Ex.7.3 Q5 Coordinate Geometry Solution - NCERT Maths Class 10

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Question

You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for \(\Delta ABC\) whose vertices are \(A(4, -6),\) \(B(3, -2)\) and \(C(5, 2)\)

 Video Solution
Coordinate Geometry
Ex 7.3 | Question 5

Text Solution

Reasoning:

Let \(ABC\) be any triangle whose vertices are  \(A(x_1, y_1)\), \(B(x_2, y_2)\) and \(C(x_3, y_3).\)

Area of a triangle

\[\begin{align}=\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) + \\  x_2 \left( y_3 - y_1 \right) + \\ x_3 \left( y_1 - y_2 \right) \end{bmatrix} \end{align}\]

What is known?

The \(x\) and \(y\) co-ordinates of the vertices of the triangle.

What is unknown?

To verify that a median of a triangle divides it into two triangles of equal areas.

Steps:

From the figure,

Given,

  • Let the vertices of the triangle be \(A\)\((4, -6),\) \(B\)\((3, -2),\) and \(C\)\((5, 2).\)
  • Let \(M\) be the mid-point of side \(BC\) of \(\Delta ABC.\)

Therefore, \(AM\) is the median in \(\Delta ABC.\)

Coordinates of point \(M\)

\[\begin{align}{\text{}}\;{{}} &= \left[ {\frac{{3 + 5}}{2} + \frac{{ - 2 + 2}}{2}} \right] \\ &= [4,0]\end{align}\]

Area of a triangle

\[\begin{align}=\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) + \\  x_2 \left( y_3 - y_1 \right) + \\  x_3 \left( y_1 - y_2 \right) \end{bmatrix}\;\;\dots(1) \end{align}\]

By substituting the values of vertices, \(A, B, M\) in the Equation \((1)\)

Area of \(\Delta ABM\)

\[\begin{align} &= \frac{1}{2} \begin{bmatrix} (4)\{ ( - 2) - (0)\} \\ + (3)\{ (0) - ( - 6)\} \\ + (4)\{ ( - 6)\} - ( - 2)\} \end{bmatrix} \\ &= \frac{1}{2}[ - 8 + 18 - 16]\\& = 3\;\text{ Square units }\end{align}\]

By substituting the values of vertics, \(A,D,C\) in the Equation \((1)\)

Area of \(\Delta ABD\)

\[\begin{align} &= \frac{1}{2} \begin{bmatrix} (4)(( - 2) - (0)) + \\(3)((0) - ( - 6))+\\((4){ ( - 6)) - ( - 2)} \end{bmatrix} \\&= \frac{1}{2}[- 8 + 18 - 16]\\&= 3{\text{ Square units }}\end{align}\]

However, area cannot be negative. Therefore, area of \(\Delta AMC\) is \(3\, \rm{square \,units}.\)

Hence, clearly, median \(AM\) has divided \(\Delta ABC\) in two triangles of equal areas.

  
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