The rectangle at the corner measures 10 cm * 20 cm
.
The right bottom corner of the rectangle is also a point on the circumference of the circle.
What is the radius of the circle in cm? Is the data sufficent to get the radius of circle?
The rectangle at the corner measures 10 cm * 20 cm
.
The right bottom corner of the rectangle is also a point on the circumference of the circle.
What is the radius of the circle in cm? Is the data sufficent to get the radius of circle?
Hint: with a coordinate system at the center of the circle, the point of intersection of the circle with the rectangle is $(10-r,r-20)$, so $ (10-r)^2+(r-20)^2=r^2. $
Note also, that to be in the situation imposed by the diagram, you must have $r>20$.
Let $R$ denote the radius, and let $w$ and $h$ be the width and the height of the rectangle.
Consider right triangle, formed by the center of the circle $O$, point where the rectangle touches the circle $A$ and the point $B$ - projection of $A$ on the horizontal diameter.
Then, by Pythagorean theorem: $ \begin{eqnarray} R^2 &=& (R-w)^2 + (R-h)^2 \\ R^2 &=& 2 R^2 - 2 R(w+h) + w^2 + h^2 \end{eqnarray} $ It remains to solve this quadratic equation, and choose the appropriate root (considering the special case of a square, when $w=h$, helps): $ R = w + h + \sqrt{2 w h} $
In the circle shown above the triangles $\triangle AGT$ and $\triangle TKX$ are similar.
We know $BC=10$ and $AG=20$
Let $CK=y$ and radius of the circle $BX=R$
In the similar triangles $\triangle AGT$ and $\triangle TKX$ we have,
$\frac{AG}{GT}=\frac{TK}{KX}$
$\frac{20}{10+y}=\frac{R-20}{R-(10+y)}$
i.e. $CK=y=10$
$GT=BK=BC+CK=10+10$
$AT=\sqrt{{AG}^2+{GT}^2}=20\sqrt{2}$
${TK}^2+{KX}^2={TX}^2$
$(R-20)^2+(R-20)^2={TX}^2$
TO BE CONTINUED
$r^2=x^2+y^2 \tag{1}$ $r=y+20$ and $r=x+10$ therefore $y+20=x+10 \quad \mbox{then } \quad y=x-10 \tag{2}$ Substitute $(2)$ into $(1)$ $(r+10)^2=x^2+(x-10)^2$ $x^2+20x+100=x^2+x^2-20x+100$ $X^2=40x$ $x=40$ Then substitute $x=40$ into $(2)$ $y=40-10$ $y=30$ substitute $x=40$ and $Y=30$ into $(1)$ $r^2=(40)^2+(30)^2$ $r=50\rm{cm}.$