The diagram shows a square inscribed in an isosceles triangle with side lengths 10, 10 and 12. Find a.
Please reply quick!
Call the leg of the right triangle formed at the base of the isosceles triangle x. The we know that
$\displaystyle 12 = 2x + a$
Now, we know the base angle of the isosceles triangle, call this angle $\displaystyle \theta$. Then we also know that
$\displaystyle tan(\theta) = \frac{a}{x}$
Thus
$\displaystyle 12 = 2x + x tan(\theta)$
Solve for x, then you can get a.
-Dan
Here is one way, by proportion.
Similar triangles are proportional
In the given figure, draw the perpendicular bisector of the 12 units long base. It will pass through the apex and it will bisect the whole isosceles triangle into two equal right triangles. It will also bisect the inscribed square.
In any of the two equal large right triangles, the smaller right triangle formed by the vertical side "a" and the hypotenuse and horizontal leg of the large right triangle is similar to the large right triangle.
We have:
Large right triangle,
vertical leg = sqrt(10^2 -6^2) = 8
horizontal leg = 12/2 = 6
Smaller right triangle,
vertical leg = a
horizontal leg = 6 -a/2
So, by proportion,
vert.leg/hor.leg = vert.leg/hor.leg
8/6 = a/(6 -a/2)
Cross multiply,
8(6 -a/2) = 6(a)
48 -4a = 6a
48 = 6a +4a
a = 48/10 = 4.8 units long ----------answer.
Hello, DivideBy0!
The diagram shows a square inscribed in an isosceles triangle with side lengths 10, 10 and 12.
Find $\displaystyle a$, the side of the square..
Consider half of the diagram.Code:A * /| / | / | 10 / | D / | * - - + 8 /| | / |a | / | | B * - * - - * C E ½a :- - 6 - -:
We have right triangle $\displaystyle ACB$ with $\displaystyle AB = 10,\:BC = 6$
. . Hence: .$\displaystyle AC = 8$
In the square: .$\displaystyle DE = a,\:EC = \frac{1}{2}a$
.Hence: .$\displaystyle BE \,= \,6 - \frac{1}{2}a$
Since $\displaystyle \Delta DEB \sim \Delta ACB\!:\;\;\frac{BE}{DE} \:=\:\frac{BC}{AC}$
. . and we have: .$\displaystyle \frac{6-\frac{1}{2}a}{a} \:=\:\frac{6}{8}$
Solve for $\displaystyle a\!:\;\;\boxed{a \:=\:4.8}$