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velikii [3]
3 years ago
6

The height of AXYZ is the

Mathematics
2 answers:
jeka943 years ago
8 0

Answer:

10 units squared

Step-by-step explanation:

Area of a triangle = 1/2 × base × height

Use the <u>distance between two points</u> formula to find the measure of the height and base of ΔXYZ.

<u>Distance between two points</u>

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points}

<u>Height of Triangle</u>

Define the points:

  • \textsf{Let }(x_1,y_1)=(2,1)
  • \textsf{Let }(x_2,y_2)=(4,5)

Substitute the points into the distance formula to find the height of the triangle:

\begin{aligned}\implies \sf height & =\sqrt{(4-2)^2+(5-1)^2}\\& = \sqrt{2^2+4^2}\\& = \sqrt{20}\end{aligned}

<u>Base of Triangle</u>

Define the points:

  • \textsf{Let }(x_1,y_1)=(0,2)
  • \textsf{Let }(x_2,y_2)=(4,0)

Substitute the points into the distance formula to find the height of the triangle:

\begin{aligned}\implies \sf base & =\sqrt{(4-0)^2+(0-2)^2}\\& = \sqrt{4^2+(-2)^2}\\& = \sqrt{20}\end{aligned}

<u>Area of Triangle</u>

\begin{aligned}\implies \sf Area & = \dfrac{1}{2} \times \sf base \times height\\& = \dfrac{1}{2}\sqrt{20}\sqrt{20}\\& = \dfrac{1}{2}(20)\\& = 10 \sf \:\:units^2 \end{aligned}

kvasek [131]3 years ago
3 0

Answer:

areas of AXYZ=10 square units1

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