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2 years ago

In an isosceles triangle the length of the base is 10 cm

Mathematics

1 answer:

lawyer [7]2 years ago

5 0

The computation shows the radius of the circle that is inscribed in the isosceles triangle will be 3.33cm.

<h3>How to calculate the radius?</h3>

From the information given, the isosceles triangle the length of a base is 10 cm and the length of a leg is 13 cm.

Let A = area of the triangle

Let S = semi perimeter of the triangle.

The radius will be: = A/S

where,

$S = \dfrac{(a + b + c)}{2} = \dfrac{(13 + 13 + 10)}{2} = 18$

The radius will be:

$=\dfrac{(\sqrt{18} - \sqrt{13})(\sqrt{18} - \sqrt{13})(\sqrt{18} - \sqrt{10})} { 18}$

= 3.33cm

In conclusion, the radius is 3.33cm.

Learn more about triangles on:

brainly.com/question/17335144

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Step-by-step explanation:

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love history [14]

Answer:

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Steps:

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2 years ago

Find the area of the triangle with the given base and height. b = 3 m and h = 10 1/2 m

svetoff [14.1K]

Statement:

The base of a triangle is 3m and its height is $10 \frac{1}{2}$ m.

To find out:

The area of the triangle.

Solution:

Given, base = 3m, height = $10 \frac{1}{2}$ m
We know,

$\sf \: area \: \: of \: \: a \: \: triangle = \frac{1}{2} \times base \: \times height$

Therefore, the area of the triangle

$\sf = (\frac{1}{2} \times 3 \times 10 \frac{1}{2} ) {m}^{2} \\ \sf = ( \frac{1}{2} \times 3 \times \frac{21}{2} ) {m}^{2} \\ = \sf \frac{63}{4} {m}^{2} \\ = \sf 15\frac{3}{4} {m}^{2}$