Question

The height of AXYZ is the
distance from point Y to XZ
Find the area of the
triangle. Round your answer
to the nearest tenth, if
necessary. y(4,5) a(2,1) x(0,2) z(4,0)
pls help quick:(

Answer 1

Answer:

areas of AXYZ=10 square units1

Answer 2

Answer:

10 units squared

Step-by-step explanation:

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

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

Distance between two points

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

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

Height of Triangle

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:

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

Base of Triangle

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:

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

Area of Triangle

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