 2016
#6
+2493
+5

Thanks CPhill !

Solveit  Jan 13, 2016

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