enter image description here

My attempts:

Attempt 1:

Radius of the circle = $5$

Angle $AOB = 45$ degrees

hence area of shaded part between A and B = ${1\over 8} \times \pi \times 25 - {1\over 2} \times 5^{2}{1\over \sqrt{2}}$ Then, by multiplying this area by 8, we get the answer as 7.9, which doesn't match any answer.

Attempt 2:

Now, if I consider the quadrilateral OHAB, the area will be

${1\over 4} \pi 25 - {1\over 2}5^{2} \sin 90$

Then, by multiplying the result with 4, we get $25 \pi - 50 = 28.5$, which is way beyond the answer choices.

Official Explanation:

enter image description here

Can someone please explain the flaws in my attempts and how exactly the solution's author calculated the area of triangles? I can't seem to draw the diagram they used to calculate the areas of triangles.


2 Answers 2


First of all. The book is wrong when it says the octagon is regular. A regular octagon with radius $5$ can't pass through points $(-3,4)$ and $(3,4)$. If it's lined up to have vertices on the axis, it'd have to have vertices $(-5\sqrt 2, 5\sqrt 2)$ and $(5\sqrt 2, 5\sqrt 2)$. If it isn't aligned on the axis and point $H = (x=-3, y=4)$, then point $A$ will be some $(x>0, 4<y< 5)$ and $B$ would have to be $(4,3)$.

So... first off the octagon is not regular and the angles are not congruent and they are not $45^\circ$.

Okay. You are right that the radius of the circle is $5$. And there are two triangles. Take $\triangle OAB$. You can declare any side to be a base so declare the line $\overline {OA}$ to be the base. That base is $5$ as it is a radius of the circle. Now what is the altitude of the triangle? That is the perpendicular distance from the base, $\overline{OA}$ to the point $B$. Well, $\overline{OA}$ is a perpendicular line straight up and down along the $y$-axis. And $B$ is the point $(x=3, y=4)$. So the distance that is from at $y$-axis is $3$. And so the altitude is the horizontal line from $(x=3,y=4)$ to $(x=0,y=4)$ of length $3$ and the base is the vertical line of $\overline{OA}$ which has length $5$.

Got it?

So the area of the triangle is $\frac 12 \cdot 5 \cdot 3$.

We can do the same for triangle $\triangle OBC$. with the base $\overline{OC}$.

The book doesn't state but I think we have to assume $F=(-3,4)$ and $D=(3,-4)$. Otherwise we have no way of figuring out the rest of it. I guess the book meant "symmetrical over the $x$-axis" when it said "regular". But IMHO the book is wrong. But it's calculation of the area for that one quadrant is correct.


EDIT: PostScript.

Oh... the answer key does state the figure is symmetric in all quadrants. So it's methodology and answer is correct. Still the statement of the problem was dead wrong when it stated the octagon was regular. It needed to state instead the octagon was reflexively symmetric across the the $x,y$-axes.


EDIT: Double post-script.

Two alternatives to figuring out the area of the triangles:

One: $A=(0,5); B=(3,4)$ so ${AB}= \sqrt{(3-0)^2 + (4-5)^2}=\sqrt{10}$. The midpoint of $\overline{AB}= (1\frac 12, 4\frac 12)$. $O= (0,0)$ so $OM=\sqrt{(\frac 32)^2 + (\frac 92)^2}= \sqrt{\frac {90}4} = \frac 32\cdot \sqrt{10}$. (As as $\triangle AOB$ is isosceles $\overline{OM}$ is the perpendicular bisector of $\overline AB$. So the area of $\triangle AOB = \frac 12 \cdot \sqrt{10}\cdot \frac 32\cdot \sqrt{10}= \frac 34\cdot 10=\frac {15}2$.

Two $m\angle AOB \ne 45^\circ$ but if we let $D=(0,4)$ then $\triangle DOB$ is a right triangle with sides $OD=4; DB=3; OB=5$ so $\theta m\angle AOB = m\angle DOB =\arctan \frac 34 =\arcsin 35=\arccos \frac 45$ so the area of $\triangle AOB = \frac 12\cdot 5 \cdot (5\sin \theta)=\frac 12\cdot 5\cdot (\sin(\arcsin 35)) =\frac 12 5\cdot 5\cdot \frac 35...$. Oh, wait. This is exactly the same way as above, isn't it.

  • $\begingroup$ Upvoted ... I have a question - Do the centers of octagon and circle need to coincide in this problem ? $\endgroup$
    – Fin27
    Commented May 27 at 5:49
  • 1
    $\begingroup$ It's not that they "have to". It's that they do. The octagonal is symmetrically reflected over then axis so its center is the origin $(0,0)$ which is also the center of the circle. In theory we don't have to use the center but it'd be a lot messier and require a lot of trig if we didn't. $\endgroup$
    – fleablood
    Commented May 27 at 6:03
  • $\begingroup$ Thankk you soo much @fleablood for the explanation 😊😊 $\endgroup$
    – Vasu Gupta
    Commented May 27 at 7:39

You attempt 1 fails because the angle AOB is not 45 degrees. It is $\arcsin\frac{3}{5}$.

Similarly, the angle BOC is $\arcsin\frac{4}{5}$. Notice that angle AOC is 90 degrees.

  • 2
    $\begingroup$ Alternative we can reason that if the angle was $45$ it would not got through the points $(\pm 3, \pm 4)$. It would go through the points $(\pm 5\sqrt 2, \pm 5 \sqrt 2)$. We don't actually need the angle. But we absolutely need to realize the triangles are not congruent. $\endgroup$
    – fleablood
    Commented May 27 at 6:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .