Given: LMNB is a square, LM = 20cm, P∈ LM , K ∈ PN , PK = 1 5 PN, LP = 4 cm Find: Area of LPKB
1 answer:
Answer:
80 cm²
Step-by-step explanation:
Trapezoid LPKB has area ...
A = (1/2)(b1 +b2)h = (1/2)(4 +20)(20) = 240 . . . . cm²
Triangle BPN has area ...
A = (1/2)bh = (1/2)(20)(20) = 200 . . . . cm²
Triangle BKN has a height that is 4/5 the height of triangle BPN, so will have 4/5 the area:
ΔBKN = (4/5)(200 cm²) = 160 cm²
The area of quadrilateral LPKB is that of trapezoid LPNB less the area of triangle BKN, so is ...
240 cm² - 160 cm² = 80 cm²
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Answer:
Step-by-step explanation:
Given
Required
Determine the percentage of the shaded cubes
First, we calculate the number of boxes:
Express as multiplication
This implies that there are 20 cubes in the box.
The percentage of the shaded cubes is:
<em>30% were shaded</em>
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