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jek_recluse [69]
3 years ago
15

What is the measure of EFG in 0 0 below?

Mathematics
2 answers:
const2013 [10]3 years ago
5 0

Answer:

C. 300

Step-by-step explanation:

EFG and EG is the total distance around the circle

EFG + EG = 360 degrees

EFG + 60 = 360

Subtract 60 from each side

EFG +60-60 = 360-60

EFG = 300

Gennadij [26K]3 years ago
5 0

Answer:

C. \widehat{GFE}=300^{\circ}

Step-by-step explanation:

We have been given an image of a circle. We are asked to find the measure of major arc EFG for our given circle.

First of all, we will find measure of arc GE.

We know that the measure of central arc is equal to its subtended arc. The measure of arc GS will be equal to measure of central angle GOE.

Since measure of central angle GOE is 60 degree, so measure of arc GE is 60 degrees as well.

We know that the circumference of circle is equal to 360 degrees. So we can set an equation as:

\widehat{GE}+\widehat{GFE}=360^{\circ}

60^{\circ}+\widehat{GFE}=360^{\circ}

60^{\circ}-60^{\circ}+\widehat{GFE}=360^{\circ}-60^{\circ}

\widehat{GFE}=300^{\circ}

Therefore, the measure of arc GFE is 300 degrees and option C is the correct choice.

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kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

step 1

Find the distance AB

substitute in the formula

d=\sqrt{(5-2)^{2}+(2-0)^{2}}

d=\sqrt{(3)^{2}+(2)^{2}}

dAB=\sqrt{13}\ units

step 2

Find the distance BC

substitute in the formula

d=\sqrt{(7-5)^{2}+(-1-2)^{2}}

d=\sqrt{(2)^{2}+(-3)^{2}}

dBC=\sqrt{13}\ units

step 3

Find the distance AC

substitute in the formula

d=\sqrt{(7-2)^{2}+(-1-0)^{2}}

d=\sqrt{(5)^{2}+(-1)^{2}}

dAC=\sqrt{26}\ units

step 4

Compare the length sides

dAB=\sqrt{13}\ units

dBC=\sqrt{13}\ units

dAC=\sqrt{26}\ units

dAB=dBC

therefore

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Applying the Pythagoras Theorem

(AC)^{2} =(AB)^{2}+(BC)^{2}

substitute

(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}

26=13+13

26=26 -----> is true

therefore

Is an isosceles right triangle

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