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bNicholson23
20.11.2019 •
Mathematics
On a coordinate plane, triangle a b c has points (1, 1), (4, 0), (3, 5).what is the area of triangle abc?
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Ответ:
7 square units.
Step-by-step explanation:
On the coordinate plane, triangle ABC has vertices at (1,1), (4,0) and (3,5).
So, the area of the triangle ABC is given by
We know the formula that area of triangle having vertices at
,
, and
is given by
Δ =![\frac{1}{2} | x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})|](/tpl/images/0383/4601/e8471.png)
Ответ:
Area of Triangle ABC =
unit² or 7.32 unit²
Step-by-step explanation:
Given coordinates of Triangle are
A = ( 1 , 1)
B = ( 4 , 0)
C = ( 3 , 5)
So , The measure of side AB is
AB =![\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}](/tpl/images/0383/4601/cda00.png)
or, AB =![\sqrt{(4-1)^{2}+(0-1)^{2}}](/tpl/images/0383/4601/3ba32.png)
Or, AB =![\sqrt{(3)^{2}+(-1)^{2}}](/tpl/images/0383/4601/697c6.png)
∴ AB =![\sqrt{10}](/tpl/images/0383/4601/60644.png)
Again ,
The measure of side BC is
BC =![\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}](/tpl/images/0383/4601/cda00.png)
or, BC =![\sqrt{(3-4)^{2}+(5-0)^{2}}](/tpl/images/0383/4601/88a03.png)
Or, BC =![\sqrt{(-1)^{2}+(5)^{2}}](/tpl/images/0383/4601/5d590.png)
∴ BC =![\sqrt{26}](/tpl/images/0383/4601/5f9f7.png)
Similarly ,
The measure of side CA is
CA =![\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}](/tpl/images/0383/4601/cda00.png)
or, CA =![\sqrt{(3-1)^{2}+(5-1)^{2}}](/tpl/images/0383/4601/e7325.png)
Or, CA =![\sqrt{(2)^{2}+(4)^{2}}](/tpl/images/0383/4601/36974.png)
∴ CA =![\sqrt{20}](/tpl/images/0383/4601/2ca86.png)
Now, let D be the mid points of side BC
So, Points ( D ) =
, ![\frac{(y_1+y_2)}{2}](/tpl/images/0383/4601/d5dc0.png)
I.e points d =
, ![\frac{(0+5)}{2}](/tpl/images/0383/4601/bc51f.png)
Or, points D =
, ![\frac{(5)}{2}](/tpl/images/0383/4601/181d3.png)
Now Measure of line AD is
AD =![\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}](/tpl/images/0383/4601/cda00.png)
AD =
+ ![(\sqrt{\frac{5}{2}-1})^{2}](/tpl/images/0383/4601/ac326.png)
Or, AD =(
)² + (
)²
Or, AD =![\sqrt{\frac{33}{4} }](/tpl/images/0383/4601/7b58e.png)
Now, Area of Triangle ABC =
× length × base
or, Area of Triangle ABC =
× AD × BC
or, Area of Triangle ABC =
×
× ![\sqrt{26}](/tpl/images/0383/4601/5f9f7.png)
Or, Area of Triangle ABC =
unit² or , 7.32 unit² Answer
Ответ:
nfhf
Step-by-step explanation: