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

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3 Answers
Apr 17, 2018

#"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#

Apr 17, 2018

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"#.

Apr 17, 2018

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#