# Question f9011

Jul 28, 2016

Area of $\Delta A Q P = \frac{1}{15}$.

#### Explanation:

We will solve it using Co-ordinate Geometry.

Let us take $A \left(0 , 0\right)$, line $A B$ as $X$-Axis, and line $A D , Y$-Axis.

Also, $\vec{A B} \mathmr{and} \vec{A D}$ are taken as the $+ v e$ directions on

the resp. axes.

Since lengths $A B = B C = C D = D A = 1$,

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

Clearly, $M \left(1 , \frac{1}{2}\right) \mathmr{and} N \left(1 , \frac{1}{4}\right)$

We note that diagonal $B D$ has both intercepts $1$, so, its eqn. is,

$B D : x + y = 1. \ldots \ldots \ldots \ldots \ldots \left(1\right)$.

With $A \left(0 , 0\right) \mathmr{and} M \left(1 , \frac{1}{2}\right)$ on line $A M$,

the eqn. of line $A M : y = \frac{1}{2} x \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right)$.

Similarly, eqn. of line $A N : y = \frac{1}{4} x \ldots \ldots \ldots . \left(3\right)$.

Now, $A M \cap B D = \left\{P\right\}$; solving $\left(1\right) \mathmr{and} \left(2\right)$, we get $P$.

$\therefore P = P \left(\frac{2}{3} , \frac{1}{3}\right)$.

Likewise, $Q = Q \left(\frac{4}{5} , \frac{1}{5}\right)$

Thus, in $\Delta A Q P , A \left(0 , 0\right) , Q \left(\frac{4}{5} , \frac{1}{5}\right) \mathmr{and} P \left(\frac{2}{3} , \frac{1}{3}\right)$.

Therefore, Area of $\Delta A Q P = \frac{1}{2} | D ' |$, where,

$D ' = | \left(0 , 0 , 1\right) , \left(\frac{4}{5} , \frac{1}{5} , 1\right) , \left(\frac{2}{3} , \frac{1}{3} , 1\right) | = \frac{4}{15} - \frac{2}{15} = \frac{2}{15}$.

Hence, Area of $\Delta A Q P = \frac{1}{2} | \frac{2}{15} | = \frac{1}{15}$, as "suggested" by Lee
B., Sir!

Aug 18, 2016

Area of $\Delta A Q P = \frac{1}{15}$.

#### 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 $\Delta A Q P = \frac{1}{2} \cdot A P \cdot A Q \cdot \sin \angle P A Q \ldots \ldots \ldots \left(\star\right)$.

In $r i g h t \Delta M B A , m \angle M B A = {90}^{\circ}$
$\Rightarrow M {A}^{2} = M {B}^{2} + B {A}^{2} = \frac{1}{4} + 1 = \frac{5}{4} \Rightarrow M A = \frac{\sqrt{5}}{2}$
Similarly, from $r i g h t \Delta N A B$, we get, $N A = \frac{\sqrt{17}}{4}$

We observe that, diagonal $B D$ of square $A B C D$ bisects the $\angle C B A$. This means that, in $\Delta M B A , B D$ bisects the $\angle M B A$, intersecting the side $A M$ (opp. to $\angle M B A$) in pt. $P$.

Hence, by the $\angle$-bisector Theorem, $\frac{A P}{P M} = \frac{A B}{B M} = \frac{1}{\frac{1}{2}} = 2$
$\Rightarrow \frac{A P}{A P + P M} = \frac{2}{2 + 1} = \frac{2}{3} \Rightarrow \frac{A P}{A M} = \frac{2}{3}$
$\Rightarrow A P = \frac{2}{3} \cdot M A = \frac{2}{3} \cdot \frac{\sqrt{5}}{2} , i . e . , \frac{\sqrt{5}}{3.} \ldots \ldots \ldots . \left(1\right)$
Arguing on the same lines, $A Q = \frac{4}{5} \cdot N A = \frac{4}{5} \cdot \frac{\sqrt{17}}{4} = \frac{\sqrt{17}}{5.} . . \left(2\right)$.

In the light of $\left(\star\right) ,$ now, it remains to find $m \angle P A Q$.

We have, from $r i g h t \Delta M B A$,
tan/_MAB=(MB)/(AB)=(1/2)/1=1/2, &, tan /_NAB=1/4.
$\therefore \tan \angle M A N = \tan \angle P A Q = \tan \left(\angle M A B - \angle N A B\right)$
$= \frac{\frac{1}{2} - \frac{1}{4}}{1 + \frac{1}{2} \cdot \frac{1}{4}} = \frac{1}{4} \cdot \frac{8}{9}$, i.e., $\tan \angle P A Q = \frac{2}{9}$.
$\therefore \sin \angle P A Q = \frac{2}{\sqrt{85.}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(3\right)$.

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