# PLZ HELP ASAP SIMMILIAR AND CONGRUENT TRIANGLES

Question

Answer:

Area of triangle ABC: A=?Formula to calculate the area of a triangle: A=(1/2)bh

In this case, we can take:

Base of the triangle: b=AC=5

Height of the triangle: h=CB=?

Triangles ABC and ACD are similars, because they have two congruent angles:

Angle DAC is a common angle and

Angle ADC = 90° = Angle ACB (the angle ADC is 90° because the angle BDC is 90°)

Then, the triangles ABC and ACD must have proportional sides:

CB/AC=CD/AB

AC=5; CD=?; AD=2

We can find CD in triangle ACD using the Pythagoras Theorem:

CD=sqrt (AC^2-AD^2)

CD=sqrt (5^2-2^2)

CD=sqrt (25-4)

CD=sqrt (21)

Now we can find CB:

CB/AC=CD/AD

Replacing the known values:

CB/5=sqrt(21)/2

Solving for CB. Multiplying both sides of the equation by 5:

5(CB/5)=5[sqrt(21/2)]

CB=5sqrt(21)/2

Now we can calculate the area of the triangle ABC:

A=(1/2)(AC*CB)

A=(1/2)(5)(5sqrt(21)/2)

A=[1*5*5sqrt(21)]/(2*2)

A=25srt(21)/4

A=25(4.582575695)/4

A=28.64109809

Rounded to the nearest hundredth

A=28.64

Answer: The area of triangle ABC is 28.64 square units

solved

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