Problems on Properties of Triangle

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 circum-radius 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 circum-radius = 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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