# How do I show that the area of the triangle OAB is 36 1/8 ?

Apr 17, 2018

$\text{see explanation}$

#### Explanation:

$\text{area of "triangleOAB=1/2xx"base"xx"height}$

$\textcolor{w h i t e}{\times \times \times \times \times x} = \frac{1}{2} \times O B \times O A$

$\text{we require to find the coordinates of A and B}$

$\text{obtain the equation of the tangent line AB}$

$\text{remembering that the tangent is at right angles to}$
$\text{the radius}$

$\text{the slope of the radius between" (0,0)" and "(1,4)" is}$

${m}_{\text{radius}} = \frac{\Delta y}{\Delta x} = \frac{4}{1} = 4$

$\text{hence the slope of the tangent line is}$

${m}_{\text{tangent}} = - \frac{1}{m} = - \frac{1}{4}$

$\Rightarrow y = - \frac{1}{4} x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$

$\text{to find b substitute "(1,4)" into the partial equation}$

$4 = - \frac{1}{4} + b \Rightarrow b = 4 + \frac{1}{4} = \frac{17}{4}$

$\Rightarrow y = - \frac{1}{4} x + \frac{17}{4} \leftarrow \textcolor{red}{\text{equation of tangent line}}$

$x = 0 \Rightarrow y = \frac{17}{4} \Rightarrow A = \left(0 , \frac{17}{4}\right)$

$y = 0 \Rightarrow - \frac{1}{4} x + \frac{17}{4} = 0 \Rightarrow x = 17 \Rightarrow B = \left(17 , 0\right)$

$\text{area of } \triangle O A B = \frac{1}{2} \times 17 \times \frac{17}{4}$

$\textcolor{w h i t e}{\times \times \times \times \times \times} = \frac{289}{8} = 36 \frac{1}{8} {\text{ units}}^{2}$

Apr 17, 2018

Please refer to the Explanation.

#### Explanation:

Name the point of contact $C \left(1 , 4\right)$.

Let, $r$ be the radius of the circle in question, say $S$.

Since, $O \left(0 , 0\right)$ is the centre of $S$, and $C \in S$, we have,

$C {S}^{2} = {r}^{2} = {\left(1 - 0\right)}^{2} + {\left(4 - 0\right)}^{2} = 17$.

Clearly, the eqn. of $S$ is given by, $S : {x}^{2} + {y}^{2} = {r}^{2} = 17$.

Recall that, the eqn. of tgt. $t$ to $S$ at $\left({x}_{0} , {y}_{0}\right)$ is given by,

$x \cdot {x}_{0} + y \cdot {y}_{0} = {r}^{2}$.

Hence, $t$ at $C$, is, $t : x + 4 y = 17 , \mathmr{and} , \frac{x}{17} + \frac{y}{\frac{17}{4}} = 1$.

This eqn. of $t$ exibits that $A = A \left(0 , \frac{17}{4}\right) , \mathmr{and} , B = B \left(17 , 0\right)$.

$\therefore \text{ The Area of right-"Delta AOB=} \frac{1}{2} \cdot O A \cdot O B$,

$= \frac{1}{2} \cdot \frac{17}{4} \cdot 17 = \frac{289}{8} = 36 \frac{1}{8} \text{ sq. unit}$.

Apr 17, 2018

See the explanation below.

#### Explanation:

Let point C(1, 4) be the point of tangency then the radius OC:
$O C = \sqrt{{\left(1 - 0\right)}^{2} + {\left(4 - 0\right)}^{2}} = \sqrt{17}$

The OC is also perpendicular to AB, draw two lines from C perpendicular to OB and OA at points D and and E respectively:
$C D = 4$
$C E = 1$

We can calculate the following angles:
OAD = arctan(1/4)=14 degrees
OCE = 90 - 14 = 76 degrees

Now consider two right triangles AEC and DCB:
Angle EAC and DCB are equal also angles ECA and DBC are equal, we can calculate the following angles:
EAC = DCB = 90 - 14 = 76 degrees
ECB = DBC = 90 - 76 = 14 degrees

We can calculate the following lengths:
$D B = 4 \cdot \tan 76 = 16$
$A E = \frac{1}{\tan} 76 = \frac{1}{4}$
then:
$O A = 4 + \frac{1}{4} = 4 \frac{1}{4}$
$O B = 16 + 1 = 17$
Finally the area of OAB is:
$A r e a = \frac{1}{2} \cdot 17 \cdot 4 \frac{1}{4} = 36 \frac{1}{8}$