Question

1. Segment AE is 7 inches and AH is 7^/3 inches. What is the length of EH? Round to the nearest tenth.

Answer 1

Answer:

B

Step-by-step explanation:

Answer 2

Answer:

1.) 9.9

2.) 15.6

Step-by-step explanation:

1.) Consider triangle AEH

AEH is a right angle triangle as ∠AEH = 90°

AH is the hypotenuse of the triangle.

Applying Pythagorean theorem

$AE^{2} + EH^{2} = AH^{2}$

substituting values as given in the question:

$7^{2} +EH^{2}=(7\sqrt{3} )^{2}\\EH^{2}=(7\sqrt{3} )^{2}-7^{2} \\EH^{2}=98\\EH=\sqrt{98}\\EH=7\sqrt{2}$

∴ EH≈9.9

2.) Consider triangle CDF

CDF is a right angle triangle as ∠CDF = 90°

CF is the hypotenuse of the triangle.

Applying Pythagorean theorem

$CD^{2} + DF^{2} = CF^{2}$

substituting values as given in the question:

$11^{2} +DF^{2}=(11\sqrt{3} )^{2}\\DF^{2}=(11\sqrt{3} )^{2}-11^{2} \\DF^{2}=242\\DF=\sqrt{242}\\DF=11\sqrt{2}$

∴ DF≈15.6