Question

Question #f9011

Answer 1

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!

Answer 2

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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