Question

PLEASE HELPP

Answer 1

(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

$\frac{AB}{BC} = \frac{AE}{DE} \\\\\frac{6.32}{10} = \frac{AE}{5} \\\\AE = 3.16$

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

$\frac{height \ of \ ADE}{base \ ADE} = \frac{height \ of \ ACB}{base \ ACB}\\\\\frac{h}{3.16} = \frac{9.5}{6.32} \\\\h = 4.75$

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: brainly.com/question/11899908

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