Triangle ABC-rectangular (angle C = 90), BC = 6cm, the projection of the leg AC on the hypotenuse is 5cm. Find the area of triangle ABC.
Given: a triangle, where ∠ACB = 90 °, BC = 6 cm, the projection of the leg AC on the hypotenuse is 5 cm. It is required to find: the area of the triangle ABC.
Let us draw the height CD to the hypotenuse AB. Since the height drawn to the hypotenuse is a perpendicular drawn to it, the legs are inclined, and the segments of the hypotenuse into which its height divides are the projections of the legs onto the hypotenuse of a right triangle. In the triangle ABC shown in the figure, AD is the projection of the leg AC onto the hypotenuse AB, similarly, BD is the projection of the leg BC onto the hypotenuse.
Let’s use the properties of the legs of a right-angled triangle. Namely, the property: “The leg of a right-angled triangle is the geometric mean between the hypotenuse and the projection of this leg onto the hypotenuse.” We have: BC ^ 2 = AB * BD.
Since AB = AD + BD and AD = 5 cm, then BD = AB – AD = AB – 5 cm.
Therefore, (6 cm) 2 = AB * (AB – 5 cm), whence AB ^ 2 – 5 * AB – 36 = 0. This quadratic equation has two roots: AB = 9 and AB = –4 (secondary root).
Now, applying the Pythagorean theorem, we get: AC ^ 2 + BC ^ 2 = AB ^ 2 or AC ^ 2 = AB ^ 2 – BC ^ 2 = (9 cm) ^ 2 – (6 cm) ^ 2 = 81 cm2 – 36 cm2 = 45 cm2, whence AC = √ (45 cm2) = 3 * √ (5) cm.
We find the required area (S) triangle ABC by multiplying its legs and dividing the result in half: S = ½ * AC * BC = ½ * (3 * √ (5) cm) * (6 cm) = 9 * √ (5) cm2.
Answer: 9 * √ (5) cm2.
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