Math, asked by chethan4757, 10 months ago

The area of the triangle whose vertices are (5,0), (8,9) and (8,4) is​

Answers

Answered by dibashdutta09
4

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The Ans is ;

Area of triangle =

 =   \frac{1}{2} | x_{1}(y_{2} -  y_{3}) + x_{2}(y_{2} +  y_{3}) +  x_{3} (y_{1} + y_{2}) |  \\  =  \frac{1}{2}  | 5(9 - 4) + 8(4 - 0) + 8(0 - 9) |  \\  =  \frac{1}{2} |  5 \times 5 + 8 \times 4 + 8 \times  - 9  | \\  =  \frac{1}{2}  | 25 + 32  - 72 | \\  =  \frac{1}{2}  \times 15 \\  =  \frac{15}{2} units

HOPE IT HELPS

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Answered by Anonymous
18

Answer:

  • Area of triangle = 7.5 cm².

Step-by-step explanation:

Given:

  • Vertices of triangle are (5, 0), (8, 9) and (8, 4).

To find:

  • Area of triangle

We know that,

\longrightarrow \sf Area\;of\;triangle=\dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]

Where,

  • x₁ = 5, x₂ = 8, x₃ = 8
  • y₁ = 0, y₂ = 9, y₃ = 4

Now, put the values in the formula, we get

\longrightarrow \sf Area\;of\;triangle=\Bigg|\dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;of\;triangle=\Bigg|\dfrac{1}{2}[5(9-4)+8(4-0)+8(0-9)]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;of\;triangle= \Bigg|\dfrac{1}{2}[25+32-72]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;of\;triangle=\Bigg|\dfrac{1}{2}[57-72]\Bigg|\\ \\ \\ \longrightarrow \sf Area\;of\;triangle=\Bigg|\dfrac{1}{2}\times (-15)\Bigg|

\longrightarrow \sf Area\;of\;triangle=\Bigg|-7.5 \Bigg|\\ \\ \\ \longrightarrow \sf Area\;of\;triangle= 7.5\;cm\;sq.

Hence, Area of triangle = 7.5 cm².

#answerwithquality

#BAL

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