Answers to Puzzle #70
Showing 1 to 50 of 816 answers:
Answer #1:
From: 7891
School: 1046
The area of the rectangle is 12 units^2.
The problem is asking for the area of the rectangle. To find the
area of the rectangle, I need to find the base and height of the
rectangle. A right triangle is inscribed inside the rectangle. The
hypotenuse of this right triangle is the base of the rectangle. The
measurements of the other two legs are 3 and 4. I used the
Pythagorean Theorem to find the length of the hypotenuse.
3^2 + 4^2 = b^2
9 + 16 = ^2
b^2 = 25
b = 5 units
The base of the rectangle is 5 units.
To find the height of the rectangle, I drew a line that was
perpendicular to segment DC and it intersected point E. This line
made a 90 degree angle on the line segment AB. The point where this
line that I just drew and segment AB intersects will be point F.
Line segment EF is the height of the rectangle because it is
congruent to line segment DA. To find the length of segment EF, I
need to use trigonometry. This makes a new triangle AFE. So AE is
3, I need to find EF, which I will give the variable of h, for
height. I will use sine form point A.
sin A = h/3
h = sin A(3)
I need to find the value of sin A. To do so, I will have to look at
the 3-4-5 triplet right triangle to find the value of sin A. On the
3-4-5 right triangle, sin A is equivalent to:
sin A = 4/5
So I will substitute 4/5 for sin A in the first equation:
sin A = h/3
h = sin A(3)
h = (4/5)(3)
h = 12/5 units.
The height of the rectangle is 12/5 units.
So I need to multiply the base of the rectangle by the height of the
rectangle.
A = bh
A = (5)(12/5)
A = 12 units^2
The area of the rectangle is 12 units^2.
I have found another completely different way to finds the area
of the rectangle. If you take a look at the right triangle that is
inscribed in the rectangle, and the line segment EF that I drew, you
can tell that triangle ABE is exactly half the area of the entire
rectangle. To prove this, You can see that triangle DAE is congruent
to triangle AFE. Triangle AFE is inside the inscribed triangle that
the problem gave me, and triangle is outside of the inscribed
triangle that they gave me, but it is in the rectangle. It is the
same for the other part of the rectangle. Triangle EFB is congruent
to triangle ECB. Triangle EFB is inside the inscribed triangle that
the problem gave me and triangle is not inside the inscribed triangle
that the problem gave me, but it is still in the rectangle. So with
this you can see that the inscribed triangle that the problem gave me
is equivalent to half of the rectangle. So I can just find the area
of the triangle and multiple that by two to find the area of the
rectangle. Okay, now to start finding the area of the triangle.
Instead finding the base and height of the triangle to find the area
of the triangle, I will use Heron’s Formula to find the area of a
triangle. Here it is:
A = Sqrt(s(s-a)(s-b)(s-c))
Sqrt = square root
s = semiperimeter
a = first side of the triangle
b = second side of the triangle
c = third side of the triangle
This is how my work looks like:
s = (a+b+c)/2
s = (3+4+5)/2
s = 12/2
s = 6
A = Sqrt(s(s-a)(s-b)(s-c))
A = Sqrt(6(6-3)(6-4)(6-5))
A = Sqrt(6(3)(2)(1))
A = Sqrt(36)
A = 6 units^2
Area of the rectangle = 2A.
2A = 12 units^2
The answer checks and is the same as the one that I got with the
other way. These are my two completely
Answer #2:
From: 10454
School: 2961
see attachment
see attachment
Answer #3:
From: 9175
10145
School: 47
The Area of Rectangle ABCD is 12 units square
In order to find the area of the rectangle we needed to have two
dimensions, the length and height. We quickly saw the length as 5,
since it was the hypotenuse of the right triangle with legs of 4 and
3 units. Here's that first equation:
3^2+4^2=5^2
root 5^2 = 5
We then saw to find the other side, we would have to simply find the
altitude of the right triangle onto this hypotenuse, and that would
be equal to the side we needed. We knew that a leg of a right
triangle is the mean proportional between the hypotenuse and the
projection of the leg on the hypotenuse, so we simply filled in the
existing data into this theorem. Heres that equation:
(3^2)/5=1.8
This gave us the short leg length, 1.8, of the Right triangle formed
by line AE and the altitude. Since we knew a leg and the hypotenuse
of this triangle, we easily found the length of that altitude
included in this triangle. We simply used pythagoras:
3^2-1.8^2=2.4^2
root 2.4^2= 2.4
With the altitude of triangle AEB now known, 2.4, we only needed to
multiply the altitude by line AB.
5*2.4= 12units^2
Answer #4:
From: 14338
School: 4263
The final answer is 12.
First, we set LineDE as Y, LineDA as X, therefore LineEC will be (5Y).
We use Pythugorean Theorem to find the lenght of LineAB.
3^2+4^2=C^2, we get C=5,which is the lenght of the recangle/Triangle.
In order to fine LineDE, the equation will be: 3^2=X^2+Y^2, another
equation will be: 4^2= X^2=(5-Y)^2 -> 16=X^2+25-10Y+Y^2.
LineDE will be: 9=X^2+Y^2
and we solve the equation.
16=X^2+25-10Y+Y^2
We get Y=1.8, which means LineDE will be 1.8, plug this answer to the
equation of 9=X^2+Y^2, and we get X=2.4. Since the equation of
rectangle is wide*lenght, so 2.4*5 will equal 12.
Answer #5:
From: 13637
School: 243
The area of rectangle ABCD is 12 units.
We all know that the equation for the area of a triangle is:
A = .5bh
or in other word the area of the triangle is half the area of the
rectangle it is in. So, (and I'm not sure why I relized this) relized
that 3 x 4 (12) was the area of the rectangle. I still wasn't sure
however so I checked it by saying 12/5 (the hypotenuse of the
triangle) it was 2.4. I then used 2.4 with 3 & 4, to figure out the
lengthes of the DE and CE respectivly. What I mean by is that, for
example with 3, I did:
3^2 = 2.4^2 + x^2
I then solved for x, and got a length of DE (x). I then did the same
thing to find CE. The lengthes added together to make five.
Answer #6:
From: 10508
School: 2980
12
Drop the perpendicular EH from point E to base AB.
From pisagor relation:
3^2+4^2=AB^2 Hence AB=5
AEB is a right-angled triangle. From Euclid relations
3^2=AH. AB
4^2=BH. BA
AH=9/5
BH=16/5
EH=AH.HB
EH^2=9/5.16/5=(12/5).(12/5)
EH=BC=AD=12/5
Area of the rectangle=AB.BC=5. 12/5=12
Answer #7:
From: 8214
School: 243
I got 12 units squared as the answer to Area of a Rectangle.
First I looked at the triangle, and I figured out that it is a right
triangle.
Second, I realized that the triangle's area was 1/2 of the
rectangle's area.
Third, I found the area of the triangle.
a=bh/2
a=4*3/2
a=12/2
a=6
Fourth, I doubled the triangle's area to find the area of the
rectangle.
a=6*2
a=12
12 units squared
Answer #8:
From: 14232
School: 636
The area of the rectangle is 12 units sq.
Find the area of the triangle and multiply that by 2 since a triangle
has half of the area of a square of rectangle. Use the equation A=
(bh)/2 to find the area of a triangle, where b=3 and h=4 where the
area of the triangle will equal 6 units sq. Next, 6 would be
multiplied by 2 to find the area of the rectangle, which will end up
equaling 12 units sq.
Answer #9:
From: 13713
School: 3041
The area of the triangle is 12 units.
I used trigonometry and geometric mean to solve this problem. The
first thing I did was to use the pythagorean theorem to find the
length. AE squared + EB squared= AB squared. so, 9+16=25, the square
root of 25 is 5, therefore the length is 5. Then I drew a altitude
from E to AB. I used this formula to find the two segments of ABleg(squared)= (Adjacent leg)(hypotenuse) So EA= AB(x). x being the
adjacent leg. x comes out to be 1.8. subtract 1.8 from 5 and so the
other segment = 3.2. This formula helped me to find the altitudealtitude(squared)=(segment1)(segment2), or altitude squared=(1.8)(3.2)
so the altitude, and the width came out to be 2.4. Multiply 5 by 2.4
and there's the area.
Answer #10:
From: 13703
School: 3041
The area of the rectangle is 12 squared units
First you should use pythagoream therom to find the measure of AB.
a(sq)+ b(sq)= c(sq) 3(sq) + 4 (sq) = AB(sq)
9 + 16 = AB(sq)
25 = AB(sq)
5 = AB
Then solve for angle EAF.
Since the information that we are given are opposite and adjacent I
used tangen.
Tan EAF = 4/3 = 1.33 therefor <EAF = 53.13 degrees
Next drop down an altitude fro E to AB. Using the measure of the
angle
we just found, using sine we can find EF.
Sin 53.13 = EF/3
7.99 = EF/3
2.4 = EF
(ABCD) = AB x AD = 5 x 2.4 = 12 units squared
The other way to solve is geometric means
THe geometric mean of EB (sq) = FB x AB
4 (sq) = FB x 5
FB = 3.2
AF = 5 - 3.2 = 1.8
The geometric mean of EF (sq) = AF x FB = 1.8 x 3.2 EF= 2.4
DA = EF = 2.4
Area of rectangle ABCD = DA x AB = 2.4 x 5 = 12 units squared
Answer #11:
From: 13706
School: 3041
Rectangle ABCD is 11.95 units squared.
I used two different ways to find the area. In the first one, I found
side AB using the Pythagorean Theorem. It came out to be 5. Then I
found the measure of angle EAB using tangent. It came out to be 53.1.
Using that, I found the altitude of triangle EAB using sine. It was
2.39. The width of the rectangle is the same as the altitude of the
triangle, and I already knew the length was 5, so I used the area
formula to find 11.95.
In the second way, I also used the Pythagorean Theorem to find the
width of the rectangle, 5. Then I used geometric mean to find the two
segments of line AB, 1.8 and 3.2. I found the altitude of the triangle
using that information and geometric mean, because (1.8)(3.2)=(h)(h)
I square-rooted that and got 2.4 for h, which is the width of the
rectangle and again used the area formula to get my answer.
Answer #12:
From: 13718
School: 3041
The area of the rectangle is 12 units squared.
First I used the Pythagorean Theorum to find the length of side AB.
Side AE (3) squared plus side EB (4) squared equals side AB squared.
3 squared + 4 squared equals 5 squared so side AB equals five. Then I
decided to find the measure of angle EAB by using sine for triangle
AEB. I found the ratio of side EB over AB which is 4/5. Then I
converted the ration into the angle for angle EAB, the measure of
angle EAB is 53.13 so I found the measure of its compliment and that
was 36.87 I used cosine to find the ratio of the adjacent side to the
hypotenuse in triangle DAE. I then solved the problem and found that
side AD equals 2.4. So I multiplied the sides and got 12.
Answer #13:
From: 13899
School: 3041
The area of the rectangle is 12 units squared.
1. To solve this problem I first used the pythagoram thereom. I
found out that 3 squared + 4 squared equals hypotonuse squared. 3
squared plus 4 squared equals 25. The square route of 25 is 5, so the
hypotonuse is 5. The length of the rectangle is 5. To find the width
of the rectangle I used the formula leg squared = hypotonuse times
adjacent segment. So 3 squared =5 times AF. since 3 squared =9 we
can divide each side by 5 and find out that AF= 1.8. I
subtracted 1.8 from 5 to find FB = 3.2. Since the altitude
is the same as the width I found the altitude by doing altitude
squared eequals segment one times segment two. By doing that formula
I found the altitude EF equals 2.4. Therefore the width is 2.4. Now
to find the area I did length times width. 5 times 2.4 equals 12.
The area is 12 units squared.
2. To find it the second way I found the hypotonuse AB the same
way. It found the width through sohcahtoa. I know that tan ofangle
EAB equals ratio 4/3. After I do that I got that angle EAB quals
53.13. After I have that angle I can find the altitude of the
triangle which is equal to the width. The sin of 53.13 equals the
altitude(EF) over 3. I solve for EF and the altitude and width equals
2.4.
THen you multiply 5 by 2.4 to find the area = 12 units squared.
Answer #14:
From: 14096
School: 4173
The area of
rectangle ABCD is 15 square units.
Since AEB is a right triangle, we can use Pythagoras's Theorem to
determine the length of of BA by 3^2 + 4^2 = BA^2.
BA = 5 since this is a common Pythagorean triple.
Let the
Let the
Let the
because
length of side AD = Y
length of segment DE = X
length of segment EC = 5 - X since the length of DC is 5,
in rectangles opposite sides are of equal length.
In rectangles all angles are right angles; thus,
triangle ADE and triangle BCE are both right triangles.
Using Pythagoras's theorem again we have
X^2 + Y^2 = 8^2
(5-X)^2 + Y^2 = 9^2
X^2 + Y^2 = 9
25 -10X +X^2 + Y^2 = 16
Multiplying the top equation by negative one allows elimination of
terms with the second:
-X^2 - Y^2 = -9
25 -10x + X^2 + Y^2 = 16
Adding the two equations together produces desired elimination:
25 - 10X = 7
-10X = -18
X = 10
We now have the length of segment DE which is 1.8
Using Pythagoras's theorem for last time gives
(1.8)^2 + Y^2 = 3^2
3.24 + Y^2 = 9
Y^2 = 9
Y =3
Since area of a rectangle A = LW
and L = 3 and W =5
Area of rectangle ABCD = 15
Answer #15:
From: 11422
School: 3244
The area of ABCD is 12.
I)Applying the Pythagorean theorem on ABE:
AB^2=3^2+4^2
AB=sqrt(9+16)=sqrt(25)=5
AB is a right triangle so its area (A(ABE)) is EAxEB/2=(3x4)/2=6.
If we divide the double of the area of a triangle by a side, we get
the height related to that side, so (drawing EH_|_AB):
C____E____D
|
/|\
|
| / | \ |
| / | \ |
|/___|___\|
A
H
B
EH=2A(ABE)/AB=(6x2)/5=12/5
The area of ABCD (A(ABCD)) is:
A(ABCD)=ABxEH=5(12/5)=12
II)If we look at the above picture, AE and EB are the diagonals (axis
of simmetry) of the rectangles AHEC and HBDE, so:
A(AHE)=A(AEC) and A(HBE)=A(BDE)
adding member to member:
A(AHE)+A(HBE)=A(AEC)+A(BDE)
but A(AHE)+A(HBE)=A(ABE) and A(AEC)+A(BDE)=A(ABCD)-A(ABE), so:
A(ABE)=A(ABCD)-A(ABE)
2A(ABE)=A(ABCD)
but A(ABE)=(3x4)/2=6, so:
A(ABCD)=12
Answer #16:
From: 13680
School: 2689
The area of the rectangle is 12.
I used similar triangle method to solve this problem, though I was
unable to find the other one, due to the limited ability of mine in
geometric field.
First, I started off by finding the segment AB, which is 5, since
it's a "special triangle" with the length of 3-4-5.
Then, I started to find the relations between triangle ADE, ECB,
and AEB. They are all similar triangles, at the first glance, then
the work began.
First, I compare segment AE to segment AB, which each representing
the hypotnuse of their own distinct triangle. Then, I found out that
the ratio of the length of triangle ADE and AEB is 3/5, since the
hypotnuse's ratio is 3:5. Then, I compared segment EB to AD. Since
the ratio of the triangle ADE and AEB is 3/5, I multiplied 3/5 by 4,
therefore gave me 12/5, and that is the length of AD.
And to further prove that the width of the rectangle is 12/5, or
2.4, I compared AEB to ECB. The ratio of triangle ECB to AEB is 4/5
since segment EB and AB each represent the hypotnuse, and EB = 4 and
AB = 5. Then, I multiplied 4/5 by 3 since the base of BC is the width
of the rectangle and 4/5(AE) = BC. BC now is 12/5, or 2.4
Therefore, the width is 2.4.
Then, we multiply 2.4 by 5 = 12.
The area is thus 12.
Answer #17:
From: 11337
School: 1646
The area of the recangle is 12.
First, I looked at triangle AEB and saw that it was a 3-4-5
triangle. Therefore, AB = 5. Then I drew an altitude from vertex E
of the right triangle to AB. Now we can say that 3 is the geometric
mean between the two segments the altitude splits the hypotenuse
into. If we call the altitude segment EF, we can say 3/AF = 5/3.
From there we can say that 9 = 5 x AF, or that AF = 9/5. Next oI
looked at triangle DEA and saw that it is a right triangle with one
leg having a measure of 9/5 and the hypotenuse having a measure of
3. From the pothagorean theorem, I was able to say that
3(squared) = 9/5(squared) + AD(squared), or 9 = 3.24 + AD(squared).
I then found AD = 5.76. since this is the other side of the
rectangle, the area = 5.76 x 5 which = 12.
Answer #18:
From: 7911
School: 2563
The area of ABCD is 12 unit sqr.
Here are two ways to solve this problem:
1) Since the base of the triangle ABE is the length of the rectangle
ABCD and the height of ABE (with respect to AB) is equal to the width
Of ABCD it follows that {area of ABCD}=2*(area of ABE}=3*4=12.
2) Here I produce the dimensions of the rectangle ABCD. The base of
the right triangle ABE is equal to 5 (5^2=4^2+3^2)which is also the
length of ABCD. Now let EH be the height of ABE(H lies on AB). Since
the triangles ABE and HEB are similar(Both are right triangles and
share angle B, which means that their respective 3rd angles have to be
the same), 3/5=EH/4 or EH=2.4. So the size of ABCD is 2.4 by 5 and
therefore its area is 12.
Answer #19:
From: 9768
School: 11403
I think that the area of the rectangle is 12 units^2.
Line AB is 5 units long because it's a right triangle that is a
Pythagorean triple. So the area of triangle EAB is 6 units^2 because
the area of a triangle is 1/2 times 4(base) times 3(height). Since
the hypotenuse of BCE is 4, the shorter leg, CB, is 2 because the
hypootenuse is supposed to be twice the length as the shorter leg.
And CE has a length of 4.5 units because DE has a length of 2.5
units. The reason why DE is 2.5 units long is because the hypotenuse
of triangle DAE is 3 units long. And the hypotenuse is twice the
length of the shorter leg. So the area of triangle DAE is 2.5 units
ans the area of the triangle CEB is 4.5 units. And all the areas add
up to 10 units. 2.5 + 4.5 + 5 = 12 units^2.
Answer #20:
From: 7292
School: 664
The area or the rectangle is 10 units squared.
First I used the pythagoream theorm to fing the legth of the
rectangle. The theorm is a squ.+b squ= c squ. The answer I got was
5. To find the width of the rectangle I used the 30-60-90 conjecture.
This conjecture is the shortest side is x. Then, the hypotenuse is
2x. The other side is x times the square root of 3. Using the
conjecture to find the x of the triangle on the right. I got the
answer of 2. Next, I multiplied 2 times 5 to get the area, which was
10 units squared.
Answer #21:
From: 11172
School: 1978
The area of ABCD is 12 units^2.
Using the Pythagorean triple one finds the hypotenuse or length of
rectangle to be 5. Then I used the theorem: If the altitude is drawn
to the hypotenuse of a right triangle, then the measure of a leg of
the triangle is the geometric mean between the measures of the
hypotenuse and the segment of the hypotenuse adjacent to that leg.
Therefore, I drew altitude EZ. I then gave the segment adjacent to
the 4-unit leg the value of X. The other segment I called 5-X. Then:
SQRT (5*X) = 4
SQRT (5X) = 4
5X = 16
X = 3.2
5-X = 1.8
The altitude of this triangle would be perpendicular to the bottom
segment of the rectangle. Since the top and bottom of the rectangles
are parallel, this means that the altitude is equal to the width of
the rectangle because they are also perpendicular to the top and
bottom segments. I now use this theorum: If the altitude is drawn to
the hypotenuse of a right triangle, then the geometric mean between
the measure of the two segments of the hypotenuse is the length of
that altitude. SO:
SQRT (1.8*3.2) = Altitude
SQRT (5.76) = Altitude
2.4 = Altitude
That is the measurement of the width of the rectangle also. So using
the area formula I find the area of the rectangle
Area = L*W
Area = 5*2.4
AREA = 12 units^2
Answer #22:
From: 7252
School: 664
The area of rectangle ABCD is 12 units squared.
To solve this problem, I found the area of the triangle within the
rectangle. I did this by using the given information about it's base
length and it's height, 3 units and 4 units, and multiplying them
together and then dividing by two (according to the triangle area
formula). Since I knew that a triangle is always half of a rectangle,
I simply multiplied my answer, 6 units squared, by two to come up with
my answer, 12 units squared.
Answer #23:
From: 12766
School: 24
Area = 12.5 units
T arrive at my answer I took some paper and traced the lengths of the
rectangle and then the lengths of the triangle. Then I noticed that
line segment AD is a little smaller than line segment AE so I
concluded that it equalled 2.5. Then I noticed that line segment DC is
a little larger than EB but resembled a whole unit so I said that it
equalled 5. Then I multiplied the two numbers together ( A = 2.5 * 5)
and got my answer A = 12.5 units.
Answer #24:
From: 13888
School: 24
My answer would be the area of ABCD would be 48.
i just added 3+3 b/c that was the short side of the triangle, it
looked equivalent to the short side of the rectangle, and the long
side of the triangle which is 4, i added 4+4 b/c it looked equivalent
to the longer side of the rectangle so i got 6 and 8 as my two sides.
Then i multiplied 6*8 b/c that is the two lengths, so i multiplied
them to get the area, which equals 48.
Answer #25:
From: 14343
School: 4266
The area of the rectangle ABCD = 12 units²
There are a few solutions to that problem:
1) - please refer to picture Nr. 1)
A right-angled triangle is in a rectangle. The side AE = 3 and
BE = 4. To calculate the area of the rectangle ABCD you need to
the lengths of the sides a and b. With the help of
Pythagoras' theorem it's pretty easy to get b:
b² = 3² + 4² = 25
--> b = 5.
Now we have to figure out the length of the side a. But how ?
Well, firstly you need to mark the altitude of the triangle ABE
altitude = a).
By doing that, the side b is diveded in(to) two sides: q and p.
current relations of q and p are:
p + q = 5
--> p = 5 - q
q = 5 - p
Both triangles AFE and FBE are right-angled. Because of that we
use Euklids' theoreme:
4² = p * b
--> p = 16/b = 16/5 = 3,2
It follows from that that
q = 5 - p = 5 - 3,2
q = 1,8
Right now you can calculate the altitude (the side a), by using
altitude theorem:
h² = p * q = 1,8 * 3,2
h² = 5,76
--> h = a = 2,4
We know that the area of a rectangle is a*b
==> A = a * b
A = 2,4 * 5
A = 12 units²
know
(the
The
can
the
2) - please refer to picture Nr. 2)
ABCD is a rectangle. In that rectangle is a right-angled triangle ABE
The side AE = 3 and BE = 4.
With these details (3 and 4 units) you can calculate the side b by
using Pythagoras' theorem:
b² = 3² + 4² = 25
--> b = 5
After you have done that you need to mark the angles by Alpha,etc.
(if you haven't done that already). The angle <EAB is Alpha , <EBA is
Beta , <EAD is Beta , <AED is Alpha , <EBC is Alpha , <BEC is Beta
--> Alpha + Beta = 90° !
Because of the current relations in a right-angled triangle, you can
calculate Alpha (and Beta), by:
sin(Alpha) = 4 / b = 4 / 5 = 0,8
--> Alpha = 53,13°
so Beta = 90° - Alpha = 90° - 53,13° = 36,87° !
Right now you can calculate the length of the side a, by :
sin(Beta) = a / 4
--> a = sin(Beta) * 4 = sin(36,87°) * 4
--> a = 2,4
So the area of the rectangle ABCD is
A[ABCD] = a*b = 2,4 * 5
A[ABCD] = 12 units²
So the area of the triangle ABE is half as much as A[ABCD]
A[ABE] = 0,5 * A[ABCD]
A[ABE] = 6 units²
3) - please refer to picture Nr.1)
You have the rectangle ABCD. Divide it in(to) two rectangles by
drawing the altitude of the rect-angled triangle.
Right now there are two rectangles (AFED and FBCE) and in both ones
are diagonals. The one diagonal AE is 3 units long, the other one BE
is 4 units long. It is known that a diagonal divides a rectangle into
two congruent triangles. It follows from that that the area of the
triangle ABE is the half of the area of the rectangle ABCD! The area
of the triangle ABE is
A[ABE]= (3*4) / 2 = 6 units²
===> The area of the rectangle is
A[ABCD] = 2 * A[ABE] = 2 * 6 units² = 12 units²
Answer #26:
From: 13293
School: 2152
The area of the ABCD is 12.
From the pictute we can see that the area of ABCD is 2 times bigger
then area of triangle ABE.
A1...area of ABCD
A2...area of ACE
A1=2*A2
We have 2 sides of the triangle so we can calculate the 3rd
with Pitagora's formula.
AB...c
sqr(c)=sqr(3)+sqr(4)
sqr(c)=9+16
sqr(c)=25
c=sqrt(25)
c=5
side
Now we can calculate the area of the triangle with Heron's formula.
s=(a+b+c)/2
s=12/2
s=6
A2=sqrt((s*(s-a)*(s-b)*(s-c))
A2=sqrt(6*2*3*1)
A2=sqrt(36)
A2=6
A1=2*A2
A1=2*6
A1=12
The area of the ABCD is 12.
Answer #27:
From: 14348
School: 3237
The area of the rectangle is 12.
Drop an altitude from
is congruent to EXA.
Therefore the area of
rectangle. Since the
area of the rectangle
point E to imaginary point X. Then triangle ADE
Also triangle ECB and BXE are congruent.
triangle AEB is one half the area of the
area of the triangle AEB is 0.5(3)(4) = 6, the
is (6)(2)=12.
Answer #28:
From: 13356
School: 3933
The area of the rectangle is 12 sqaure units.
Automatically since triangle AEB is a right angled triangle, AB being
the hypotenuse must be 5 (square root of 3^2 + 4^2). Also, since it is
a rectangle, DC must be 5, and AD and BC must be the same length.
Letting x rep DE and y rep AD, we know 3 things:
1: EC=5-x
2: x^2+y^2=9
and 3: EC^2+y^2=16
solving,
(1) x^2+y^2=9
(2) x^2+y^2 +25-10x=16 (by subbing 5-x in for EC)
replacing x^2+y^2 in (2) with 9,
34-10x=16
10x=18
x=9/5
subbing x into (1),
(9/5)^2 + y^2=9
y^2=225/25 - 81/25
y^2=144/25
y=12/5
since y is the height and the length (AB) was previously determined to
be 5, the area is simply the product or 12 sqaure units.
Answer #29:
From: 13635
School: 11403
i worked it out two different ways and got 12 units squared for both answers.
On the first one i just cut the triangle that you gave us the numbers for at
the point e and figured that each half of the triangle was one half of the
rectangle and i got 12 out of that. Then i did x2 plus z2 =4 squared ninus x2
plus y2=3 squared then i got z2 - y2=7 so then i factored it out and i got z
plus y =5 and z-y =7/5 so then i minus that and got -2y= -3.6 and i square
rooted it and got 1.8.
Then i just used pythagream theoream and figured the
sides out and got 12 to be the area.
Answer #30:
From: 12981
12980
School: 1329
We used the pythagorean thereom on the triangle and rectangle.
Okay to get the answer 5=c, we used the pythagorean thereom which is
a2+b2=c2. Then subsituted 3 for a, 4 for b, and whe had to find c. So
we squared both 3 and 4 and got 9+16=c2. Then added 9 and 16 and got
25=c2 so then I square-rooted 25 and got an answer of 5=c.
and how I got 10x=18. I took 4 squared in which then I got (5x)squared+y2+4squared. Then I foiled it all out to get 25-5x-5x+
xsquared+ysquared=16. Then I added all the like terms to get 25-10x+x2+
y2=16.
For 3 squared, It was x2+y2=3squared, and got x2+y2=9
then I took 10x+x2+y2=9 and came up with my final answer 10x-18.
Answer #31:
From: 8582
13022
School: 1329
For the area of the rectangle we got 12 sq. units.
To find the area of the triangle I used the pathagorean Theorem.
also used foil. I would explain it all, but my hands are tired.
I
Answer #32:
From: 12976
School: 1329
The area of this rectangle can be figured by using the pathagorean
theorum.
In order to figure out the area of the rectangle you need to figure out
the measure of each side.
You will need to use the pathagorean theorum (a2 + b2 = c2)
3^2 + 4^2 = c2
9 + 16 = c2
square root of 25 = 5
The bottom side of the rectangle equals 5
You will need to use this measurement to find the length of each
segment on the top side.
Segment DE will be Y
EC = 5-Y
DA = X
(5-y)2 +x2 =4^2
Foil out 5-y
This is your new equation
25 - 10y +y2 + x2 = 16 Then subtract 25 from both sides
Now you can subtract your first equation (x2 + y2 = 9)
This leaves you with -10y = -9-9
-10y/-10 = -18/-10
x2 + 1.8^2 = 9
x2 + 3.24 = 9
-3.24 -3.24
y = 1.8
Then square root x2 and 5.76
X = 2.5
Answer #33:
From: 14135
School: 54
The area of the rectangle is 12 units squared.
The first thing I realized was that the two shapes shared a side
and angle E was on the opposite side. So this means that the triangle
AEB equals 1/2 the total area of the rectangle ABCD. I multiplied 3
times 4 which equals 12, and then I divided 12 by 2 to find the area
of the triandle. Since I know that 6 is 1/2 the area of the rectangle
I multiplied 6 by 2. The answer was 12, and since I am finding area it
is 12 units squared.
Another way to solve the problem:
You can find the perimeter of the rectangle and solve the problem that
way. If you use the pythagorean theorm you can find the length og one
side. 3 squared + 4 squared = C squared. C refers to the line segment
AB. If you know that the area is 12 already and you want to find the
height, you simple figure what multiplied by 5 equals 12. If you
multiply 5 by 2.4 then you get 12.
Answer #34:
From: 10951
School: 2689
The area of the rectangle is 12 square units.
I did the Pythagorean Theorem to find the length of AB.
3^2+4^2=x^2
9+16=x^2
25=x^2
5=x
AB is 5 units. DC is also 5 units. Since DA=BC, I made a new
diagram which would be easier to solve. I used the Pythagorean
Theorem to find the sides of the triangle, and I got these two
equations.
4^2-y^2=x^2 and 3^2-(5-y)^2=x^2
square root 16-y^2=x and square root 9-(25-10y+y^2)=x
I substituted the x for the x.
square root 16-y^2=square root 9-(25-10y+y^2)
^2
^2
16-y^2=-16+10y-y^2
+y^2
+y^2
16=-16+10y
+16 +16
32=10y
/10 /10
y=3.2
Now that I know that y=3.2, all I need to know is x so that I can
find the area, so I used the Pythagorean Theorem to find x.
4^2-(3.2)^2=x^2
16-10.24=x^2
5.76=x^2
2.4=x
Now that I have x, all I have to do is multiply x by 5.
2.4*5=12
The area of the rectangle is 12 square units.
Answer #35:
From: 13020
School: 1329
The area of the rectangle is 12 sq. units.
By using the pthagorean theorem, I figured this problem out.
(a^2 +
b^2 =c^2) First, I filled in the letters with numbers as follows and
using the big triangle in the middle to figure out the width of the
rectangle. I filled the numbers in as follows: 3^2 + 4^2 =c^2, 9+16=
c^2, 25=c^2, c=5. 5 is the width of the rectangle. Then I needed to
figure out what segment DE and DA were. We used the pythagorean
theorem. I named segment DE, X. I named segment DA, Y. I used the
pythagorean theorum again. I got x^2 + y^2 =3^2 or 9. That was the
area of the little triangle in the left corner. Then I took 5-x^2+y^2=
4^2. (5-x)(5-x), 25-5x-5x+x^2=4^2, 25-10x+x^2+y^2=16,From this answer
I subtracted x^2+y^2=9 Then I got 25-10x=7, I subtracted 25 from both
sides, -10x=-18, I divided by 10 on both sides, x=1.8 .
this was to
figure out what segment DE was. DE = 1.8
The nI needed to figure out what Da was. I used the pythagorean
theorem agaim. 1.8^2 +y^2=3^2, 3.24+y^2 =9, y^2=5.76, y=2.4
After that I figured out that segment EC would have to be 5-x because
5 was the whole width of the rectangle.
To figure out the area of this problem i needed to multiply legnth
times width. A=5x2.4
Area = 12 square units.
Answer #36:
From: 7300
School: 664
The area of rectangle ABCD is 12 units squared.
Since the triangle takes up half the area of the rectangle you find
the area of the triangle by base times height and dividing in half and
then doubling it because it takes up half the are of the rectangle.
3*4/2=6*2=12units squared.
Answer #37:
From: 7288
School: 664
The area of ABCD is 12 sq.
First you do the height times the width and you get 12 then divide it
by two because it is only halp of a rectangle. then cut the picure in
half and realize that the rectangle is half of the total area. Then
you times it by two and you get the area of 12.
Answer #38:
From: 7298
School: 664
The area of the rectangle ABCD is 12 square units.
since the triangle is half the area of the rectangle, you just don't
divde the triangle area by two and then you get the rectangle's area.
Answer #39:
From: 13027
13028
School: 1329
The area of the rectangle is 12.
I found the answer to this by finding the measures of the sides by
using the pathagorean theorem. I then used those answers and took the
length times the width and that was 2.4 times 5 and that equaled the
answer 12.
Answer #40:
From: 12868
School: 2535
The area of the rectangle is 12 units.
Well, since the area of a rectangle is LxW, I came the the conclusion
of 3X4. This would make the answer 12. Since there is no indicated
units, my answer is 12 units. The right angles are irrelevant.
Answer #41:
From: 7499
School: 2490
The area of the rectangle is approximately 12 square units.
Using Pythagorean's theorem, I discovered that the length of the
longest side is 5 units. (a^2 + b^2 = c^2
3^2 + 4^2 = c^2
9 +
16 = 25 = c^2 {c = 5}). Then I used some basic trigonometry to
discover the measure of angle EBA. It is approximately equal to
36.9º units. Angles BEC, EBA, and EAD are the same, so all of them
equal 36.9º units. 90º - 36.9º = 53.1º So, angles DEA, EAB, and ECB
are all approximately equal to 53.1º. Again using trigonometry, the
lengths of EC, ED, DA, and CB can be discovered. DA = CB 2.4 units;
EC = 3.2 units; ED = 1.8 units (all of these numbers are
approximate). To find the area of the rectangle, we multiply base
times height to get [5 * 2.4 = 12*]. So the area of the rectangle is
approximately 7.4 square units.
*NOTE: It appears I added instead of multiplying on that last
equation. I have fixed that(-:
Answer #42:
From: 9728
School: 11403
The area of the triangle is 12 units ^2.
First, I thought about how the Pythagorean Throrem says that a^2 + b^2 = c^2.
Since a=3 and b=4, c^2 = 25. If you take the square root of both c^2 and 25
you get c=5. So one of the sides is 5. But you still don't know the length of
the other side. You could find the side by finding the length of cb or da by
trying to find the numbers that satisfy the equations that would find the
area for ade and ecb. So then I thought about how if you find the area of the
triangle by using the formula a=1/2 bh but don't multiply by 1/2 then you have
the formula for a rectangle. ( The reason you divid by two is beacause two
congruent triangles can always form a rectangle.) So if I plugged in the
lenghts of the right triangle, aeb, then the formula would look like this:
a=1/2(3)(4)x 2. If you multiply the formula out you would get 12 units ^2. And
since you know that one side is 5 units you know that the other side is 2.4
units.
Answer #43:
From: 12764
School: 24
The area of rectangle ABCD is 13.75 .
I found this by printing out the problem on a sheet of paper. I traced only
the triangle on a piece of tracing paper and rotated it using point A as a
fixed point. By rotating it I found that the side that measured 4 was
shorter than side DC and the side measuring 3 was longer than side AD. I
estimated that side DC was 5 units and it turned out to be right if i split
the side into 5 sections. to find side AD i made a proportion: 5/4 = x/3 ( x
being side AD) when i cross multiplied i got that side AD equals 2.75. since
i had my length and width now all i had to do is multiply. I multiplied 2.75
by 5 and got 13.75.
Answer #44:
From: 9637
School: 2404
The area of rectangle ABCD is 12 square units.
First I found side AB to be equal to 5 units using the pythagorean
theorem.
I will use this information later.
I next decided to use trigonometry to find side AD.
Angle EAB = invsin (4/5) by definition of sin Angle EAB =
opposite/hypotenuse.
Because ABCD is a rectangle, Angle DAB is 90 degrees. This makes
Angle DAE = 90-invsin (4/5).
Trying to find side AD, I used
cos Angle DAE = Adjacent/ hypotenuse=AD/3
Solving for AD and substituting for angle DAE, I got
AD= 3*(cos(90-invsin(4/5)))
The area of rectangle ABCD is length times width or AB*AD.
Substituting I get area of rectangle ABCD equal to
5 * 3*(cos(90-invsin(4/5))) =12 square units.
Answer #45:
From: 14351
School: 4272
it is just 3*4=12 because the rectangle is made up of 2 equivalant
triangles with area .5 b*h with b=4 and h=3. You could also drop a
perpendicular from E and it would be the height of the recangle,
knowing that DC is 5, using pythagorian thm make a system of eqns to
solve for that line from E, because it makes up one side of the
triangle that 4 is the hypotnuse of and one side of the triangle that
3 is the hypotnuse of. The other sides can be written as, like DE=x
and the EC=5-x. Then put those into the pythagorian thm and solve for
x and then plug it into the other equation which is now in terms of a
single variable, and then you have the height and multiply it by the
base=5 to get 12
oops, its above
Answer #46:
From: 8063
School: 2447
The area of rectangle ABCD is 12.
Since AE=3 and EB=4, AB must equal 5 because of the Pythagorean
Theorem. Since AB=5, then DC must equal 5 because it's a rectangle.
Then, I made DE=x and EC=5-x. I made DA and CB equal to y. For
triangle EDA, I figured out that y2=9-x2. for triangle BCE, I
figured out that y2=16-(5-x)2. Since they both equal y2, I set them
equal to each other. I solved to find out that x=1.8. I plugged
that back into the y2=9-x2 equation and found that y=2.4. Now that I
know that the sides are 2.4 and 5, I can find the area by multiplying
them together and getting 12.
Another way to solve....
Anytime you have a rectangle, and you form a triangle inside it so
that one side of the triangle is the same as one side of the
rectangle, and the third vertex of the triangle lies on the opposite
side of the rectangle, then the area of the triangle is half the area
of the rectangle. In the diagram for this problem, it doesn't matter
where point E is on segment CD, as long as it is on that segment. If
you drop an altitude from E to the base AB, that altitude of the
triangle will be equal in length to the other sides of the rectangle
(AD and BC). Since the area of the triangle is .5*AB*h, and the area
of the rectangle is AB*AD, and h and AD are equal, then the area of
ABCD is 2 times the area of triangle ABE. Then, the area of triangle
ABE is easy to find, because it's a right triangle (you don't need to
use the altitude from E to AB). So, the area of triangle ABE
is .5*3*4=6. The area of the rectangle ABCD is 12.
Answer #47:
From: 14357
14356
School: 3484
10 is the area of the rectangle
the pythagorean triples showed us that the base is 5 because the other
2 sides are 3 & 4. A 30 60 right triangle is the same product as a
pythagorean triple so if 4 divided by 2 is 2 5 times 2 equals 10!
Answer #48:
From: 14353
School: 3484
The area of rectangle ABCD is twelve square units.
First, I drew a line segment from point E so that it was perpendicular
to side AB. The intersection with side AB is a new point, point F.
This segment divides the rectangle into two pairs of congruent
triangles. Since triangles AEF and ADE are congruent, and triangles
BEF and BCE are congruent, triangles AEF and BEF combine to form half
the area of the rectangle. Triangles AEF and BEF together make
triangle AEB, so the area of triangle AEB is one half of the area of
the rectangle. Since a base (3) and a height (4) of triangle AEB are
given, the area of triangle AEB can be calculated: 3(4)/2. The area
of triangle AEB is six square units. Since triangle AEB makes up half
the area of the rectangle, you multiply the area of triangle AEB by
two:6(2)=12. Therefore, the area of rectangle ABCD is twelve square
units.
Answer #49:
From: 14362
14363
School: 574
We found that the solution equaled 12.
Using the Pythagorean Theorem, we solved the problem. Since we know
that a2+b2=c2, then we took tri. DEA and placed it into tri. AEB.
Next, we also placed tri. BCE into it also. Since line AE = 3 and
line BE = 4, 3x4=12. Since it is a rectangle, we did not divide it by
2, because the smaller triangles fit perfectly into the large
triangle. So once again, are answer is 12.
Thanks.
Answer #50:
From: 14373
14374
School: 574
The area of the rectangle is 12 square units.
We know that the two smaller triangles that are outside of the the
square fit perfectly inside of the bigger triangle. So the area of the
whole rectangle is twice the size of the larger triangle. Since we
know that the area of the larger triangle is Bh/2, and the B= 3 and
the h=4. So the area of one triangle is six, multiplied by two it is
12 because there are in essence two large triangles.
Thank You for your time.
Love, Katie and Lolly
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