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larisa86 [58]
3 years ago
10

Isosceles trapezoid ABCD is inscribed in ⊙O with radius 5. AD=6 and the median of ABCD has length 7. Find the distance from AD t

o BC. this was the only info given!
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
5 0

Answer:

The distance from AD to BC is 7

Step-by-step explanation:

The information given are;

The type of inscribed quadrilateral ABCD = Isosceles trapezoid

The radius of the circle = 5

Segment AD of ABCD = 6

The median of the trapezoid ABCD = 7

Given the trapezoid theorem, the median is equal to half the length of the two bases added together, we have;

(AD + BC)/2 = 7

Which gives;

(6 + BC)/2 = 7

BC = 7×2 - 6 = 8

Therefore the distance from AD to BC is given by the distance from BC to the median line added to the distance from AD to the median line given as follows;

The distance from BC to the median = √(Radius² - (BC/2)²) = √(5² - (8/2)²) = 3

The distance from BC to the median = 3

The distance from AD to the median = √(Radius² - (AD/2)²) = √(5² - (6/2)²) = 4

Which gives;

The distance from AD to BC = 3 + 4 = 7

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8 0
3 years ago
Prove:<br> tan 70° = tan 20° + 2tan 50°
Natali5045456 [20]
As we all know,
According to the trigonometric identity,
tan70=tan(20+50)
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Also tan70tan20=tan70cot70=1
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tan70-tan50=tan20+tan50
So tan70=tan20+2tan50
Hope this helps :) :)
7 0
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Read 2 more answers
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