circlesgeometry
Two side-by-side squares are inscribed in a semicircle. The diameter of the semicircle is $16$. What is the sum of the two squares' areas?
This is a bonus question from my online math class that I've been trying to solve but haven't gotten very far. I named $CD$ $x$ and $HD$ $y$, then drew $OB$ and $OF$, then did the Pythagorean theorem to try to get $x^2+y^2$ but I didn't have much luck with that.
Is my solution correct?
Construction : Join $PS$ and $OR$.
$\angle ORQ = 90^o$ (Radius is perpendicular to tangent at the point of contact)
In $\triangle ORQ$,
$ OQ^2 = OR^2 + RQ^2 $ (using Pythagoras Theorem)
$\implies 13^2 = 8^2 + RQ^2$
$\implies RQ = \sqrt{105}$ $...(\mathtt i)$
Now, $SR=RQ$ (perpendicular from the center of the circle to a chord, bisects the chord)
$\implies SR = \sqrt{105}$ $...(\mathtt{ii})$
Adding $(\mathtt i)$ and $(\mathtt{ii})$,
$SR+RQ=2 \sqrt{105}$
$\implies SQ =2 \sqrt{105} $ $...(\mathtt{iii})$
Now, $\angle PSQ= 90^o$ (angle subtended by a diameter)
In $\triangle PSQ$ ,
$PQ^2=PS^2+SQ^2$ (using Pythagoras Theorem)
$\implies 26^2 = PS^2 + (2\sqrt{105})^2$ ($\because SQ= 2\sqrt{105}$, using $(\mathtt{iii})$ )
$\implies 676 = PS^2 + 420$
$\implies PS^2 = 676-420$
$\implies PS = \sqrt{256}$
$\implies PS = 16$ $...(\mathtt{iv})$
Now, In $\triangle PSR$,
$PR^2 = PS^2 + SR^2$ (using Pythagoras Theorem)
$\implies PR^2 = 16^2 + (\sqrt{105})^2$ ($\because SR= \sqrt{105}$, using $(\mathtt{ii})$ )
$\implies PR^2 = 256 + 105$
$\implies PR^2 = 361$
$\implies PR = \sqrt{361}$
$\implies PR = 19 cm$
$\angle XIZ = 180^0 – \dfrac {X + Z}{2}$
Also, $(X + Z) = 180^0 – Y$
Result follows after eliminating X+ Z from the above.
Note that the location of I is fixed and comes directly from the triangle XZ and Y. It is not allowed to set Y as arbitrary after that.
Best Answer
Let $OD=c$ and the side lengths of the two squares $a$ and $b$. From right triangles OAB and OHF,
$$r^2=a^2+(a-c)^2;\>\>\>\>\> r^2=b^2+(b+c)^2$$
Eliminate $c$ to get,
$$a-\sqrt{r^2-a^2}=\sqrt{r^2-b^2}-b$$
Square both sides,
$$a\sqrt{ r^2-a^2} =b\sqrt{r^2-b^2}$$
Square again and rearrange,
$$r^2(a^2-b^2)=a^4-b^4$$
Thus, the sum of the two areas is
$$a^2+b^2=r^2=64$$