How to find the area of a square ABC D
1 answer:
Answer:
The answer to your question is 13 u²
Step-by-step explanation:
We know that the small triangle is surrounded by right triangles so we can use the Pythagorean theorem to find the lengths of the small triangle
AD² = 3² + 2²
Simplify
AD² = 9 + 4
AD² = 13
AD =
Find the area of the square
Area = side x side
Area = AD x AD
Area =
Area = 13 u²
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We have that
A(-2,-4) B(8,1) <span>
let
M-------> </span><span>the coordinate that divides the directed line segment from A to B in the ratio of 2 to 3
we know that
A--------------M----------------------B
2 3
distance AM is equal to (2/5) AB
</span>distance MB is equal to (3/5) AB
<span>so
step 1
find the x coordinate of point M
Mx=Ax+(2/5)*dABx
where
Mx is the x coordinate of point M
Ax is the x coordinate of point A
dABx is the distance AB in the x coordinate
Ax=-2
dABx=(8+2)=10
</span>Mx=-2+(2/5)*10-----> Mx=2
step 2
find the y coordinate of point M
My=Ay+(2/5)*dABy
where
My is the y coordinate of point M
Ay is the y coordinate of point A
dABy is the distance AB in the y coordinate
Ay=-4
dABy=(1+4)=5
Mx=-4+(2/5)*5-----> My=-2
the coordinates of point M is (2,-2)
see the attached figure
A(-2,-4) B(8,1) <span>
let
M-------> </span><span>the coordinate that divides the directed line segment from A to B in the ratio of 2 to 3
we know that
A--------------M----------------------B
2 3
distance AM is equal to (2/5) AB
</span>distance MB is equal to (3/5) AB
<span>so
step 1
find the x coordinate of point M
Mx=Ax+(2/5)*dABx
where
Mx is the x coordinate of point M
Ax is the x coordinate of point A
dABx is the distance AB in the x coordinate
Ax=-2
dABx=(8+2)=10
</span>Mx=-2+(2/5)*10-----> Mx=2
step 2
find the y coordinate of point M
My=Ay+(2/5)*dABy
where
My is the y coordinate of point M
Ay is the y coordinate of point A
dABy is the distance AB in the y coordinate
Ay=-4
dABy=(1+4)=5
Mx=-4+(2/5)*5-----> My=-2
the coordinates of point M is (2,-2)
see the attached figure