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Citrus2011 [14]

3 years ago

6

What is the length of side AC of the triangle?

2 answers:

Romashka-Z-Leto [24]3 years ago

8 0

By converse of isosceles triangle theorem, 5x+5=7x-1

5x+5=7x-1
-5x+5x+5=7x-5x-1
5=2x-1
5+1=2x-1+1
6=2x
6/2=2/2x
X=3

AC=6x+7 when x=3

6x+7=6 (3)+7=18+7=25

AC=25

saul85 [17]3 years ago

8 0

6x+7+5x+5+7x-1=180 combine all like terms 6x+5x+7x=18x 7+5-1=11 rewrite 18x+11=180 subtract 11 from both sides 169 rewrite 18x=169 divide 18 from both sides witch is 9 x=9 plug 9 in for x 6(9)+7=54+7=61 AC=61
Hopes this helps and i hope this is right have a nice evening

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Answer: Area of ΔABC is 2.25x the area of ΔDEF.

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$\frac{A_{1}}{A_{2}}= (\frac{\sqrt{3}.1.44a^{2} }{4})(\frac{4}{\sqrt{3}.0.64a^{2} } )$

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<u>The area of ΔABC is </u><u>2.25x</u><u> greater than the area of ΔDEF.</u>

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