Urgent 5 perimeter and area questions

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July 19th 2007, 06:33 PM
sanee66

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Urgent 5 perimeter and area questions

Could you take a look at the attachment and let me know what to do?

Also would this be correct?
[FONT='Arial','sans-serif']NDBA is a rectangle with parallelogram GDFA enclosed. If[/FONT][FONT='Arial','sans-serif'] AN = NG, BF = DB, GD = 11, and DF = 8√2. What is the area of GDFA?[/FONT]
Area = height *base
Base=GD=AF=11
DBF is an isosceles right triangle so DB=FB. To get DF take x2+x2=([FONT='Arial','sans-serif']8√2)2=2x2=128 so x=8[/FONT]
[FONT='Arial','sans-serif']Would this make the area 88?[/FONT]
July 19th 2007, 06:52 PM
malaygoel

1st question

You got AB right. It is 24.
in DBC, BC is 24, hence the sum of other two sides is 50-24=26.
Sincs DBC is isosceles the other two sides are 13 and 13.
Hence the answer is 24+13=37

Keep Smiling
Malay
July 19th 2007, 06:57 PM
malaygoel

2nd question

The are two distinct sides in parallelogram.
One you got right..it is 12.
To find the other, we use perimeter of parallelogram
P=2(a+b)
60=2(12+b)
b=18.

Keep Smiling
Malay
July 19th 2007, 06:59 PM
Ranger SVO

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Can I try #1? Triangle ABC is an equalateral so each side is 24. Triangle DBC is an isosceles with base BC which is 24. The perimeter of Triangle DBC is 50

We have 2 unknown sides + the base which equals the perimeter

So 2*a + 24 = 50 What we should get is a = 13

So AB + BD = 37 because 24 + 13 = 37
July 19th 2007, 07:00 PM
malaygoel

3rd question

You need to know the following thing to solve the question

If O is the mid-point of KN, it impies KO=ON.

Keep Smiling
Malay
July 19th 2007, 08:12 PM
Soroban

Hello, Sanee!

Quote:

C. In the trapezoid KLMN, OP joins the midpoints of two sides, and $KL +MN \:=\:32$

If $KO + PM \:=\:11$ , which is the perimeter of the trapezoid?

. . $a)\;43\qquad b)\;44\qquad c)\;54\qquad d)\;58$

Code:

K L • * • * * * * * * O• * * * * * •P * * * * * * • * * * * * * * * * • N M

They gave us: . $KL + MN \:=\:32$
. . We have the total length of the two parallel sides.

They gave us: . $KO +PM\:=\:11$

Since $O$ and $P$ are midpoints: . $\begin{array}{ccc}KO \,= \,\frac{1}{2}KN & \Rightarrow & KN \,= \,2\!\cdot\!KO \\PM \,= \,\frac{1}{2}LM & \Rightarrow & LM \,= 2\!\cdot\! PM\end{array}$

Hence: . $KN + LM \:=\:2\!\cdot\!KO + 2\!\cdot\!PM \:=\:2(KO + PM) \:=\:2(11) \:=\:22$
. . And we have the total length of the two nonparallel sides.

Therefore, the perimeter is: . $32 + 22 \:=\:54\;\;(c)$
July 19th 2007, 08:26 PM
sanee66

[FONT='Times New Roman','serif']Area of the of square P is 36 square inches and the perimeter of the rectangle R is 144 inches. If a width of the rectangle is two times a length of the square, which is the length of a length of the rectangle?
[FONT='Times New Roman','serif']70,60,45,30[/FONT]
[/FONT]
[FONT='Times New Roman','serif']If the area is 36 then each side is 9 square inches. the perimeter is 144 with a width of 2*9.[/FONT]
[FONT='Times New Roman','serif']Then 144=2(18+b) gets to be b=54 Where am I wrong at?[/FONT]
July 19th 2007, 09:04 PM
Soroban

Hello, Sanee!

Quote:

Area of the of square $P$ is 36 in²
and the perimeter of the rectangle $R$ is 144 inches.
If a width of the rectangle is two times a length of the square,
which is the length of a length of the rectangle?
. . $a)\;70\qquad b)\;60\qquad c)\;45\qquad d)\;30$

If the area is 36, then each side is 9 inches. . ← Here!

The area is 36 in², so the side is six inches.
July 20th 2007, 01:57 AM
sanee66

So the perimeter of the rectangle is 144=2(6+b) 144=12+2b 132=2b b=66 which is still not an answer I can choose from, what now?
July 20th 2007, 02:19 AM
sanee66

duhh
144=2(12+b) 144=24+2b 120=2b b=60:)