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PolarNik [594]
3 years ago
7

In the diagram below DE is the midsegment of AABC. Find the length of DE and BC.

Mathematics
1 answer:
baherus [9]3 years ago
4 0

Answer:

DE = 5

BC = 10

Step-by-step explanation:

ΔAED is a right triangle with legs of 4 and 6/2 = 3

Use the Pythagorean theorem to find DE

4^{2}  + 3^{2} = DE^{2}

16 + 6 = 25 = DE^{2}

DE = 5

The midsegment of a triangle is 1/2 the length of the corresponding side.

Therefore, BC = 2 DE = 2(5) = 10 = BC

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Answer:

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Step-by-step explanation:

APRB form a cyclic trapezoid

∠APO  = x° (Base angle of an isosceles triangle)

∠OPR  = ∠ORP (Base angle of an isosceles triangle)

∠ORB = ∠OBR (Base angle of an isosceles triangle)

∠APO + ∠OPR + ∠OBR = 180° (Sum of opposite angles in a cyclic quadrilateral)

Similarly;

∠ORB + ∠ORP + x°  = 180°

Since ∠APO = x° ∠ORB = ∠OBR and ∠OPR  = ∠ORP we put

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∠OPR = ∠AOP = ∠BOR (Alternate interior angles of parallel lines)

Hence 2·x° + ∠AOP = 180° (Sum of angles in a triangle) = 2·∠OBR + ∠BOR

Therefore, 2·x° = 2·∠OBR, x° = ∠OBR = ∠ABC.

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3 years ago
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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(08.03 LC) Factor completely: x2 + 10x + 24 (5 points) Prime (x + 12)(x + 2) (x + 3)(x + 8) (x + 6)(x + 4)
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Answer:

Option C is correct

Step-by-step explanation:

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