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3 years ago

Find the values of X and Y that makes these triangles congruent by the HL theorem

Mathematics

1 answer:

muminat3 years ago

8 0

Answer:

C. x = 3, y = 2

Step-by-step explanation:

If both triangles are congruent by the HL Theorem, then their hypotenuse and a corresponding leg would be equal to each other.

Thus:

x + 3 = 3y (eqn. 1) => equal hypotenuse

Also,

x = y + 1 (eqn. 2) => equal legs

✔️Substitute x = y + 1 into eqn. 1 to find y.

x + 3 = 3y (eqn. 1)

(y + 1) + 3 = 3y

y + 1 + 3 = 3y

y + 4 = 3y

y + 4 - y = 3y - y

4 = 2y

Divide both sides by 2

4/2 = 2y/2

2 = y

y = 2

✔️ Substitute y = 2 into eqn. 2 to find x.

x = y + 1 (eqn. 2)

x = 2 + 1

x = 3

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3 years ago

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neonofarm [45]

Answer:

No, because the distance from the origin to point (7 , -7) is greater than the radius of the circle ⇒ 3rd answer

Step-by-step explanation:

From the graph of the circle

∵ The center of the circle F is at the origin

∴ F is (0 , 0)

∵ The circle passes through point (7 , 0)

- The length of the radius of the circle is the distance between

the center of the circle and a point on the circle

∴ r is the distance between points (0 , 0) and (7 , 0)

∴ $r=\sqrt{(7-0)^{2}+(0-0)^{2}}=\sqrt{49}=7$

∴ The length of the radius of the circle is 7 units

Let us find the distance between point (7 , -7) and the origin (0 , 0)

If the distance is equal to the radius of the circle, then the point is on the circle
If the distance is greater than the radius, then the point is outside the circle
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∵ The distance = $\sqrt{(7-0)^{2}+(-7-0)^{2}}=\sqrt{49+49}=\sqrt{98}=7\sqrt{2}$

∵ r = 7

∵ $7\sqrt{2}$ > 7

∴ The distance is greater than the radius of the circle

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The correct answer is:

No, because the distance from the origin to point (7 , -7) is greater than the radius of the circle

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