Triangle question

What is the maximum perimeter of an obtuse-angled triangle with area 10cm210cm^2 and the longest side being 10cm10cm?

Note by Sumukh Bansal
3 years, 6 months ago

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Hint 1: area of a triangle, Area=10=12absinC \text{Area} = 10= \dfrac12 ab \sin C .

Hint 2: Cosine rule and double angle identities, c2=102=a2+b22abcosCc^2 = 10^2 = a^2 + b^2 - 2ab \cos C , with cosC=2cos2C21 \cos C = 2\cos^2 \frac C2 - 1 .

Hint 3: Connect the two 2 equations above.

Hint 4: Lagrange multipliers.

Pi Han Goh - 3 years, 6 months ago

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@Pi Han Goh I understood the first 3 hints but what do you mean by Lagrange Multipliers.I am not able to understand and How to use them?

Sumukh Bansal - 3 years, 6 months ago

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@Sumukh Bansal You want to maximize the subject constraint a+b+10 a + b + 10 . You know the relationship between a a and bb. Lagrange multipliers is needed.

Pi Han Goh - 3 years, 6 months ago

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@Pi Han Goh Thanks.

Sumukh Bansal - 3 years, 6 months ago

@Pi Han Goh @Pi Han Goh I am still not able to find the answer as I am not able to eliminate the variables.Could you help?

Sumukh Bansal - 3 years, 6 months ago

@Pi Han Goh By cosC=2cos2C21 \cos C = 2\cos^2 \frac C2 - 1 you mean cosC=2cos2(C21) \cos C = 2\cos^2 (\frac C2 - 1)

Sumukh Bansal - 3 years, 6 months ago

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@Sumukh Bansal If I put cosC=2cos2C21 \cos C = 2\cos^2 \frac C2 - 1 . Then how will I eliminate cosC \cos C

Sumukh Bansal - 3 years, 6 months ago
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