We will solve different types of problems on properties of triangle.
1. If in any triangle the angles be to one another as 1 : 2 : 3, prove that the corresponding sides are 1 : √3 : 2.
Solution:
Let the angles be k, 2k and 3k.
Then, k + 2k + 3k = 180°
⇒ 6k = 180°
⇒ k = 30°
So, the angles are 30°, 60° and 90°
Let x, y, and z denote the sides opposite to these angles.
Then, x/sin 30° = y/sin 60° = c/sin 90°
⇒ x : y : z = sin 30° : sin 60° : sin 90°
⇒ x : y : z = ½ : √3/2 : 1
⇒ x : y : z = 1 : √3 : 2.
2. Find the lengths of the sides of a triangle, if its angles are in the ratio 1 : 2 : 3 and the circumradius is 10 cm.,
Solution:
According to the problem, the angles of the triangle are in the ratio 1 : 2 : 3 hence, we assume that the angles are k, 2k, and 3k
i.e., A = k, B = 2k and C = 3k.
Now, A + B + C= 180°
⇒ k + 2k + 3k = 180°
⇒ 6k = 180°
⇒ k = 30°
Therefore, the angles of the triangle are:
A = k = 30°, B = 2k = 60° and C = 3k = 90°
Again, the circumradius = R = 10 cm.
Therefore, if the lengths of the sides of the triangle be a, b,c then
A = 2R sin A = 2 ∙ 10 ∙ sin 30° = 10 cm.;
B = 2R sin B= 2 ∙ 10 ∙ sin 60° = 10√3 cm.; and
C = 2R sin C = 2 ∙ 10 ∙ sin 90° = 20 cm.
3. If a : b : c = 2 : 3 : 4 and s = 27 inches, find the area of the triangle ABC.
Solution:
Since, a : b : c = 2 : 3 : 4
Let us assume, a = 2x, b = 3x and c = 4x.
Therefore, a + b + c = 2x + 3x + 4x = 9x
Therefore, 9x = 2s
⇒ 9x = 2 × 27, [Since, a + b + c = 2s]
⇒ x = 6
Therefore, the lengths of the three sides are 2 × 6 inches, 3 × 6 inches and 4 × 6 inches i.e., 12 inches, 18 inches and 24 inches.
Therefore, the area of the triangle ABC
= √(s(s  a)(s  b) (s  c))
= √(27.(27  12)(27  18) (27  24)) sq. inches.
= √(27 ∙ 15 ∙ 9 ∙ 3) sq. inches.
= 27√15 sq. inches.
11 and 12 Grade Math
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