Question #f9011

2 Answers
Jul 28, 2016

Area of #Delta AQP=1/15#.

Explanation:

We will solve it using Co-ordinate Geometry.

Let us take #A(0,0)#, line #AB# as #X#-Axis, and line #AD, Y#-Axis.

Also, #vec(AB) and vec (AD)# are taken as the #+ve# directions on

the resp. axes.

Since lengths #AB=BC=CD=DA=1#,

we have #B(1,0), C(1,1) & D(0,1)#

Clearly, #M(1,1/2) and N(1,1/4)#

We note that diagonal #BD# has both intercepts #1#, so, its eqn. is,

#BD : x+y=1................(1)#.

With #A(0,0) and M(1,1/2)# on line #AM#,

the eqn. of line #AM : y=1/2x..................(2)#.

Similarly, eqn. of line #AN : y=1/4x..........(3)#.

Now, #AM nn BD ={P}#; solving #(1) and (2)#, we get #P#.

#:. P=P(2/3,1/3)#.

Likewise, #Q=Q(4/5,1/5)#

Thus, in #DeltaAQP, A(0,0), Q(4/5,1/5) and P(2/3,1/3)#.

Therefore, Area of #Delta AQP=1/2|D'|#, where,

#D'=|(0,0,1),(4/5,1/5,1),(2/3,1/3,1)|=4/15-2/15=2/15#.

Hence, Area of #Delta AQP=1/2|2/15|=1/15#, as "suggested" by Lee
B., Sir!

Aug 18, 2016

Area of #DeltaAQP=1/15#.

Explanation:

I have solved this problem with the help of Co-ordinate Geometry. Now, I have been able to find its solution using only Geometry and a little bit of Trigonometry.

We know from Trigo. that,
the Area of #DeltaAQP=1/2*AP*AQ*sin/_PAQ.........(star)#.

In #rightDeltaMBA, m/_MBA=90^@#
#rArr MA^2=MB^2+BA^2=1/4+1=5/4rArr MA=sqrt5/2#
Similarly, from #rightDeltaNAB#, we get, #NA=sqrt17/4#

We observe that, diagonal #BD# of square #ABCD# bisects the #/_CBA#. This means that, in #Delta MBA, BD# bisects the #/_MBA#, intersecting the side #AM# (opp. to #/_MBA#) in pt. #P#.

Hence, by the #/_#-bisector Theorem, #(AP)/(PM)=(AB)/(BM)=1/(1/2)=2#
#rArr (AP)/(AP+PM)=2/(2+1)=2/3rArr(AP)/(AM)=2/3#
#rArrAP=2/3*MA=2/3*sqrt5/2, i.e., sqrt5/3...........(1)#
Arguing on the same lines, #AQ=4/5*NA=4/5*sqrt17/4=sqrt17/5...(2)#.

In the light of #(star),# now, it remains to find #m/_PAQ#.

We have, from #rightDeltaMBA#,
#tan/_MAB=(MB)/(AB)=(1/2)/1=1/2, &, tan /_NAB=1/4#.
#:. tan/_MAN=tan/_PAQ=tan(/_MAB-/_NAB)#
#=(1/2-1/4)/(1+1/2*1/4)=1/4*8/9#, i.e., #tan/_PAQ=2/9#.
#:. sin/_PAQ=2/sqrt85......................(3)#.

Therefore, by #(star),(1),(2), &, (3)#,
Area of #DeltaAQP=1/2(sqrt5/3)(sqrt17)/5)(2/sqrt85)=1/15#, as was derived earlier!

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