A median of a triangle divides it into two triangles of equal areas. Verify this result for $△ A B C$ whose vertices are $A (4, - 6) B (3, - 2) C (5, 2)$

Open in App

Solution

D is midpoint of BC
D $= (3 + \frac{5}{2}) = (- 2 + \frac{2}{2}) = (4, 0)$
Area of ABD
$= \frac{1}{2} | 4 (- 2 - 0) + 3 (0 + 6) + 4 (- 6 + 2) |$
$= \frac{1}{2} | - 8 + 18 - 16 |$
$= \frac{1}{2} | - 6 | = 3$
area of ADC
$= \frac{1}{2} | 4 (2 - 0) + 5 (0 + 6) + 4 (- 6 - 2) |$
$= \frac{1}{2} | 8 + 30 - 32 |$
$= \frac{1}{2} | 6 |$
$= 3$ .
Hence we are done.

Suggest Corrections

Similar questions

You have studied in class 10 that a median of triangle divides in two triangle of equal area . verify this result for triangle ABC whose vertices are A(4,-6) B(3,-2) C(5,2)

Q. You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for

△ A B C

whose vertices are

A (4, - 6), B (3, - 2)

and

C (5, 2)

Q. For ∆ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2), verify that a median of the triangle divides it into two triangles of equal areas.

If A (4, –6), B (3, –2) and C (5, 2) are the vertices of triangle ABC, then verify the fact that a median of a triangle ABC divides it into two triangle of equal areas

Q. Question 5
You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for

Δ

ABC whose vertices are A (4, - 6), B (3, - 2) and C (5, 2).

Join BYJU'S Learning Program

A median of a triangle divides it into two triangles of equal areas. Verify this result for △ABC whose vertices are A(4,−6)B(3,−2)C(5,2)

D is midpoint of BCD =(3+52)=(−2+22)=(4,0)Area of ABD=12|4(−2−0)+3(0+6)+4(−6+2)|=12|−8+18−16|=12|−6|=3area of ADC=12|4(2−0)+5(0+6)+4(−6−2)|=12|8+30−32|=12|6|=3.Hence we are done.

A median of a triangle divides it into two triangles of equal areas. Verify this result for $△ A B C$ whose vertices are $A (4, - 6) B (3, - 2) C (5, 2)$