Question

If A (0,5), B(6,11) C (10,7)are vertices of a triangle ABC .D and E are midpoints of AB and AC respectivelly. Then find the area of Triangle ADE

Answer 1

Answer:

6 units

☸Given☸

• A ΔABC with coordinate of vertices A(0,5), B(6,11) and C(10,7)
• D and E are midpoints of AB and AC respectively

☛ To Find:

• Area of ΔADE

$\rule{200}{2}$

Mid-Point Formula

$\setlength{\unitlength}{0.9 cm}}\begin{picture}(12,4)\thicklines\put(6,6){\line(1,0){5.5}}\put(5.6,5.9){ P }\put(11.7,5.9){ Q }\put(5.4,5.5){ (x_1\,,\,y_1) }\put(11.4,5.5){ (x_2\,,\,y_2) }\put(8.5,6){\circle*{0.2}}\put(8.3,6.3){ R }\put(8,5.5){ (x\,,\,y) }\put(7.2,6.3){ 1 }\put(9.9,6.3){ 1 }\put(11.7,5.9){ Q }\end{picture}$

If R divides the line segment joining points P and Q, internally in the ratio 1 : 1 . i.e., if R is the mid-point of PQ , then

$\sf{(x,y)=\bigg(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2}\bigg)}$

$\rule{200}{2}$

Diagram:

$\setlength{\unitlength}{1.05 cm}}\begin{picture}(12,4)\thicklines\put(6,6){\line(1,0){5}}\put(6,6){\line(2,3){2.52}}\put(11,6){\line(-2,3){2.52}}\put(7.35,8){\line(1,0){2.3}}\put(8.4,9.9){ A }\put(5.6,5.9){ B }\put(11.2,5.9){ C }\put(9.9,7.9){ E }\put(6.9,7.9){ D }\put(8.7,9.9){ (0,5) }\put(5.5,5.5){ (6,11) }\put(10.7,5.5){ (10,7) }\put(6,7.9){ (a\,,\,b) }\put(10.2,7.9){ (c\,,\,d) }\put(11.2,5.9){ C }\end{picture}$

✺Solution✺

Let coordinate D be (a,b) and E be (c,d)

It is given that D and E are the mid-points of AB and AC respectively

So, by applying Mid-Point Formula, we get

$\longrightarrow\sf{(a,b)=\bigg(\dfrac{0+6}{2},\dfrac{5+11}{2}\bigg)}$

$\longrightarrow\sf{(a,b)=\bigg(\dfrac{6}{2},\dfrac{16}{2}\bigg)}$

$\longrightarrow\sf{(a,b)=(3,8)}$

Similarly,

$\longrightarrow\sf{(c,d)=\bigg(\dfrac{0+10}{2},\dfrac{5+7}{2}\bigg)}$

$\longrightarrow\sf{(c,d)=\bigg(\dfrac{10}{2},\dfrac{12}{2}\bigg)}$

$\longrightarrow\sf{(c,d)=(5,6)}$

Now, In ΔABC

Let

• A(0,5)⇒(x₁,y₁)
• D(3,8)⇒(x₂,y₂)
• E(5,6)⇒(x₃,y₃)

Here,

x₁=0, x₂=3, x₃=5

y₁=5, y₂=8, y₃=6

We know that,

$\sf{Area\:of\:\triangle=\Bigg|\dfrac{1}{2}\Big(x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\Big)\Bigg|}$

So,

$\sf{Area\:of\:\triangle ADE=\dfrac{1}{2}\Bigg|\Big(0(8-6)+3(6-5)+5(5-8)\Big)\Bigg|}$

$\sf{Area\:of\:\triangle ADE=\Bigg|\dfrac{1}{2}\Big(0+3(1)+5(-3)\Big)\Bigg|}$

$\sf{Area\:of\:\triangle ADE=\Bigg|\dfrac{1}{2}\Big(3-15\Big)\Bigg|}$

$\sf{Area\:of\:\triangle ADE=\Bigg|\dfrac{1}{2}\Big(-12\Big)\Bigg|}$

$\sf{Area\:of\:\triangle ADE=\mid-6\mid}$

$\sf\pink{Area\:of\:\triangle ADE=6}$, Because modulus always give absolute value and also area of triangle cannot be negative.

Hence, Area of ΔADE= 6units