In $\triangle ABC$, $AB = 5$ cm, $BC = 10$ cm, and the altitude drawn to $\overline{AB}$ is 8 cm. What is the number of centimeters in the length of the altitude to $\overline{BC}$?

Mellie Aug 22, 2015

#2**+11 **

According to Melody's diagram, the triangle formed by the origin and B and C has an area of (1/2)*(8)*(6) = 24 sq cm

And the triangle formed by the origin and A and C has an area of (1/2)*(1)*(8) = 4 sq cm

So the area formed by triangle ABC must be the difference of these = 20 sq cm

And this is given by (1/2)(10)h = 20 ....so.....→ 5h = 20 → h = 4 (cm)

This can also be seen by similar triangles.....Call the origin "D" ....and we have

BC /CD = AB/h

10/8 = 5/h → h = (5)(8)/(10) = 40/10 = 4 (cm)

CPhill Aug 23, 2015

#1**+10 **

Hi Mellie :)

$$\\In $\triangle ABC$, $AB = 5$ cm, $BC = 10$ cm, and the altitude drawn to $\overline{AB}$ is 8 cm. $\\$What is the number of centimeters in the length of the altitude to $\overline{BC}$?$$

Let h=altitude to BC

When I tried to draw the pic I realized that the altitude to AB had to lie outside the triangle.

That is, angle CAB is an obtuse angle.

Area of triangle ABC = 1/2 * **5***8 = 20cm^2 (error fixed - thanks Chris)

Area of triangle also = 1/2 * BC * altitude to BC = 1/2 * 10* h = 5h cm^2

so 5h=20

h=4

The altitude to BC is 4 cm

Melody Aug 23, 2015

#2**+11 **

Best Answer

According to Melody's diagram, the triangle formed by the origin and B and C has an area of (1/2)*(8)*(6) = 24 sq cm

And the triangle formed by the origin and A and C has an area of (1/2)*(1)*(8) = 4 sq cm

So the area formed by triangle ABC must be the difference of these = 20 sq cm

And this is given by (1/2)(10)h = 20 ....so.....→ 5h = 20 → h = 4 (cm)

This can also be seen by similar triangles.....Call the origin "D" ....and we have

BC /CD = AB/h

10/8 = 5/h → h = (5)(8)/(10) = 40/10 = 4 (cm)

CPhill Aug 23, 2015