Question

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

Answer 1

`"see explanation"`

Explanation:

`"area of "triangleOAB=1/2xx"base"xx"height"`

`color(white)(xxxxxxxxxxx)=1/2xxOBxxOA`

`"we require to find the coordinates of A and B"`

`"obtain the equation of the tangent line AB"`

`"remembering that the tangent is at right angles to"`
`"the radius"`

`"the slope of the radius between" (0,0)" and "(1,4)" is"`

`m_("radius")=(Deltay)/(Deltax)=4/1=4`

`"hence the slope of the tangent line is"`

`m_("tangent")=-1/m=-1/4`

`rArry=-1/4x+blarrcolor(blue)"is the partial equation"`

`"to find b substitute "(1,4)" into the partial equation"`

`4=-1/4+brArrb=4+1/4=17/4`

`rArry=-1/4x+17/4larrcolor(red)"equation of tangent line"`

`x=0rArry=17/4rArrA=(0,17/4)`

`y=0rArr-1/4x+17/4=0rArrx=17rArrB=(17,0)`

`"area of "triangleOAB=1/2xx17xx17/4`

`color(white)(xxxxxxxxxxxx)=289/8=36 1/8" units"^2`

Answer 2

Please refer to the Explanation.

Explanation:

Name the point of contact `C(1,4)`.

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

Since, `O(0,0)` is the centre of `S`, and `C in S`, we have,

`CS^2=r^2=(1-0)^2+(4-0)^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 `(x_0,y_0)` is given by,

`x*x_0+y*y_0=r^2`.

Hence, `t` at `C`, is, `t : x+4y=17, or, x/17+y/(17/4)=1`.

This eqn. of `t` exibits that `A=A(0,17/4), and, B=B(17,0)`.

`:." The Area of right-"Delta AOB="1/2*OA*OB`,

`=1/2*17/4*17=289/8=36 1/8" sq. unit"`.

Answer 3

See the explanation below.

Explanation:

Let point C(1, 4) be the point of tangency then the radius OC:
`OC=sqrt[(1-0)^2+(4-0)^2]=sqrt17`

The OC is also perpendicular to AB, draw two lines from C perpendicular to OB and OA at points D and and E respectively:
`CD=4`
`CE=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:
`DB = 4 * tan76=16`
`AE=1/tan76=1/4`
then:
`OA=4+1/4=4 1/4`
`OB=16+1=17`
Finally the area of OAB is:
`Area= 1/2 * 17 * 4 1/4=36 1/8`