Line XY bisects and is perpendicular to AB and CD. CD = 10 and XY = 24. What is the length of AP?

Mathematics

1 answer:

myrzilka [38]2 years ago

8 0

Answer:

AP = 13

Step-by-step explanation:

See the attached figure.

CD = 10 and XY = 24

Line XY bisects and is perpendicular to AB and CD

∴ XY bisects at Y

∴ CY = 0.5 CD = 0.5 * 10 = 5

And from the geometry of the shape

P is the midpoint of XY

∴ XP = YP = 0.5 XY = 0.5 * 24 = 12

XY is perpendicular to CD

∴ ΔPYC is a right triangle at Y

∴ PC (hypotenuse) = $\sqrt{PY^2+YC^2} =\sqrt{12^2+5^2}=\sqrt{144+25} =\sqrt{169} =13$

From the figure AP = PC = 13

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MZJ and m M are base angles of isosceles trapezoid JKLM. If mZJ = 18x + 8 and m_M=11x + 15 find m2K.

maxonik [38]

The question is incomplete. Here is the complete question.

m∠J and m∠Kare base angles of an isosceles trapezoid JKLM.

If m∠J = 18x + 8, and m∠M = 11x + 15 , find m∠K.

A. 1

B. 154

C. 77

D. 26

Answer: B. m∠K = 154

Step-by-step explanation: Isosceles trapezoid is a parallelogram with two parallel sides, called Base, and two non-parallel sides that have the same measure.

Related to internal angles, angles of the base are equal and opposite angles are supplementary.

In trapezoid JKLM, m∠J and m∠M are base angles, so they are equal:

18x + 8 = 11x + 15

7x = 7

x = 1

Now, m∠K is opposite so, they are supplementary, which means their sum results in 180°:

m∠J = 18(1) + 8

m∠J = 26

m∠K + m∠J = 180

m∠K + 26 = 180

m∠K = 154

The angle m∠K is 154°

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2 years ago

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