I have four questions:
1. The area of trapezoid ABCD is 60. One base is 4 units longer than the other, and the height of the trapezoid is 5. Find the length of the median of the trapezoid.
2. In trapezoid ABCD, line BC is parallel to line AD, angle ABD = 105 degrees, angle A = 43 degrees, and angle C = 141 degrees. Find angle CBD, in degrees.
3. The bases of trapezoid ABCD are lines AB and CD. Let M be the midpoint of line AD. If the areas of triangles ABM and CDM are 5 and 17 respectively, then find the area of trapezoid ABCD.
4. Let ABCD be a parallelogram. Extend line BC past B to F, and let E be the intersection of lines AB and DF. If the areas of triangles BEF and ADE are 1 and 9, respectively, find the area of parallelogram ABCD.
Man leave the boy alone GOD
People are always bullying these days, what you could have said s "Hey man this website is for helping each other not cheating so can you try and show us your thoughts." Not "I'm sick and tired of your cheating on our homework problems! Your account has been deactivated." DO YOU UNDERSTAND ME
How would you feel if you got bullied. Think about it.
Yes, I agree. However, HOW IS IT POSSIBLE THAT AOPS CAN TRACK YOUR COMPUTER?! They have said that they can and will see your IP address, except I'm skeptical about their ability to do that.
IKR He's Stupid
lol im questioning if the guy is even from AOPS
Go ahead & deactivate half of your users because around that much doesn't care about it on your public board, oh yeah including me.
I'm 6 years too late, but...
1. Let the median of the trapezoid be $m$. Then by the formula for the area of the trapezoid,
\[5m = 60,\]so $m = \boxed{12}$. (The fact that one base is 4 units longer than the other is irrelevant.)
2. (This is written using different variables, but I think you can figure it out) Since $\overline{XY}$ and $\overline{WZ}$ are parallel, we have $\angle WXY = 180^\circ - \angle W = 180^\circ - 43^\circ = 137^\circ$.
[asy]
unitsize(1.5 cm);
pair A, B, C, D;
A = (0,0);
B = (0.5,1);
C = (2,1);
D = (3,0);
draw(A--B--C--D--cycle);
draw(B--D);
label("$W$", A, SW);
label("$X$", B, NW);
label("$Y$", C, NE);
label("$Z$", D, SE);
[/asy]
Then $\angle YXZ = \angle WXY - \angle WXZ = 137^\circ - 105^\circ = \boxed{32^\circ}$.
Also, basing off the unprofessionalism the AoPS account displayed, it isn't actually the official account. So rest assured, no ones account will ever be deactivated. You pay for it, they aren't going to kick you.
Sure! I can help you with these questions. Let's start with the first question:
1. To find the length of the median of the trapezoid, we first need to find the lengths of the bases. Let's call the longer base x units.
Since one base is 4 units longer than the other, we can represent the shorter base as (x - 4) units.
The formula for the area of a trapezoid is A = (1/2) * (b1 + b2) * h, where A is the area, b1 and b2 are the lengths of the bases, and h is the height.
We know the area (A = 60) and the height (h = 5), so we can substitute these values into the formula:
60 = (1/2) * ((x - 4) + x) * 5
Simplifying the equation, we get:
60 = (1/2) * (2x - 4) * 5
60 = (x - 2) * 5
12 = x - 2
x = 14
Therefore, the longer base of the trapezoid is 14 units, and the shorter base is 10 units (14 - 4).
Now, to find the median of the trapezoid, we know that the median is the average of the lengths of the bases. Therefore, we have:
Median = (10 + 14) / 2
= 24 / 2
= 12 units
So, the length of the median of the trapezoid is 12 units.
Now, let's move on to the second question:
2. In trapezoid ABCD, we are given that line BC is parallel to line AD, angle ABD = 105 degrees, angle A = 43 degrees, and angle C = 141 degrees. We need to find angle CBD.
Since BC is parallel to AD, angle ABD and angle CBD are corresponding angles, and they are equal.
Therefore, angle CBD = angle ABD = 105 degrees.
So, the measure of angle CBD is 105 degrees.
Next, let's move on to the third question:
3. In trapezoid ABCD, the bases are lines AB and CD. Let M be the midpoint of line AD. We are given that the areas of triangles ABM and CDM are 5 and 17, respectively. We need to find the area of trapezoid ABCD.
Area of trapezoid ABCD = (area of triangle ABM + area of triangle CDM) + (area of triangle CDM)
= (5 + 17) + 17
= 39
Therefore, the area of trapezoid ABCD is 39.
Lastly, let's move on to the fourth question:
4. In parallelogram ABCD, line BC extends past B to point F, and line AB intersects line DF at point E. We are given that the areas of triangles BEF and ADE are 1 and 9, respectively. We need to find the area of parallelogram ABCD.
The area of a parallelogram can be found by multiplying the length of the base by the height.
Since AB and CD are parallel, the height of the parallelogram is the perpendicular distance between AB and CD.
The area of triangle ADE = 9, so the height of the parallelogram is 9.
Let's denote the length of BC as b and the length of AD as h.
The area of triangle BEF = 1, which is equal to (1/2) * b * 9, since the height is 9.
Simplifying the equation, we get:
1 = (1/2) * 9b
1 = 4.5b
b = 1 / 4.5
b = 2/9
Now, we know that the length of BC is 2/9 times the length of AD.
Since AB and CD are parallel, AD and BC have the same length.
Let's assume the length of AD is x, then the length of BC is 2/9 * x.
The area of parallelogram ABCD = base * height = AD * height = x * 9 = 9x.
Therefore, the area of parallelogram ABCD is 9x.
To find the value of x, we know that the length of BC is 2/9 times the length of AD.
So, we have 2/9 * x = x.
Solving for x, we get:
2/9 * x = x
2x = 9x
9x - 2x = 0
7x = 0
x = 0
Therefore, the area of parallelogram ABCD is 9 times the length of AD, which is 9 * 0 = 0.
Hence, the area of parallelogram ABCD is 0.