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Radical national mathematics competition
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It is known that A and B are the lengths of two sides of an isosceles triangle, which satisfies equation 2√(3a-6)+3√(2-a)=b-4. Find the perimeter and area of this isosceles triangle.

If so.

2√(3a-6)+3√(2-a)=b-4

It becomes 2√3*√(a-2) +3√(2-a)=b-4.

A-2≥0, 2-a≥0, so a=2.

Left =0, so b-4=0 and b=4.

And △ABC is an isosceles △. So the perimeter C=4+4+2= 10.

In △ABC, according to the cosine theorem, we can get

cosA=(b? +c? +a? )/2bc=(4? +4? -2? )/(2*4*4)=7/8

Sina =√ 15/ 8

s = 1/2 bcsina = 1/2 * 4 * 4 *√ 15/8 =√ 15