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Anna007 [38]
3 years ago
11

PLEASE HELP MATH CIRCLES

Mathematics
1 answer:
Burka [1]3 years ago
6 0

Answer:

The measure of DE is 12

Step-by-step explanation:

* Lets study the figure to solve the problem

- There are two intersected circles B and C

- BA is a radius of circle B and CE is a radius of circle C

- AD is a tangent to the two circles touch circle B in A and touch

 circle C in E

- The line center BC intersects the tangent AD at D

- There are two triangles in the figure Δ BAD and Δ CED

* Now lets solve the problem

∵ AD is a tangent to circles B and C

∵ BA and CE are radii

∴ BA ⊥ AD at A

∴ CE ⊥ AD at E

- Two lines perpendicular to the same line, then the two lines are

 parallel to each other

∴ BA // CE

- From the parallelism

∴ m∠ABD = m∠ECD ⇒ corresponding angles

∴ m∠BAD = m∠CED ⇒ corresponding angles

- In any two triangles if their angles are equal then the two triangles

 are similar

- In the two triangles BAD and CED

∴ m∠ABD = m∠ECD ⇒ proved

∴ m∠BAD = m∠CED ⇒ proved

∵ ∠D is a common angle of the two triangle

∴ The two triangle are similar

- There are equal ratios between their sides

∴ BA/CE =AD/ED = BD/CD

∵ BD = 50 , AD = 40 , CD = 15

∴ 40/ED = 50/15 ⇒ using cross multiplication

∴ ED(50) = 15(40)

∴ 50 ED = 600 ⇒ divide both sides by 50

∴ ED = 12

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Answer:

The coordinates of J' when rotating by 90° counterclockwise will be: J'(3, 1)

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Step-by-step explanation:

Square JKLM with vertices

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We have to determine the answer for the image J' of the point (1, -3) when we rotate the point by 90° counterclockwise, we need to switch x and y, make y negative.

In other words, the rule to rotate a point by 90° counterclockwise.

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As we are given that J(1, -3), so the coordinates of J' will be:

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Therefore, the coordinates of J' when rotating by 90° clockwise will be:  J'(-3, -1)

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