Tangential right trapezoid

Let $x$ be the radius. Then $EC=22-x$ , $FD=35-x$ , $CH=13$ , $HD=2x$ .

By congruent triangles, $CG=22-x$ and $GD=35-x$ , so $CD=57-2x$ .

By right triangle $CHD$ we have $13^{2}+(2x)^{2}=(57-2x)^{2}$

Solving gives $x=\frac{770}{57}$

And so the area sought is $\frac{1}{2} \cdot (2x) \cdot (22+35) = \boxed{770}$

Let $r$ be the radius of the circle and $E$ , $F$ and $G$ tangential points, so: $\angle OEB = \angle OGC = 90°$ . Since the center of a circle inscribed in an angle lies on its bisector:

$\angle EBO = \angle FBO = \dfrac{1}{2} \angle EBF$ and $\angle FCO = \angle GCO = \dfrac{1}{2} \angle FCG$ .

Also, since $\angle EBF + \angle FCG = 180°$ : $\angle COG = 90° - \angle GCO = 90° - \dfrac{1}{2} \angle FCG = \dfrac{180° - \angle FCG}{2} = \dfrac{1}{2} \angle EBF = \angle EBO$

So, triangles $\triangle EBO$ and $\triangle GOC$ are similar, which means: $\dfrac{CG}{OE} = \dfrac{OG}{BE}$ $\Rightarrow$ $\dfrac{22 - r}{r} = \dfrac{r}{35 - r}$ .

Solving this equation gives us $r = \dfrac{770}{57}$

Area of the trapezoid is: $\dfrac{AB + CD}{2}*AD = \dfrac{22 + 35}{2}2r = \dfrac{57*2*770}{2*57} = \boxed{770}$

Let the sloping line on the right hand side have length L and the diameter of the circle be D.

Pitot's theorem states that each pair of opposite sides of a quadrilateral with an in-circle sum to the same total! So

$D+L=57 \dots(1)$

An easy application of Pythagoras gives

$D^{2}+13^{2}=L^{2} \\ \implies (L-D)(L+D)=169$

Substitute from (1) to get

$L-D=\frac{169}{57}$

Subtract this from (1) and divide by 2 to get

$D=\frac{1540}{57}$

Which gives the area of the trapezium as

$\frac{1}{2}(22 + 35)\frac{1540}{57}=770$

Let $R$ be the radius of the circle. Let the lengths of top and bottom bases be $a \pm b$ ; in this case, $a = 28\tfrac12$ and $b = -6\tfrac12$ . Then the area of the trapezoid is $A = 2aR$ .

Choose a coordinate system with the origin $O$ at the center of the circle. The two vertices on the right then have coordinates $V_\pm = (a \pm b, \pm R)$ . The point of tangency between these vertices has coordinates $T\ (a-R+bx,Rx)$ for some $-1 < x < +1$ . The segment $OT$ satisfies two conditions:

• its length is $R$ , i.e. $(a-R+bx)^2 + (Rx)^2 = R^2;$

• it is perpendicular to the side $V_{-}V_{+}$ , i.e. the vectors $(a-R+bx,Rx)$ and $(b,R)$ are perpendicular: $(a-R+bx)b + R^2x = 0.$

We multiply the second equation by $x$ and subtract it from the first to obtain $(a-R)(a-R+bx) = R^2;$ substituting this again into the second equation, we get $(a-R+bx)(b+(a-R)x) = 0.$ This implies $x = -b/(a-R)$ , and our third equation becomes $R^2 = (a-R)^2 - b^2\ \ \ \therefore\ \ \ 2aR = a^2 - b^2\ \ \ \therefore\ \ \ R = \frac{(a+b)(a-b)}{2a}.$ Finally, the area is $A = 2aR = (a+b)(a-b) = 22\cdot 35 = \boxed{770}.$

For any right trapezoid with an inscribed circle that touches all four sides, the area is equal to the product of the two widths of the trapezoid.

22 x 35 = 770

By comparing angles, we can see that right-triangle $\triangle BAO$ is the same shape as $\triangle OCD$

We know that $OC$ and $AO$ both equal the radius $r$ , and by extension, $CD=35-r$ and $BA=22-r$ .

As the two triangles are proportionate, the ratio $BA:AO$ is the same as $OC:CD$ . From this, we can work out the radius:

$\frac{22-r}{r}=\frac{r}{35-r}$

$r^2=(22-r)(35-r)$

$r^2=770-57r+r^2$

$0=770-57r$

$57r=770$

$r=\frac{770}{57}$

And so the area is $\frac{1}{2}\times(2 \times \frac{770}{57})\times(35+22)=\boxed{770}$

This may have been completely coincidental, but I just took the two given lengths and multiplied them together. So 22(35) = 770

The area of a convex poligon can be found by S=p.r , where p represents half the perimeter and r is the radius of the inscribed circle. Moreover, the Area of any trapezoid is also A = ( b + B) .h/2, where B is the 'bigger' base, b the smaller one and h the hight, which in this case equals to twice the radius.
So : p.r = (22+35) .(2r)/2 p= 57 If we close a rectangle triangle by connecting the furthest right point of the smaller base to it's projection on the bigger base, we are gonna have the triangle with sides 13, h and √(13²+h²). With that, we have: Perimeter = 22+ 35 + h + √(13²+h²) = 2p 57 + h + √(13²+h²) = 2. 57 => √(13²+h²) = 57-h => 169 + h²= 57² - 114h + h² => 114h = 3249-169 => h = 1540/57

By using this result above, we have: S=(22+35) .h/2 => S= (57 . 1540)/ (2. 57) => S= 1540/2 => S= 770

$(57 - 2r)^2 = (2r)^2 + 169 \implies 3249 -228r = 169 \implies r = \dfrac{770}{57} \implies$

Tha area of the trapezoid $A_{e} = 57r = \boxed{770}$ .

Let us consider the radius of the circle to r. Let the parallel sides of the trapezium be a and b.

We can use Harmonic mean (HM) to find the radius. HM = r = ab/a+b

Area of the trapezium being 1/2 * h * (a+b) where we can equate h = 2r

Solving by substituting the value of a we get Really of eight tangential trapezoid to be the product of the parallel sides (i.e) ab

Hence A = 22*35 = 770

Above is someone else's work for solving for the radius of an inscribed circle for a right trapezoid/trapezium.

Our radius in this case is (22*35)/57

Next: The formula for the area of a trapezoid is 0.5*(b 1+b 2)*h.

b_1 = 22

b_2 = 35

h = 2*r = 2*(22*35)/57

Therefore: A = 0.5*(57)*2*(22*35)/57

Since 0.5*2 and 57/57 are both 1, our area is just 22*35 = 770.

It can be proved that the area equals the product of the lengths of the parallel sides. I wonder if there exists a visual proof...