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# 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 l

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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
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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
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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
+78618
+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

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