Unknown Area Problems on the Coordinate Plane

Unknown Area Problems on the Coordinate Plane exercise

• Explain how to calculate the area of a composite shape on a coordinate plane.

  Hints

  You can determine each needed length by subtracting coordinates of the given points.

  For example, the radius of the circle is given by

  $r=\frac{10-2}2$

  Taking the formula for the area of a rectangle:

  $A=lh$

  If $h=12$ and $l=8$ we still have to plug in those values.

  $A=(12)(8)=96$ units squared.

  Solution

  Here you see the separation of the given shape in (from top to the bottom) a semicircle, a rectangle, and a triangle.

  So we need the corresponding formulas for the different shapes, as well as the values to plug in those formulas.

  Let's start with the rectangle. First, we determine the length $l$ as well as the height $h$ using the coordinates of the given points

  • $l=10-2=8$
  • $h=13-5=8$

  Oh, that's a square. The formula for the area of a square is given by:

  $A=s^2$

  where $s$ is the length of the side of the square. Here $s=8$, which we plug in the formula to get

  $A=(8)^2=64$ units squared.

  Let's establish the area of the triangle using the formula

  $A=\frac{1}{2}bh$

  • $b=l=8$ the length of the rectangle.
  • $h=5-0=5$

  Plugging those values in the formula above leads to

  $A=\frac{1}{2}(8)(5)=20$ units squared.

  There's just the semicircle left. For this, we need the following formula for area of a circle:

  $A=\pi r^2$

  and multiply it by $\frac{1}{2}$ because a semicircle is half of a circle. Then, we plug $r=4$

  $A\approx\frac{1}{2}(3.14)(4^2)\approx(8)(3.14)\approx 25.12$ units squared.

  Lastly, we have add all those results to get the total area:

  $64+20+25.12=109.12$ units squared.

  The wanted area is $109.12\cdot 100~m^2=10,912~m^2$.

• Determine the area of the composite shape.

  Hints

  Both lengths $b_1$ as well as $b_2$ are given as the difference of the x-coordinates of the given points.

  You could also count the units by the number of boxes.

  The sum of $b_1$ and $b_2$ is $20$.

  Notice there is a height of the trapezoid and a different height of the triangle.

  Solution

  The given shape can be separated into one trapezoid and one triangle.

  All of the needed lengths or heights can be determined as the difference of coordinates of the given points.

  The formula for the area of a trapezoid is given by

  $A=\frac12(b_1+b_2)h$

  So, we need

  • $b_1=0-(-12)=12$
  • $b_2=-2-(-10)=8$
  • $h=10-6=4$

  Now we plug in those values to get

  $A=\frac12(12+8)(4)=40$ units squared.

  For the triangle we need the formula

  $A=\frac{1}{2}bh$

  with

  • $b=b_2=8$
  • $h=6-4=2$

  So we get the area of the triangle:

  $A=\frac12(8)(2)=8$ units squared.

  We still have to add those results to get the total area

  $40+8=48$ units squared. Last we multiply this value with $100$ to get the $m^2$

  $48\times 100=4800~m^2$

• Determine the area of the new amusement park.

  Hints

  You can break this composite shape up into a semicircle, right triangle, and rectangle.

  Recall that the area of a right triangle is half the area of a rectangle.

  Use the following formulas:

  • Circle: $A=\pi~r^2$
  • Rectangle: $A=lh$
  • Triangle: $A=\frac12bh$

  Solution

  In the image you can see how we've broken up the composite shape to 3 shapes we already know.

  • Area of the Semicircle

  We start with the purple semicircle. For this we should first determine the radius, $r$. Remember the radius is half the diameter. Therefore we can subtract $4$ from $7$ giving us $3$ (the diameter). Then divide by $2$. This gives us a radius of $1.5$.

  We plug the radius in the formula for a circle:

  $A=\pi~r^2$

  to get

  $A_1=(3.14)(1.5)^2=7.065\approx 7.1$ square units.

  • Area of the Rectangle

  Next, we determine the area of the yellow rectangle. For this we have to determine the length, $l$, and height, $h$.

  $l=10-4=6$

  $h=4-1=3$

  Now we plug those values in the formula for area of a rectangle:

  $A=lh$

  This gives us $A_2=(6)(3)=18$ square units.

  • Area of the Right Triangle

  Now we calculate the area of the blue-green right triangle using the height, $h$, and base, $b$.

  Any side of the triangle can be a base. All that matters is that the base and the height must be perpendicular. The height is the line from the opposite vertex and perpendicular to the base.

  $h=7-4=3$

  $b=10-4=6$

  Then we use the area of a triangle formula: $A=\frac12bh$

  So we get $A_3=\frac12(6)(3)=\frac12(18)=9$ square units.

  • Area of the Composite Shape

  Last, we add all results to $7.1+18+9=34.1$ square units.

• Find the errors in the calculations for the area of the composite shapes.

  Hints

  You can estimate the lengths as well as the areas using the coordinate plane.

  Use the following formulas:

  • Circle $A =\pi~r^2$
  • Square $A =s^2$
  • Trapezoid $A=\frac12(b_1+b_2)(h)$
  • Triangle $A=\frac{1}{2}bh$

  There are a total of five mistakes.

  Solution

  For composite shapes we need a few formulas. The formulas we need for these figures above are:

  • Circle $A=\pi~r^2$
  • Square $A=s^2$
  • Trapezoid $A=\frac12(b_1+b_2)(h)$
  • Triangle $A=\frac{1}{2}bh$

  Let's do the calculations from top to the bottom:

  $~$

  Square and Semicircle

  The needed lengths are

  • $r=\frac{4-1}{2}=1.5$
  • $s=4-1=3$

  This gives us

  • Semicircle $A =\pi~(1.5)^2\approx7.1$ square units
  • Square $A=3^2=9$ square units

  Adding those areas leads to $A=7.1+9=16.1$ square units for the first composite shape.

  $~$

  Trapezoid and Triangle

  The needed dimensions are for the trapezoid:

  • $b_1=3-(-2)= 3 +2=5$
  • $b_2=4-(-1)=4+1=5$
  • $h=10-6=4$

  and for the triangle:

  • $b=5$
  • $h=2$.

  Plugging those values in the corresponding formula leads to

  • Trapezoid $A=\frac12(5+5)(4)=(5)(4)=20$ square units
  • Triangle $A=\frac12(5)(2)=\frac12(10)=5$ square units

  Lastly, we add those areas together to get the total area for the composite shape $A=20+5=25$ square units.

  $~$

  Trapezoid and Semicircle

  The needed values are

  • $b_1=2-(-1)=3$
  • $b_2=4-(-2)=6$
  • $h=9-6=3$

  So we get for the trapezoid

  $A_{\text{trapezoid}}=\frac12(3+6)(3)=(4.5)(3)=13.5$ square units.

  The semicircle has the radius $r=\frac{b_2}2=3$ and thus we have

  $A_{\text{half circle}}=\pi~(3)^2\approx28.26$ square units.

  So, we still have to add those results. This gives us the total area of the composite shape

  $A=13.5+28.26=41.76$ square units.

• Find the formulas for the different shapes.

  Hints

  Take the special triangle, a right triangle. This is the half of a rectangle. So the area is the half of the product of both legs.

  A square is a special rectangle.

  The area of a rectangle is given by the formula $A=lh$,

  where $l$ is the length and $h$ the height.

  Solution

  To determine areas of composite shapes you have to memorize several formulas for areas:

  • A circle with the radius $r$ has the area $A=\pi r^2$ where $\pi\approx 3.14$.
  • A square with the side length $s$ has the area $A=s^2$.
  • A triangle with the base, $b$, and the corresponding height, $h$, has the area $A=\frac{1}{2}bh$.
  • A trapezoid has two parallel sides with base lengths $b_1$ as well as $b_2$. The height of the trapezoid is $h$. So we use the following formula to determine the area $A=\frac{1}{2}(b_1+b_2)(h)$.

• Determine how expensive the new school’s playground is going to be.

  Hints

  This composite shape is made up of a triangle and a trapezoid.

  Use the following formulas:

  • Triangle $A=\frac12(b)(h)$
  • Trapezoid $A=\frac12(b_1+b_2)(h)$

  $b_1$ and $b_2$ are the lengths of the parallel sides of the trapezoid.

  For example $b_1=4-(-1)=4+1=5$.

  The height of the trapezoid is $h=12-8=4$.

  The side length of the triangle is $b=b_1=4-(-1)=4+1=5$ and the height is $h=8-5=3$.

  Don't forget to include the correct units in your answer.

  Solution

  This composite shape is made up of a trapezoid and a triangle

  The needed lengths for the trapezoid are:

  • $b_1=4-(-1)=5$
  • $b_2=3-(-3)=6$
  • $h=12-8=4$.

  For the triangle the dimensions are:

  • $b=b_1=5$
  • $h=8-5=3$.

  Now we are able to plug those values in the correct formula

  • Trapezoid: $A=\frac12(5+6)(4)=(11)(2)=22$ square units
  • Triangle: $A=\frac12(5)(3)=\frac12(15)=7.5$ square units

  To determine the cost we have to add those results. So we get the total area $A=22+7.5=29.5$ square units.

  Each square unit corresponds to $100~ft^2$. This gives us the area

  $A=29.5\times 100=2950~ft^2$. Since each square foot costs a dollar the resulting amount the class has to collect is $ \$ 2,950.00$.