Match the correct values of 'a' for each triangle.

\sqrt{216}

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\sqrt{115}

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\sqrt{88}

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Solution

Using Pythagoras theorem,
$a^{2} + 15^{2} = 21^{2}$
$a^{2} + 225 = 441$
$a^{2} = 216$
$∴ a = \sqrt{216}$

$a^{2} + 9^{2} = 14^{2}$
$a^{2} + 81 = 196$
$a^{2} = 115$
$∴ a = \sqrt{115}$

$a^{2} + 9^{2} = 13^{2}$
$a^{2} + 81 = 169$
$a^{2} = 88$
$∴ a = \sqrt{88}$

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Question 88

Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle.

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Match the correct values of 'a' for each triangle.

Using Pythagoras theorem, a2+152=212 a2+225=441 a2=216 ∴a=√216 a2+92=142 a2+81=196 a2=115 ∴a=√115 a2+92=132 a2+81=169 a2=88 ∴a=√88

Using Pythagoras theorem,
$a^{2} + 15^{2} = 21^{2}$
$a^{2} + 225 = 441$
$a^{2} = 216$
$∴ a = \sqrt{216}$

$a^{2} + 9^{2} = 14^{2}$
$a^{2} + 81 = 196$
$a^{2} = 115$
$∴ a = \sqrt{115}$

$a^{2} + 9^{2} = 13^{2}$
$a^{2} + 81 = 169$
$a^{2} = 88$
$∴ a = \sqrt{88}$