The top three steps of a stairway are shown. The stairway has 82 steps with 82 risers that are each 6 cm high. The 82 treads are
each 10 cm wide. If all the stairs were shown in the figure, what
would be the area of the figure?
The solution shown there entailed computing the area of the first rectangle, computing how many rectangles were hidden in the stairsteps and then
multiplying the number of rectangles by the area of the first step.
Here is an alternative method.
Again, focus on the first step and see that it's a rectangle that's 6 units high by 10 units wide. It's area is clearly 60 square units.
Notice that if you snip a triangle off a corner and rotate the snipped triangle that the rectangle transforms to a trapezoid.
Rotating the triangle changed the rectangle into a trapezoid but the area of the new trapezoid and the old rectangle are the same.
You might object that it's not a trapezoid but remember that a trapezoid is a quadrilateral with at least two parallel sides. The figure above
qualifies. Watch it again if you wish.
Apply the same transformation to the top rectangle on each stair step and the stairstep figure in the problem changes into a trapezoid as shown below.
Recall the formula for the area of a trapezoid is `Area_\text(trapezoid) = h *(a+b)/2`
The distance between the two unequal parallel sides is , `h`, is computed by
`h=\text(number of steps) * \text(width of one step)`
The idea that `h` can stand for width may confuse you. When you first encountered the area of a trapezoid formula, the `h` represented the height.
This trapezoid has just been rotated 90 degrees so `h` has turned into the width. The key thing to remember is that the `h` represents in the formula is the distance between side a and side b.
The shorter side, `a`, is half as high as the orignal riser so it's found by
The half riser height tacked on at the end is the height of the red triangle.
We set up the `(a+b)/2` term by plugging in the above expressions for a and b:
` (a+b)/2 =` ` ( \text(riser height)/2 + ( \text(# of steps on the left side) * \text(riser height)) + \text(riser height)/2)/2`
Notice that we're adding half a riser height at the beginning and half a riser height at the end so we can simplify: the `(a+b)/2` term to:
` (a+b)/2 =` `( ( \text(# of steps on the left side) * \text(riser height))+ \text(riser height))/2`
Recall that
`Area_\text(trapezoid) = h*(a+b)/2` and
`h =\text(number of steps) * \text(width of one step)`
plug in the numbers:
`Area_\text(trapezoid) = (82 *10)* (82 * 6 + 6)/2`
`Area_\text(trapezoid) = 820 * 498/2`
`Area_\text(trapezoid) = 204180`
Problem 2:
A train has 25 seats of which 18 are broken. The conductor needs to fill out a form that identifies the broken seats.
If she randomly identifes the 18 broken seats, what is the probability that she identified all the correct seats?
Express your answer in scientific notation to 3 digit accuracy.
Recall that
`\text(Probability) = \text(Desiried)/\text(Total)`
In this problem the number of ways of getting the desired outcome is one. There's only 1 way to correctly mark the form.
The total number of ways marking the from is `\ _25 C_18` .
The reason it's not `\ _25 P_18` is it doesn't matter in what order she marks the broken seats. Order doesn't matter..
Pro Tip!
Use `\ _n P_r` when order matters. Use `\ _n C_r` when order doesn't matter.
The answer then is found by computing `1/(\ _25 C_18)` or `2.08*10^(-6)`
Problem 3:
Xavier and Sean are playing the Shade Game. Two standard dice are rolled, and the sum of the numbers shown is the number of blank squares
that must now be shaded in.
It is Sean‛s turn, and the grid is shown here. What is the probability that Sean will win on this turn by rolling the exact sum
needed to shade the rest of the grid? Express your answer as a common fraction.
There are 10 white squares left. That means Sean has to roll a 10 to win.
Count the number of ways 2 dice can add t0 10 to get the desired outcome count..
Imagine Sean has a green die and a red die. The first number is the green die's roll and the second is the red die's roll. He can roll:
6 and 4 or 5 and 5 or 4 and 6or
So there are exactly 3 ways he can roll a 10. The probability of rolling a 6 is 1/6 and the probability of rolling a 4 is 1/6 so the probability of rolling a
6 anda 4 is:
`p=1/6*1/6`
`p=1/36`
The other two rolls have the same probability of occurring, i.e., 1/36.
Since the three rolls are or events you add their probability to get `3/36` which you can simplify to `1/12`.
Problem 4:
What is the single discount that is equivalent to the two successive discounts of 17% off followed
by 22% off the discounted_price? Answer to the nearest hundredth of a percent.
You can pretend the price is $100 and apply the discounts. The number you have at the end is the
price you would pay after the discounts. The single discount then would be
`100 - (\text(price you would pay))`
It doesn't matter what price you choose to start with.The single discount is found by computing
`\text((the price you started with)) - ` `(\text(the price you would pay after applying the discounts to the starting price)) `
We'll compute the answer by assuming the starting price is $100.
A 17% discount on $100 is $17 so we have a new price of $83.
We then compute the second discount:
`$83*22/100 = $18.26`
`$83-$18.26 = $64.74`
$64.74 is the price after the two discounts so the single discount that would give you a final price of $64.74
if you started at $100 is
`$100 - $64.74 `
or 35.26% Notice the "%" follows the number.
Problem 5:
A nursery employee wishes to plant
6 Golden Delicious apple trees,
3 Bartlett pear trees and
2 Harcot apricot trees in one row. How many distinct arrangements are possible?
Express your answer in scientific notation to 3 digit accuracy. Show the setup in the space below.
This is a Monsoon problem.
There are 6 Goldens + 3 Bartletts + 2 Harcots or 11 total trees.
If the trees were distinct, there would be 11! ways to arrange the trees.
Since the 6 Goldens are not distinct, you have to divide 11! by the ways you can arrange the
6 Goldens or 6!.
Similarly, you divide by the number of ways you could arrange the 3 Bartletts (3!)and the 2 Harcots (2!)
Your caculator setup is as follows:
`((6+3+2)!)/(6!*3!*2!)`
The parenthesis around the denominator is necessary to ensure the 3! and 2! divide the numerator.
An alternative setup could be:
`(6+3+2)!\text(/6!/3!/2!)`
or:
`11!\text(/6!/3!/2!)`
The key point is to be sure to divide the ways that the total number of trees can be
arranged by the number of ways each kind of tree can be arranged.
