# Finding the altitude(s) of a parallelogram

In light of that problem, I would like to offer my version. It could have puzzled both the American and Russian students:-

ABCD is a parallelogram with the lengths of the 2 adjacent sides being 5 cm and 10 cm. If one the altitude is 8 cm, what could be the length of the other altitude(s)

(a) 4 cm

(b) 16 cm

(c) 4 cm or 16 cm

(d) none of the above

The area of a parallelogram is the "base" times the "height." Each base has its own height, but the area will be the same whichever base you choose.

If the altitude (height) of $8$ cm goes with the side $5$ cm, then the area is $40$ square cm and the altitude on the side $10$ cm is $4$ cm.

If the altitude (height) of $8$ cm goes with the side $10$ cm, then the area is $80$ square cm and the altitude on the side $5$ cm is $16$ cm. However, we cannot have an altitude of $8$ cm if the other side is only $5$ cm (we would have a right triangle with leg $8$ and hypotenuse $5$), so this choice is not possible.

So the answer is (a), 4 cm.

• just need to prove that both parallelograms really exist. For the first one Kiran linsuain's sketch is good enough. Oh, you were faster :-) – z100 Jan 8 '16 at 18:35
• @z100: Yes, I checked my first answer after I typed it in and saw that one parallelogram was not possible. I changed my answer before you typed your comment... but later than I should have. Better late than never! – Rory Daulton Jan 8 '16 at 18:37

4 cm is the answer, the topmost angle is equal to the bottom one, this implies that the triangles are similar.

Also, area is altitude times side length and 4x10=5x8.