Inscribed circles on the arc; improvement answer is required in full.

Mathematics

1 answer:

e-lub [12.9K]3 years ago

8 0

Answer:

The measure of an inscribed angle is half the measure of the intercepted arc

∠EFG creates Arc EDG

if ∠EFG is 1/2 arc EDG then 115° x 2 = 230°

A circle is 360°.

The remaining arc EFG is 360° - 230° = 130°

∠EDG is 1/2 of arc EFG so, 130° ÷ 2 = 65°

∠EDG = 65°

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Let's begin by listing out the information given to us:

We start out by observing that Triangles MKR & ACD are similar or proportional

$\begin{gathered} MK=21;AC=\text{?} \\ MR=24;AD=28\frac{4}{5} \\ KR=CD=\text{?} \end{gathered}$

We will solve for the missing side by using the similar triangle theorem. This is shown below:~

$\begin{gathered} \Delta MKR\approx\Delta ACD \\ \frac{MK}{AC}=\frac{MR}{AD} \\ \frac{21}{AC}=\frac{24}{28\frac{4}{5}} \\ \text{Cross multiply, we have:} \\ 24\cdot AC=28\frac{4}{5}\cdot21 \\ AC=\frac{28\frac{4}{5}\cdot21}{24}=25\frac{1}{5} \\ AC=25\frac{1}{5} \end{gathered}$

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