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Quadrilateral ABCD is inscribed in this circle.
1 answer:
Answer:
m∠A = 116°
Step-by-step explanation:
For better understanding of the solution, see the attached figure of the problem :
The quadrilateral ABCD is inscribed in a circle, Thus the quadrilateral ABCD is a cyclic quadrilateral.
Also, Sum of opposite angles of a cyclic quadrilateral is supplementary.
Now, m∠A = (x - 36)°, m∠B = 28° and m∠D = x°
But, since the sum of opposite angles is 180°
⇒ 28° + x° = 180°
⇒ x = 180 - 28
⇒ x = 152
So, m∠A = (x - 36)°
⇒ m∠A = (152 - 36)°
⇒ m∠A = 116°
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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
