if (9,7),(2,-4) and (0,0) are the mid points of the sides ab,bc,ca respectively then find area of triangle abc
Answers
Step-by-step explanation:
100 sq unit
Step-by-step explanation:
let the midpoint of AB be D = (9, -7)
let the midpoint of BC be E = (2, 4)
let the midpoint of AC be F = (0, 0)
join D E F to form a triangle
area of triangle =
\begin{gathered} = \frac{1}{2} |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| \\ = \frac{1}{2} |9(4 - 0) + 2(0 - ( - 7) + ( 0)( - 11 - ( - 3))| \\ = \frac{1}{2} |9 \times 4 + 2 \times 7 + 0 \times - 8| \\ = \frac{1}{2} |36 + 14| \\ = \frac{1}{2} \times 50 \\ = 25 \: sq \: unit\end{gathered}
=
2
1
∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
=
2
1
∣9(4−0)+2(0−(−7)+(0)(−11−(−3))∣
=
2
1
∣9×4+2×7+0×−8∣
=
2
1
∣36+14∣
=
2
1
×50
=25squnit
according to midpoint theorem
ratio of area formed by joining midpoint of the triangle to area of triangle is 1:4
25 : x :: 1 : 4
x × 1 = 25 × 4
x = 100
therefore the area of required triangle ABC is 100 sq units
alternative method
let the required triangle be ABC
let the midpoint of AB be D = (9, -7)
let the midpoint of BC be E = (2, 4)
let the midpoint of AC be F = (0, 0)
A = (x1, y1) B = (x2, y2) C = (x3, y3)
D is midpoint of AB
\begin{gathered} \frac{x1 + x2}{2} = 9 \: \: \: x1 + x2 = 18 \\ \frac{x2 + x3}{2} = 2 \: \: \: x2 + x3 = 4 \\ \frac{x3 + x1}{2} = 0 \: \: \: x3 + x1 = 0 \\ adding \: the \: above \: \\ 2(x1 + x2 + x3) = 22 \\ x1 + x2 + x3 = 11 \\ x1 + 4 = 11 \: \: \: \: \: \: x1 = 11 - 4 = 7 \\ x2 + 0 = 11 \: \: \: \: \: \: x2 = 11\\ x3 + 18 = 11 \: \: \: \: x3 = - 7 \\ \\ \frac{y1 + y2}{2} = - 7 \: \: \: \: y1 + y2 = - 14 \\ \frac{y2 + y3}{2} = 4 \: \: \: \: \: \: \: y2 + y3 = 8 \\ \frac{y3 + y1}{2} = 0 \: \: \: \: y3 + y1 = 0 \\ adding \: the \: above \\ 2(y1 + y2 + y3) = - 6 \\ y1 + y2 + y3 = - 3 \\ y1 + 8 = - 3 \: \: \: \: y1 = - 11 \\ y2 + 0 = - 3 \: \: \: \: y2 = - 3 \\ y3 - 14 = - 3 \: \: \: y3 = 11\end{gathered}
2
x1+x2
=9x1+x2=18
2
x2+x3
=2x2+x3=4
2
x3+x1
=0x3+x1=0
addingtheabove
2(x1+x2+x3)=22
x1+x2+x3=11
x1+4=11x1=11−4=7
x2+0=11x2=11
x3+18=11x3=−7
2
y1+y2
=−7y1+y2=−14
2
y2+y3
=4y2+y3=8
2
y3+y1
=0y3+y1=0
addingtheabove
2(y1+y2+y3)=−6
y1+y2+y3=−3
y1+8=−3y1=−11
y2+0=−3y2=−3
y3−14=−3y3=11
A = (x1, y1) = (7, -11)
B = (x2, y2) = (11, -3)
C = (x3, y3) = (-7, 11)
area of ∆ABC
\begin{gathered} = \frac{1}{2} |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| \\ = \frac{1}{2} |7(11 - ( - 3)) + 11(11 - ( - 11)) + ( - 7( - 11 - ( - 3)))| \\ = \frac{1}{2} |(7 \times - 14) + (11 \times 22) + ( - 7 \times - 8)| \\ = \frac{1}{2} | - 98 + 242 + 56| \\ = \frac{1}{2} \times ( - 98 + 298) \\ = \frac{1}{2} \times 200 \\ = 100 \: sq \: units\end{gathered}
=
2
1
∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
=
2
1
∣7(11−(−3))+11(11−(−11))+(−7(−11−(−3)))∣
=
2
1
∣(7×−14)+(11×22)+(−7×−8)∣
=
2
1
∣−98+242+56∣
=
2
1
×(−98+298)
=
2
1
×200
=100squnits
Answer:
arman song in the world