What type of triangle has side lengths of 4√5, √145 and 19?

Mathematics

1 answer:

tester [12.3K]3 months ago

5 0

a, b, c - side lengths (a ≤ b ≤ c)

If $a^2+b^2 < c^2$ , it forms an Obtuse triangle.

If $a^2+b^2=c^2$ , it constitutes a Right triangle.

If $a^2+b^2 > c^2$ , it describes an Acute triangle.

$a=4\sqrt5,\ b=\sqrt{145},\ c=19\to a < b < c$

Verify that the sum of the lengths of the first two sides exceeds that of the third.

$a+b=4\sqrt5+\sqrt{145}\approx9+12=21 > 19=c\\\\CORRECT$

$a\neq b\neq c\neq a$ , thus identifying it as a Scalene triangle.

$a^2=(4\sqrt5)^2=4^2(\sqrt5)^2=16(5)=80\\\\b^2=(\sqrt{145})^2=145\\\\c^2=19^2=361\\\\a^2+b^2=80+145=225 < 361=c^2\\\\a^2+b^2 < c^2$

It qualifies as Obtuse triangle.

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