SSAT Middle Level Math : How to find a rectangle on a coordinate plane

Study concepts, example questions & explanations for SSAT Middle Level Math

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Example Questions

Example Question #1 : How To Find A Rectangle On A Coordinate Plane

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In the above figure Dr. Robinson's shed is represented by the red square, and the shed has a perimeter of \displaystyle \small 12 yards. Find how much area Dr. Robinson has in his backyard--excluding the area that the shed occupies.   

Possible Answers:

\displaystyle \small 110 \displaystyle \small yards^2

\displaystyle \small 120 \displaystyle \small yards^2

\displaystyle \small 129 \displaystyle \small yards^2

\displaystyle \small \small 111 \displaystyle \small yards^2

Correct answer:

\displaystyle \small \small 111 \displaystyle \small yards^2

Explanation:

To find how much area Dr. Robinson has around his shed in his backyard, first apply the formula: \displaystyle \small \small P=4S, where \displaystyle \small S represents one side of the red shed. Given that \displaystyle \small \small P=12
\displaystyle \small S must equal \displaystyle \small 3 because \displaystyle \small 4\times3=12.

Now you have enough information to find the area of the red shed.
Apply the formula: \displaystyle \small A=S^2
\displaystyle \small A=3^2=3\times3=9

Since the rectangular fence has a width of \displaystyle \small 15 yards and a length of \displaystyle \small 8 yards, apply the formula: \displaystyle \small Area=width\times length in order to find the area of the entire backyard. 
Thus, 
\displaystyle \small Area=15\times8=120

Then simply find the difference between the area of the entire yard and the area of the shed. 

Thus the final solution is:
\displaystyle \small 120-9=111
  

Example Question #2 : How To Find A Rectangle On A Coordinate Plane

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Identify the correct set of coordinates for rectangle \displaystyle \small ABCD.

Possible Answers:

\displaystyle \small (-11,3),(-11,7),(-3,-3),(3,7)

\displaystyle \small (-11,3),(11,7),(3,3),(-3,7)

\displaystyle \small (-11,3),(-11,7),(3,3),(3,7)

\displaystyle \small (11,3),(-10,7),(3,3),(3,7)

Correct answer:

\displaystyle \small (-11,3),(-11,7),(3,3),(3,7)

Explanation:

To identify the correct set of coordinate points for rectangle \displaystyle \small ABCD, note that the correct set must have two sets of matching \displaystyle \small x coordinates and two sets of matching \displaystyle \small y coordinates. The only answer choice that meets these specifications is:\displaystyle \small \small (-11,3),(-11,7),(3,3),(3,7)

Example Question #223 : Ssat Middle Level Quantitative (Math)

Dr. Robinson recently put a rectangular fence around his backyard. The fence has a width of \displaystyle \small 15 yards and a length of \displaystyle \small 8 yards. If Dr. Robinson paid \displaystyle \small \small \$ 12.50 for every yard of fence, how much did the fence cost? 

Possible Answers:

\displaystyle \small \$525

\displaystyle \small \$575

\displaystyle \small \$625

\displaystyle \small \$375

Correct answer:

\displaystyle \small \$575

Explanation:

To find the cost of the fence, apply the formula \displaystyle \small P=2(width)+2(length) in order to first find the length of the perimeter of the fence.

Then multiply the perimeter by \displaystyle \small \$12.50.

Thus, the solution is:

\displaystyle \small P=2(15)+2(8)
\displaystyle \small P=30+16
\displaystyle \small P=46

\displaystyle \small 46\times12.5=575 

Example Question #224 : Ssat Middle Level Quantitative (Math)

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The rectangle shown above has a width of \displaystyle \small 14 and a length of \displaystyle \small 4. Find the area and perimeter of the rectangle. 

Possible Answers:

\displaystyle \small \small A=56 units^2, P=36

\displaystyle \small \small A=50 units^2, P=34

\displaystyle \small \small A=43 units^2, P=26

\displaystyle \small \small A=36 units^2, P=36

Correct answer:

\displaystyle \small \small A=56 units^2, P=36

Explanation:

To solve this problem apply the formulas: \displaystyle \small Area=width\times length & \displaystyle \small Perimeter=2(w)+2(l)

Thus, the solution is:

\displaystyle \small A=14\times4=56
\displaystyle \small P=2(14)+2(4)=28+8=36

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