side BC =√52
side BC is the bottom of triangle ABC
Now if you call segment AD = x
Now you have 2 equations:
(AB)^2 + (√52)^2 = (6 + AD)^2
4^2 + (AD)^2 = (AB)^2 Both by pythagorean theorem.
Remember AB is side AB. And AD is segement AD. Now you have 2 equations and 2 unknowns. Rearrange:
(AB)^2 + (√52)^2 = (6 + AD)^2
4^2 + (AD)^2 - (AB)^2 =0 If you add both equations notice side (AB)^2 cancels out.
Can you now solve for AD?
]]>Bobby
4^2 + 6^2 = c^2 = 52 (c should be 10)
4^2 + x^2 = s^2 (4 + x^2 = s^2 <--- should be this)??
I know, I ask so many questions, sorry my teachers told me too.
I hate this question. Help me, Please,
Find triangle BDC by pythagorean theorem.
4^2 + 6^2 = c^2 = 52
Side BC = √52
Now find the other triangle:
Side AB = s, Side AC = 6 + x, x = AD
s^2 + (√52)^2 = (6 +x)^2
4^2 + x^2 = s^2
Add the two equations up.
68 + x^2 = ( 6 + x )^2
x = 8 / 3
So Side AC is 6 + 8 / 3 = 26 / 3
]]>Let AB = x
10² + x² = (6 + √(x² - 4²))²
Solve for 'x'
]]>EXPLAIN PLEASE.THANKS
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