In triangle ABC, the angle A is 45, BC = 13, and the height BD cuts off the segment DC
In triangle ABC, the angle A is 45, BC = 13, and the height BD cuts off the segment DC equal to 12 cm on the AC side. Find the area of triangle ABC.
1. It is known that
a) the area of the triangle is equal to half the product of its base by the height.
b) by the Pythagorean theorem, the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the legs.
2. In a right-angled triangle DBC, DC = 12 cm is known, BC = 13 cm, so you can find the height of the triangle BD.
BD = (BC ^ 1/2 – DC ^ 1/2) ^ 1/2 = (13 ^ 1/2 – 12 ^ 1/2) 1/2 = (169 – 144) ^ 1/2 = 5 cm.
3. In a right-angled triangle ABD, the angle A = 45 *, and tg 45 * = 1.
DB: AD = tg A;
AD = DB: tg A = 5: 1 = 5 cm.
4. Determine the base of the triangle AC.
AC = AD + DC = 5 cm + 12 cm = 17 cm.
5. Let’s calculate the area S of the triangle ABC.
S = AC * BD: 2 = 17cm * 5cm: 2 = 85: 2 = 42.5cm ^ 2.
Answer: The area of a triangle is 42.5 square centimeters.
