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Geometry Level 3

The sides of a non-degenerate triangle are all integers . If the perimeter of this triangle is 8 and its area can be expressed as a b a\sqrt b , where a a and b b are positive integers with b b square-free, find a + b a+b .


The answer is 4.

Ayush G Rai by Ayush G Rai , 20, India

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2 solutions

Triangle is isosceles because possible sides are 3 , 3 , 2 3\ , 3\ , 2 .

Using Heron's Formula. ⇒ s ( s − a ) ( s − b ) ( s − c ) \Rightarrow \sqrt{s(s-a)(s-b)(s-c)} , where s s is semi-perimeter and a , b , c a\ , b\ , c are sides.

4 ( 4 − 3 ) ( 4 − 3 ) ( 4 − 2 ) = 2 2 \sqrt{4(4-3)(4-3)(4-2)}=2\sqrt{2} .

∴ a + b = 2 + 2 = 4 \therefore a+b=2+2=\boxed{4}

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it is the same thing which I did.

Ayush G Rai - 4 years, 10 months ago
Ayush G Rai
Jul 17, 2016

The only possible triangle will be of side lengths 3 , 3 , 2. 3,3,2. Semi-perimeter S = 4. S=4.
By Heron's formula,area of triangle = 4 ( 4 − 3 ) ( 4 − 3 ) ( 4 − 2 ) = 2 2 . =\sqrt{4(4-3)(4-3)(4-2)}=\boxed{2\sqrt2}.
Therefore a = 2 a=2 and b = 2. b=2. So a + b = 4 . a+b=\boxed 4.

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