Question

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

Answer 1

Answer:

8.3 cm

Step-by-step explanation:

An isosceles triangle has two sides being equal. These two sides are 13 cm each and its base is 20 cm. The formula for the area of a triangle is 1/2*the base of the triangle*the height of the triangle. In order to find the height if the triangle, you have to use the Pythagorean Theorem which states that

a² + b² = c²

where

a      =      base of the triangle

b = side of  triangle

c = hypotenuse

The height of the triangle is not known but according to the Pythagorean theorem it will become our b.

The base of the triangle is 20 cm. However the height is the point from the apex to the base so in this case we will have to divide the base which is 20 cm into half. Therefore we will end up with 10 cm which will become our a.

The hypotenuse is the slanted side which is also 13 cm which is our c.

To find the height we must replace the values.

We will therefore get

10²+b²=13²

Now all we do is to simplify this equation.

100+b²=169

b²=169-100

b²= 69

Now at this point we find the square root of both sides.

Our answer will therefore be 8.3066 cm

This is just the height. Now I said earlier that the area of a triangle is $1/2×b×h$

The weird A above is the multiplication sign.

The area will therefore become 1/2×20×8.3

The final answer becomes 83 cm²

Thanks

Answer 2

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Here is your required answer••••

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