The heights AD and CE of the acute-angled triangle ABC intersect at point O, OA = 4cm

| July 26, 2021 | education

The heights AD and CE of the acute-angled triangle ABC intersect at point O, OA = 4cm, OD = 3cm, BD = 4cm. Find the distance from point O to the speaker side.

Let’s draw the third height BH, which also passes through the point O, since the heights in the triangle intersect at one point.

Consider a right-angled triangle OBD, and by the Pythagorean theorem, determine the length of the leg OB.

ОВ ^ 2 = ВD ^ 2 + ОD ^ 2.

OB ^ 2 = 4 ^ 2 + 3 ^ 2 = 16 + 9 = 25.

OB = 5 cm.

Let us prove that triangles AOH and BOD are similar. Both triangles are rectangular, with right angles H and D, and the angle AOH = BOD as vertical angles, then right-angled triangles are similar in acute angle.

Then:

OH / OD = OA / OB.

OH / 3 = 4/5.

OH = 3 * 4/5 = 2.4 cm.

Answer: The distance from point O to the AC side is 2.4 cm.



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