Question

Find the area of triangle with vertices A (3,0) B (7,0) C8,4)

Answer 1

GIVEN :–

• Vertices of triangle A (3,0) , B(7,0) & C(8,4).

TO FIND :–

• Area of triangle = ?

SOLUTION :–

• If Vertices are $\bf A(x_1 , y_1 ) , B(x_2 , y_2 )\:\:and\:\:C(x_3 , y_3 )$ then Area –

$\\ \bf \implies Area = \left|\begin{array}{ccc} \bf \: x_1 & \bf\:y_1& \bf1 \\ \\\bf \: x_2 & \bf\:y_2& \bf1 \\ \\ \bf \: x_3 & \bf\:y_3& \bf1\end{array}\right|$

• Here –

$\\ \bf \:\: {\huge{.}} \:\:\: x_1 = 3$

$\\ \bf \:\: {\huge{.}} \:\:\: y_1 = 0$

$\\ \bf \:\: {\huge{.}} \:\:\: x_2 = 7$

$\\ \bf \:\: {\huge{.}} \:\:\: y_2 = 0$

$\\ \bf \:\: {\huge{.}} \:\:\: x_3 = 8$

$\\ \bf \:\: {\huge{.}} \:\:\: y_3 = 4$

• Now put the values –

$\\ \bf \implies Area = \left|\begin{array}{ccc} \bf \:3 & \bf\:0& \bf1 \\ \\\bf \: 7& \bf\:0& \bf1 \\ \\ \bf \:8& \bf\:4& \bf1\end{array}\right|$

$\\ \bf \implies Area = 3 \{(1)(0) - (4)(1)\} - 0 \{ (7)(1) - (8)(1)\} + 1 \{(7)(4) - (8)(0) \}$

$\\ \bf \implies Area = 3 \{0-4\} - 0 \{ 7 - 8\} + 1 \{28-0\}$

$\\ \bf \implies Area = 3 \{-4\} - 0 \{ - 1\} + 1 \{28\}$

$\\ \bf \implies Area = - 12- 0 + 28$

$\\ \bf \implies Area =28 - 12$

$\\ \bf \implies \large{ \boxed{ \bf Area =16 \: \: sq. \: unit}}$

Answer 2

${\sf{\underline{\underline{\pink{GIVEN :–}}}}}$

ᴠᴇʀᴛɪᴄᴇꜱ ᴏꜰ ᴛʀɪᴀɴɢʟᴇ ᴀ (3,0) , ʙ(7,0) & ᴄ(8,4).

${\sf{\underline{\underline{\pink{To \ FIND :–}}}}}$

•$\sf\green{Area \ of \ triangle = ?}$

${\sf{\underline{\underline{\pink{SOLUTION :–}}}}}$

ɪꜰ ᴠᴇʀᴛɪᴄᴇꜱ ᴀʀᴇ

$\bf A(x_1 , y_1 ) , B(x_2 , y_2 )\:\:and\:\:C(x_3 , y_3 )$

$\sf\green{then \ Area –}$

$\begin{gathered}\\ \bf \implies Area = \left|\begin{array}{ccc} \bf \: x_1 & \bf\:y_1& \bf1 \\ \\\bf \: x_2 & \bf\:y_2& \bf1 \\ \\ \bf \: x_3 & \bf\:y_3& \bf1\end{array}\right|\\\end{gathered}$

${\sf{\underline{\underline{\pink{Here-}}}}}$

$\begin{gathered}\\ \bf \:\: {\huge{.}} \:\:\: x_1 = 3 \\\end{gathered}$

$\begin{gathered}\\ \bf \:\: {\huge{.}} \:\:\: y_1 = 0 \\\end{gathered}$

$\begin{gathered}\\ \bf \:\: {\huge{.}} \:\:\: x_2 = 7 \\\end{gathered}$

$\begin{gathered}\\ \bf \:\: {\huge{.}} \:\:\: y_2 = 0 \\\end{gathered}$

$\begin{gathered}\\ \bf \:\: {\huge{.}} \:\:\: x_3 = 8\\\end{gathered}$

$\begin{gathered}\\ \bf \:\: {\huge{.}} \:\:\: y_3 = 4 \\\end{gathered}$

$\sf\green{ Now \ put \ the \ values –}$

$\begin{gathered}\\ \bf \implies Area = \left|\begin{array}{ccc} \bf \:3 & \bf\:0& \bf1 \\ \\\bf \: 7& \bf\:0& \bf1 \\ \\ \bf \:8& \bf\:4& \bf1\end{array}\right|\\\end{gathered}$

$\begin{gathered}\\ \bf \implies Area = 3 \{(1)(0) - (4)(1)\} - 0 \{ (7)(1) - (8)(1)\} + 1 \{(7)(4) - (8)(0) \}\\\end{gathered}$

$\begin{gathered}\\ \bf \implies Area = 3 \{0-4\} - 0 \{ 7 - 8\} + 1 \{28-0\}\\\end{gathered}$

$\begin{gathered}\\ \bf \implies Area = 3 \{-4\} - 0 \{ - 1\} + 1 \{28\}\\\end{gathered}$

$\begin{gathered}\\ \bf \implies Area = - 12- 0 + 28\\\end{gathered}$

$\begin{gathered}\\ \bf \implies Area =28 - 12\\\end{gathered}$

$\begin{gathered}\\ \bf \implies \large{ \boxed{ \bf Area =16 \: \: sq. \: unit}}\\\end{gathered}$