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(a) one triangle has side lengths 19, 16, and 14. compute the sum of th…

Question

(a) one triangle has side lengths 19, 16, and 14.
compute the sum of the squares of the shorter lengths.
$14^{2}+16^{2}=\square$
compute the square of the longest length.
$19^{2}=\square$
what kind of triangle is it?
acute triangle
right triangle
obtuse triangle

(b) the other triangle has side lengths 12, 20, and 16.
compute the sum of the squares of the shorter lengths.
$12^{2}+16^{2}=\square$
compute the square of the longest length.
$20^{2}=\square$
what kind of triangle is it?
acute triangle
right triangle
obtuse triangle

Explanation:

Step1: Calculate \(14^{2}+16^{2}\)

First, calculate \(14^{2}=14\times14 = 196\) and \(16^{2}=16\times16 = 256\).
Then, \(14^{2}+16^{2}=196 + 256=452\)

Step2: Calculate \(19^{2}\)

\(19^{2}=19\times19 = 361\)
Since \(452>361\) (i.e., \(a^{2}+b^{2}>c^{2}\), where \(a = 14\), \(b = 16\), \(c = 19\)), the triangle is acute.

Step3: Calculate \(12^{2}+16^{2}\)

Calculate \(12^{2}=12\times12 = 144\) and \(16^{2}=16\times16 = 256\).
Then, \(12^{2}+16^{2}=144+256 = 400\)

Step4: Calculate \(20^{2}\)

\(20^{2}=20\times20=400\)
Since \(12^{2}+16^{2}=20^{2}\) (i.e., \(a^{2}+b^{2}=c^{2}\), where \(a = 12\), \(b = 16\), \(c = 20\)), the triangle is right - angled.

Answer:

(a) \(14^{2}+16^{2}=452\), \(19^{2}=361\), Acute triangle.
(b) \(12^{2}+16^{2}=400\), \(20^{2}=400\), Right triangle.