(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.
Similar triangles
Triangle ABC is similar to triangle AED because angle BAC is congurent to angle EAD and angle ADE is also congurent to angle ACB.
Distance from C and to the line through AB
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
Length of AE
[tex]\frac{AB}{BC} = \frac{AE}{DE} \\\\\frac{6.32}{10} = \frac{AE}{5} \\\\AE = 3.16[/tex]
Angle BAC
BAC = 180 - (36.87 + 71.565)
BAC = 71.565⁰
also, since AC = BC, angle B = angle A = 71.565⁰
Shortes distance from C to E
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
Height of triangle ADB
[tex]\frac{height \ of \ ADE}{base \ ADE} = \frac{height \ of \ ACB}{base \ ACB}\\\\\frac{h}{3.16} = \frac{9.5}{6.32} \\\\h = 4.75[/tex]
Area of the whole figure
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: https://brainly.com/question/11899908
#SPJ1
