Question

Find the area of the triangle whose vertices are (a, b + c), (a, b
c.And ( a, c).

Answer 1

Answer:

Centre of mass at y

= [(πR₂²/2)(4R₂/3π) - (πR₁²/2)(4R₁/3π)] / [(πR₂²/2) - (πR₁²/2)]

= [(2/3)R₂³ - (2/3)R₁³] / [π/ 2(R₂² - R₁²)]

= [4(R₂-R₁)(R₂² + R₁² + R₁R₂)] / [3π(R₂-R₁)(R₂+R₁)]

= [4(R₂² + R₁² + R₁R₂)] / [3π (R₁ + R₂)] 

[4(R₂² + R₁² + R₁R₂)] / [3π (R₁ + R₂)] above the centre.

Hope it helps!

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Answer 2

Question:-

Find the area of the triangle whose vertices are (a, b + c), (b, c+a )

And ( c ,a+ b).

Answer:-

Area of the given triangle is "0".

Step - by - step explanation:-

Using property :-

Area of triangle of three points are given

$(x_1 \: y_1) \: </em><em>,</em><em> </em><em>\</em><em>:</em><em>(x_2 \: y_2) \: and \: (x_3 \: y_3) \\ \\ area \: of \triangle \: \\ = \frac{1}{2} \bigg(x_1 \: (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigg)$

Given that:-

Let these ,

$x_1 = a \: \: </em><em>,</em><em>\: y_1 = b + c \\ x_2 \: = b \: \: </em><em>,</em><em> \: \: y_2 = c + a \\ x_3 = c \: \:</em><em>,</em><em> \: y_3 = a + b$

Solution:-

Now ,

Using the given property of area of triangle .

Let area of triangle is A ,

Then,

Area of the given triangle A ↓

$= \frac{1}{2} \bigg(a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a ) \bigg) \\ \\ = \frac{1}{2} \big(ac \: - ab \: + ab - bc + cb - ca \big) \\ \\ = \frac{1}{2} \times (0) \\ \\ </em><em>A</em><em>rea \: </em><em>A</em><em> \: = 0$

Area of the given triangle of "0".

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