Help me with number 5 rn i need help its due soon.
2 answers:
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Answer:
7 1/17
Step-by-step explanation:
A figure can be helpful.
The inscribed semicircle has its center at the midpoint of th base. It is tangent to the side of the isosceles triangle, so a radius makes a 90° angle there.
The long side of the isosceles triangle can be found from the Pythagorean theorem to be ...
BC² = BD² +CD²
BC² = 8² +15² = 289
BC = √289 = 17
The radius mentioned (DE) creates right triangles that are similar to ∆BCD. In particular, we have ...
(long side)/(hypotenuse) = DE/BD = CD/BC
DE = BD·CD/BC = 8·15/17
DE = 7 1/17 ≈ 7.059
Answer: SAS
Step-by-step explanation:
Here ABC is a triangle, In which angle BAC is bisected by line AR where, .
Thus in triangles ABR and ACR,
AB= AC ( given)
Therefore
( by the property of angle bisector)
And, (reflexive)
Thus by SAS postulate,
△ABR ≅ △ACR.
Therefore, Option D is correct.
