How do you solve this geometric problem?
If the characters weren't so callous, I'd know how to do it.
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A parallel to the base side at a distance of 2 or x helps for understanding.
Then expand the figures accordingly and calculate them.
Do you think I'm supposed to make a tip on the short length?
correct
That's what someone suggested, and it doesn't seem to work like that.
Below is a calculation according to the set of beams.
View image 2. The orange triangle.
This thus represents a ray set.
The purple dimensions are easy to calculate in the head.
Red measurements are given.
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Task 11a
x1 / 6 = 7 / 21
x1 = 7 / 21 * 6
x1 = 7 / 21 * 6
x1 = 2
x = x1 + 2
x = 2 + 2
x = 4
a) Drag a parallel to the base side (7+14) which is applied to the upper point of the right perpendicular (2). Then you have an auxiliary structure to which you can apply the ray set. From this you can calculate x-2 and you have x.
(b) is similar, you calculate 10-x and thus also x.
so it is
y/2 = (y+7)/x
Now you've slipped.
An equation with two unknowns… that will be nothing.
You should have betrayed that you can set up a second equation that includes the 8 .
Then you have two equations with two unknowns.
The two tasks can be solved by extending the hypothenuse and the ancathete by the value y until they intersect.
Task a)
y/2=(y+14+7)/8|×8
4y=y+21
y=7
x/(7+7)=2/7|×14
x=4
(b)
x/y=10/(y+12)
x/y=20/(y+16+12)=20/(y+28)
10/(y+12)=20/(y+28)
(y+12)/10=(y+28)/20|×20
2y+24=y+28
y=4
x/4=10/(4+12)
x/4=10/16|×16
4x=10
x=2.5