If {1,2} {3,4} and {5,7} are the mid points of the sides of a triangle, then the area of the triangle is
Answers
Answer:4 square units
Step-by-step explanation: Area of the triangle = 4 × Area of the triangle formed by joining the midpoints of the sides
= formula to calculate the area of the triangle= [(frac{1}{2} [x_{1} (y_{2} - y_{3})+x_{2} (y_{3} - y_{1})+x_{3} (y_{1} - y_{2})]
So, area of the triangle
frac{1}{2} [1 (4 - 7)+3 (7 - 2)+5 (2 - 4)
= frac{1}{2} [(- 3)+15+(- 10)}
= 1 square unit
Hence, area of the given triangle = 4 × 1 = 4 square units.
Answer
Let A(x
1
,y
1
),B(x
2
,y
2
),C(x
3
,y
3
) be the vertices of △ABC
Let D(2,1) , F(-1,-3) and E(4,5) be the mid points of AB,AC and BC.
D and F are mid points pf ABC and AC
∴DF∥BE
E and F are mid points pf BC and Ac
∴EF∥BD
∴ DBEF is a parallelogram.
The diagonals of a parallelogram bisect each other i.e, both diagonals have same mid point
i.e, Midpoint BF=Midpoint of DE
(
2
x
2
+(−1)
,
2
y
2
+(−3)
)=(
2
2+4
,
2
1+5
)
∴
2
x
2
+(−1)
=
2
2+4
∴x
2
=7
Similarly
2
y
2
+(−3)
=
2
1+5
∴y
2
=9
i.e,(x
2
,y
2
)=(7,9)
D is the mid point of AB
D=(2,1)=(
2
x
1
+x
2
,
2
y
1
+y
2
)
2
x
1
+7
=2∴x
1
=−3
i.e,(x
1
,y
1
)=(−3,−7)
2
y
1
+9
=1∴y
1
=−7
F is the midpoint of AC
F=(−1,−3)=(
2
x
1
+x
3
,
2
y
1
+y
3
)
−1=
2
−3+x
3
∴x
3
=1
i.e,(x
3
,y
3
)=(1,1)
−3=
2
−7+y
3
∴y
3
=1
∴ The vertices of triangle are =(−3,−7),(7,9),(1,1)