Find the area of the shaded region as a ratio to the area of the square

[chwala]

Homework Statement

find the area of the shaded region as a ratio to the area of the square (kindly see attached diagram)

Homework Equations

The Attempt at a Solution

$A= \frac 1 2$$b×h$
$A= \frac 1 2$$×2x × 3x$

 

[Mark44]

Homework Statement

find the area of the shaded region as a ratio to the area of the square (kindly see attached diagram)

Homework Equations

The Attempt at a Solution

$A= \frac 1 2$$b×h$
$A= \frac 1 2$$×2x × 3x$

Start by figuring out as many lengths as you can. The small upper right triangle is similar to the larger lower triangle, so the corresponding sides of these two triangles.

After you find all three sides of the small upper right triangle, you can calculate the angle that the two intersecting lines make, and from that, you can calculate the area of the shaded region. This problem is mostly geometry and a bit of trig.

 

[andrewkirk]

Label the vertices of the square as A, B, C, D starting at top left and proceeding clockwise.
Label the other point on the top edge as E
Label the point of intersection of two lines inside the square as X.

Then the are of the shaded part is $\Delta ABD - \Delta AED - \Delta EXB$.

The first two of those are easily calculated.
The base of the third triangle is known, so we only need to work out the perpendicular distance h from X to the top edge.

How do you think you might go about calculating that?

 

[chwala]

then $(hX)^2= (EX)^2-(hE)^2$ is this correct going with post 3.

 

[andrewkirk]

then $(hX)^2= (EX)^2-(hE)^2$ is this correct going with post 3.

No. $h$ is a length, not a point. I think you interpreted it as meaning the point at the foot of the perpendicular from X to EB. But it is not that point, it is the height of the perpendicular. If we denote that foot by H (capital for a point, as opposed to lower case for a distance) and replace h by H in the equation you wrote, it is correct, but it doesn't lead clearly to a solution. I suggest you take h as the height mentioned, then work out the (perpendicular) distance from X to the line DC in terms of h. Then use the fact that triangle EXB is similar to CXD to write an equation in h, which you can then solve.

 

[chwala]

using similarity
$\frac {h_{1}} {x}$ = $\frac {h_{2}} {3x}$
...Ratio of shaded : square is $1:8$

 

[andrewkirk]

using similarity
$\frac {h_{1}} {x}$ = $\frac {h_{2}} {3x}$
...Ratio of shaded : square is $1:8$

Correct!

 

[chwala]

there's also an alternative method considering the square we have the equations $y=x$ and $y=-3x+9$ the point of intersection is $(9/4,9/4)$ now we have 4 points that is
$(0,0), (9/4,9/4), (2,3) and (0,0)$, using shoe lace formula, the area of shaded is $\frac 9 8$units square and area of square is $9$ square units therefore,
ratios will be $\frac 9 8$:$9$ = $\frac 1 8$:1 = $1:8$