Need help with this!

[thick bane]

I was doing a past paper until I came across this question, can anyone teach me how?

[unkempt night]

calculate the area of triangle ABC in terms of a and b then compare with the given area

I mean equate them

[thick bane]

the area calculated: a + b√5 + 2a√5 + 10b = area given: 34 + 11√5

so...

[unkempt night]

(a+10b) + (2a+b)sqrt(5) = 34 + 11sqrt(5)

[thick bane]

ohhh then you div through by √5

[unkempt night]

then you get 2 equations in terms of a and b

[unkempt night]

no

[thick bane]

so:
a + 10b = (34 + 11√5) - ((2a + b)√5)
and
(2a+b)√5 = (34+11√5) - (a+10b)

[unkempt night]

no

compare corresponding terms in LHS and RHS

terms with and without sqrt(5)

[thick bane]

so:
(2a + b - 11)√5 = 34 - a - 10b

[unkempt night]

a + 10b = 34
2a + b = 11

[thick bane]

can you show how you got that

[unkempt night]

(a+10b) + *(2a+b)*sqrt(5) = 34 + 11sqrt(5)

[thick bane]

OHHHHHH

Im sorry

understood

then solve simultaneuosly

👍🏼

[unkempt night]

yes as usual

[thick bane]

thanks a lot

[unkempt night]

you're welcome

[vivid mural]

!done

[grave raptor]

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[thick bane]

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