Analytical Geometry> In the figure given below, ABCDEF is a re...

In the figure given below, ABCDEF is a regular hexagon of side length 1, AFPS and ABQR are squares.Then the ratio Area (APQ)/ Area (SRP) equals

Vineeth , 11 Years ago

Grade 11

3 Answers

Sandeep Pathak

Last Activity: 11 Years ago

Notice length of AP = $\sqrt{2}$ = length of AQ.
Also note $\angle PAQ = 30^{\circ} = \angle RSP$
and length of SR = length of SP = 1.

so the two triangles are iscosceles with same angle between the equal sides. So the ratio of areas is equal to square of ratio of equal sides. This gives
$\frac{Area (APQ)}{Area (SRP)} = \left( \frac{AQ}{SR}\right )^2=2$

Vineeth

Last Activity: 11 Years ago

Sir , can you explain this answer with more steps.

Sandeep Pathak

Last Activity: 11 Years ago

length of AP and AQ is $\sqrt{2}$ as these are diagonals of square with side 1.
Since AP and AQ are diagonals of square ASPF and ABQR respectively
$\angle BAQ = 45^{\circ}, \angle PAF = 45^{\circ}$
Also, due to hexagon property, angle between adjacent sides is $120^{\circ}$ , so $\angle FAB = \angle FAP + \angle PAQ + \angle BAQ = 120^{\circ} \\ \Rightarrow \angle PAQ = 30^{\circ}$
Now, consider triangle RAS.
$\angle FAS = 90^{\circ}, \angle FAB = 120^{\circ} \Rightarrow \angle BAS = 30^{\circ}\\ \because \angle BAS+\angle SAR = \angle BAR = 90^{\circ}\Rightarrow \angle SAR = 60^{\circ}$
Similarly other angles can be proved to be equal and this triangle RAS is an equilateral triangle. So, RS = AS = SP = 1.
Now, in triangle RSP,
$\angle ASR+\angle RSP = \angle ASP = 90^{\circ}\Rightarrow \angle RSP = 30^{\circ}$

Hope the steps are clear to you. In future, I would suggest if something is not clear in the solution, please specify what exactly you were not able to understand so that We can provide answers that can help in a better way. Cheers!