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The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC

Solution:

Given, the vertices of a triangle ABC right angled at B are A(2, 9) B(a, 5) and C(5, 5).

We have to find the value of a and the area of the triangle ABC.

Since the triangle ABC is a right triangle with B at right angle.

By pythagoras theorem,

AC² = AB² + BC²

The distance between two points P (x₁ , y₁) and Q (x₂ , y₂) is

√[(x₂ - x₁)² + (y₂ - y₁)²]

Distance between A(2, 9) and C(5, 5) = √[(5 - 2)² + (5 - 9)²]

= √[(3)² + (-4)²]

= √(9 + 16)

= √25

= 5

Distance between A(2, 9) and B(a, 5) = √[(a - 2)² + (5 - 9)²]

= √[(a - 2)² + (-4)²]

= √[(a - 2)² + 16]

Distance between B(a, 5) and C(5, 5) = √[(5 - a)² + (5 - 5)²]

= √[(5 - a)² + 0]

= √(5 - a)²

Now, (5)² = (√[(a - 2)² + 16])² + (√(5 - a)²)²

25 = (a - 2)² + 16 + (5 - a)²

By using algebraic identity,

(a - b)² = a² - 2ab + b²

Now, 25 = a² - 4a + 4 + 16 + 25 - 10a + a²

25 = 2a² -14a + 45

2a² - 14a + 45 - 25 = 0

2a² - 14a + 20 = 0

Dividing by 2,

a² - 7a + 10 = 0

On factoring,

a² - 5a - 2a + 10 = 0

a(a - 5) - 2(a - 5) = 0

(a - 2)(a - 5) = 0

Now, a - 2 = 0

a = 2

Also, a - 5 = 0

a = 5

The values of a are 2 and 5.

When a = 5, distance BC = √(5 - 5)² = 0

So, a = 5 is not possible.

Therefore, the value of a is 2.

The area of a triangle with vertices A (x₁, y₁) , B (x₂, y₂) and C (x₃, y₃) is

1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

Here, (x₁ , y₁) = (2, 9) (x₂ , y₂) = (2, 5) and (x₃ , y₃) = (5, 5)

Area of triangle ABC = 1/2[2(5 - 5) + 2(5 - 9) + 5(9 - 5)]

= 1/2[0 + 2(-4) + 5(4)]

= 1/2[-8+20]

= 1/2[12]

= 6 square units.

Therefore, the area of the triangle ABC is 6 square units.

✦ Try This: The points A(3, 0) B(a, -2) and C(4, -1) are the vertices of triangle ABC right angled at vertex A. Find the value of a.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 17

The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC

Summary:

The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. The values of a are 5 and 2. Hence the area of ∆ABC is 6 square units

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