Question

5. You have studied in Class IX, (Chapter 9. Example 3), that a median of a triangle divides
it into two triangles of equal areas. Verify this result for A ABC whose vertices are
A(4-6). B 3.-2) and C(5.2).​

Answer 1

$\huge\mathfrak\blue{Answer:}$

Given:

A median divides a triangle into two equal triangles.

To Verify:

We need to verify the given result for △ABC whose vertices are (4,-6) (3, -2) and (5,2).

Solution:

Let us assume that D is the midpoint of BC.

Therefore, cordinates of D will be (5+3)/2 , (2-2)/2 = (8/2,0/2) = (4,0)

Now, area of △ADC

$= \frac{1}{2} \times [4(0 - 2) + 4(2 + 6 +) + 5( - 6 - 0)]$

$=\frac{1}{2} (8 + 32 - 30)$

$= \frac{1}{2} ( - 8 + 2)$

$= \frac{1}{2} ( - 6)$

$= 3square \: units \: as \: area \: cannot \: be \: negtive$

Now, area of △ABD

$= \frac{1}{2} [4( - 2 - 0) + 3(0 + 6) +4 ( - 6 + 2)]$

$= \frac{1}{2} ( - 8 + 18 - 16)$

$= \frac{1}{2} ( - 24 + 18)$

$= \frac{1}{2} ( - 6)$

$= 3square \: units \: as \: area \: cannot \: be \: negtive$

Clearly, Area of △ADC = Area of △ABD.

Therefore Median AD divides it into two triangles of equal area.

Answer 2

Answer:

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