A round ball has a diameter of 10 inches. What is the approximate volume of the ball?
1 answer:
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Let
<span>A (3, 1)
B (0, 4)
C(3, 7)
D (6, 4)
step 1
find the distance AB
d=</span>√[(y2-y1)²+(x2-x1)²]------> dAB=√[(4-1)²+(0-3)²]-----> dAB=√18 cm
step 2
find the distance CD
d=√[(y2-y1)²+(x2-x1)²]------> dCD=√[(4-7)²+(6-3)²]-----> dCD=√18 cm
step 3
find the distance AD
d=√[(y2-y1)²+(x2-x1)²]------> dAD=√[(4-1)²+(6-3)²]-----> dAD=√18 cm
step 4
find the distance BC
d=√[(y2-y1)²+(x2-x1)²]------> dBC=√[(7-4)²+(3-0)²]-----> dBC=√18 cm
step 5
find slope AB and CD
m=(y2-y1)/(x2-x1)
mAB=-1
mCD=-1
AB and CD are parallel and AB=CD
step 6
find slope AD and BC
m=(y2-y1)/(x2-x1)
mAD=1
mBC=1
AD and BC are parallel and AD=BC
and
AB and AD are perpendicular
BC and CD are perpendicular
therefore
the shape is a square wit length side √18 cm
area of a square=b²
b is the length side of a square
area of a square=(√18)²------> 18 cm²
the answer is
18 cm²
see the attached figure
<span>A (3, 1)
B (0, 4)
C(3, 7)
D (6, 4)
step 1
find the distance AB
d=</span>√[(y2-y1)²+(x2-x1)²]------> dAB=√[(4-1)²+(0-3)²]-----> dAB=√18 cm
step 2
find the distance CD
d=√[(y2-y1)²+(x2-x1)²]------> dCD=√[(4-7)²+(6-3)²]-----> dCD=√18 cm
step 3
find the distance AD
d=√[(y2-y1)²+(x2-x1)²]------> dAD=√[(4-1)²+(6-3)²]-----> dAD=√18 cm
step 4
find the distance BC
d=√[(y2-y1)²+(x2-x1)²]------> dBC=√[(7-4)²+(3-0)²]-----> dBC=√18 cm
step 5
find slope AB and CD
m=(y2-y1)/(x2-x1)
mAB=-1
mCD=-1
AB and CD are parallel and AB=CD
step 6
find slope AD and BC
m=(y2-y1)/(x2-x1)
mAD=1
mBC=1
AD and BC are parallel and AD=BC
and
AB and AD are perpendicular
BC and CD are perpendicular
therefore
the shape is a square wit length side √18 cm
area of a square=b²
b is the length side of a square
area of a square=(√18)²------> 18 cm²
the answer is
18 cm²
see the attached figure