(a) Triangle ABC is similar to triangle AED because angle BAC is congurent to angle EAD and angle ADE is also congurent to angle ACB.
(b) The distance from C and to the line through AB is 9.5.
(c) The length AE IS 3.16
(d) Angle BAC is 71.565⁰.
(e) The shortest distance from C to E is 8.99.
(f) The area of the whole figure is 37.53 sq unit.
<h3>
Similar triangles</h3>
Triangle ABC is similar to triangle AED because angle BAC is congurent to angle EAD and angle ADE is also congurent to angle ACB.
<h3>Distance from C and to the line through AB</h3>
Let the distance from C and to the line through AB = h
BC = 10
Line through C divides the angle into two, = ¹/₂ x 36.87⁰ = 18.435⁰
Angle B = 90 - 18.435⁰ = 71.565⁰
Sin71.565⁰ = h/10
h = 10 x Sin71.565⁰
h = 9.5
<h3>Length of AE</h3>

<h3>Angle BAC</h3>
BAC = 180 - (36.87 + 71.565)
BAC = 71.565⁰
also, since AC = BC, angle B = angle A = 71.565⁰
<h3>Shortes distance from C to E</h3>
The shortest distance from C to E is a vertical line that connects C and E.
Let the vertical line = h
|BA| + |AE| = 6.32 + 3.16 = 9.48
sinB = h/9.48
h = 9.48 x sinB
h = 9.48 x sin71.565
h = 8.99
<h3>Height of triangle ADB</h3>

<h3>Area of the whole figure</h3>
Total area = Area of ADE + Area of ACB
A = ¹/₂x base x height + ¹/₂ x base x height
A = ¹/₂ x 3.16 x 4.75 + ¹/₂ x 6.32 x 9.5
A = 37.53 sq unit
Learn more about similar triangles here: brainly.com/question/11899908
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