Question

9. Find the area of an isosceles triangle each of whose equal side
measures 13 cm and whose base measures 20 cm.​

Answer 1

Answer:

$10\sqrt{69}$

Step-by-step explanation:

Use Herron's formula directly. Semi perimeter is (13+13+20)/2 = 23

$\sqrt{23(23-13)(23-13)(23-20)$= 10sqrt of 69.

You can round this off to 83

Answer 2

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Your answer----

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$\rm \: here \: each \: equal\: side \: \: a = 13cm \: and \: base = 20cm \\ \\ \therefore \rm \: area \: of \: the \: triangle = \big({ \frac{1}{4} b. \sqrt{4a {}^{2} - b {}^{2} } } \big)cm {}^{2} \\ = \big( \frac{1}{4 } + 20 \times \sqrt{4 \times 169 - 20 \times 20} \big) cm {}^{2} \\ = 83.07 \: cm {}^{2}$