Question

in a triangle ABC (3,1)(5,6)and(-3,2)are the midpoints of sides AB,BC and AC respectively, find the area of triangle ABC​

Answer 1

Answer:

• formula of triangle =1/2(x^2(y2-y3)+(X2(y3-y1)+X3(y1-y2

• =1/2(3(6-(2)+5(2-1)-3(1-5)

• =1/2(12+5-12)

• ,=1/2(5)

• =10

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Answer 2

The area of ΔABC is 64 sq. units

Step-by-step explanation:

in a triangle ABC (3,1), (5,6) and (-3,2) are the midpoints of sides AB,BC and AC respectively

Let P (3,1) =$(x_1,y_1)$, Q(5,6)=$(x_2,y_2)$, R(-3,2)=$(x_3,y_3)$ are the coordinate of the triangle formed by joining the mid points of sides of triangle ABC

as we know area of a triangle is given by

$area= \dfrac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) )$

where $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ are the coordinates of the triangle

Therefore

$\text { area of } \triangle PQR = \dfrac{1}{2} (3(6-2)+5(2-1)+(-3)(1-6))\\\\\Rightarrow \text { area of } \triangle PQR= \dfrac{1}{2} (12+5+15)=\dfrac{32}{2} =16$

Now as we know the area of triangles formed by joining midpoints of a triangle is one-forth of area of given triangle

$\therefore \text{area of } \triangle PQR = \dfrac{1}{4} \text{area of } \triangle ABC\\\\\Rightarrow 16= \dfrac{1}{4} \text{area of } \triangle ABC\\\\\Rightarrow \text{area of } \triangle ABC=4 \times 16 = 64$

Hence, the area of ΔABC is 64 sq. units

#Learn more

Prove that the area of triangles formed by joining the midpoints of a triangle is 1/4 of area of the given triangle

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