MATH SOLVE

4 months ago

Q:
# Answer please. Need help

Accepted Solution

A:

Step One

Show that ΔADC ≡ ΔACB are similar Maybe ≡ this might mean congruent, I'm not sure. I want similarity.

A is common to both triangles.

<ADC = <ACB Both angles are right angles.

Conclusion

ΔADC is similar to ΔACB Angle Angle theorem which is enough to declare similarity.

Step Two

Find AB

Set up a proportion.

AD/AC = AC/AB corresponding parts of similar triangles bear the same ratio.

Substitute and solve for AB

2/5 = 5/AB Cross multiply

2AB = 25 Divide by 2

AB = 13/2

AB = 6.5

Step Three

Now you need the height of the triangle (CD)

That's just a^2 + b^2 = c^2

a = 2

c = 5

b = ???

2^2 + b^2 = 5^2

4 + b^2 = 25 Subtract 4 from both sides.

b^2 = 25 - 4

b^2 = 21

b = sqrt(21)

Step 4

Find the area.

A = 1/2 b*h

h = sqrt(21)

b = 6.5

Area = 1/2 6.5 * sqrt(21)

Area = 14.89

Show that ΔADC ≡ ΔACB are similar Maybe ≡ this might mean congruent, I'm not sure. I want similarity.

A is common to both triangles.

<ADC = <ACB Both angles are right angles.

Conclusion

ΔADC is similar to ΔACB Angle Angle theorem which is enough to declare similarity.

Step Two

Find AB

Set up a proportion.

AD/AC = AC/AB corresponding parts of similar triangles bear the same ratio.

Substitute and solve for AB

2/5 = 5/AB Cross multiply

2AB = 25 Divide by 2

AB = 13/2

AB = 6.5

Step Three

Now you need the height of the triangle (CD)

That's just a^2 + b^2 = c^2

a = 2

c = 5

b = ???

2^2 + b^2 = 5^2

4 + b^2 = 25 Subtract 4 from both sides.

b^2 = 25 - 4

b^2 = 21

b = sqrt(21)

Step 4

Find the area.

A = 1/2 b*h

h = sqrt(21)

b = 6.5

Area = 1/2 6.5 * sqrt(21)

Area = 14.89