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If AH = 8 mm, A F = 17 mm, and FC = 41 mm, what would be the distance between A and D? Round your answers to two decimal places. (had to put a bunch of spaces between a and f cuz brainly wouldnt allow it lol

If AH 8 mm A F 17 mm and FC 41 mm what would be the distance between A and D Round your answers to two decimal places had to put a bunch of spaces between a and class=

Respuesta :

Answer:

AD = 27.29

Step-by-step explanation:

Notice that you can use similar triangles to solve this problem.

The small triangle on the top right (AFH) is similar to the triangle ACD.

So you can use that the ratio between corresponding sides must be equal.

Notice also that the side AC has length = 17 mm + 41 mm = 58 mm

You can set then the following proportion:

[tex]\frac{AD}{AC} =\frac{FA}{AH} \\\frac{AD}{58} = \frac{8}{17} \\AD=\frac{58*8}{17} =\frac{464}{17}[/tex]

in decimal form, this can be rounded to: 27.29

Answer:

19.29 mm

Step-by-step explanation:

AH/AD = A F/AC

8/AD = 17/(41+17)

AD = 8 × 58/17

AD = 27.2941176472

HD = AD - AH

= 27.2941176472 - 8

= 19.2941176471