Pls answer the first question.....
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here is ur answer buddy....
In ∆ABE
AE²=AB²+BE²
AE²=AB²+BC²/4
4AE²=4AB²+BC²=(1)
IN ∆DBE
DC²=BD²+BC²
DC²=(AB²/4)+BC²
4DC²=AB²+4BC²=(2)
ADDING (1)&(2)
4DC²+4AE²=AB²+BC²+4BC²+4AB²
4(DC²+AE²)=5(AB²+BC²) =>(A)
IN ∆ABC
AC²=AB²+BC²=(B)
(B) IN (A)
4{(√40)²+(5)²}=5(AC²)
=4(40+25)=5AC²
=>4(65)=5AC²
=>AC²=260/5
=>AC²=52
=>AC=√52
HOPE IT HELPS
In ∆ABE
AE²=AB²+BE²
AE²=AB²+BC²/4
4AE²=4AB²+BC²=(1)
IN ∆DBE
DC²=BD²+BC²
DC²=(AB²/4)+BC²
4DC²=AB²+4BC²=(2)
ADDING (1)&(2)
4DC²+4AE²=AB²+BC²+4BC²+4AB²
4(DC²+AE²)=5(AB²+BC²) =>(A)
IN ∆ABC
AC²=AB²+BC²=(B)
(B) IN (A)
4{(√40)²+(5)²}=5(AC²)
=4(40+25)=5AC²
=>4(65)=5AC²
=>AC²=260/5
=>AC²=52
=>AC=√52
HOPE IT HELPS
jude0704:
the answer for the question u just asked in the same for the answer for the question in which we are comenting now
its ok sis
hey u r confusing her
sry
did u understand???
mark me as brainliest pls
hmmm mark him
yes
But if someone answers it only I can mark it as brainliest.....
i answered know!!!!
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