Thread: Area of Triangle enclosed between arcs

1. Area of Triangle enclosed between arcs

The triangle DEF, an isosceles right triangle, EF=DF=2.
Circular arcs of radius 2 centered at F and D meet the hypotenuse at G and H, respectively. Find the area inside the triangle but not enclosed between the arcs. Use real numbers for your answer and not decimal approximations.

2. Originally Posted by warriors837
The triangle DEF, an isosceles right triangle, EF=DF=2.
Circular arcs of radius 2 centered at F and D meet the hypotenuse at G and H, respectively. Find the area inside the triangle but not enclosed between the arcs. Use real numbers for your answer and not decimal approximations.
The way you've described the triangle, F is the vertex of the right angle. A circular arc centered at F would intersect the hypotenuse at ponits D and E. Did you mean to say that the circular arcs are centered at D and E?

If so, then I'm attaching a picture of what the two arcs look like. If you calculate the area of each (they are congruent) and add them together, how will the combined area differ from the area of the triangle overall?

If you can see that, the answer to your question should be fairly simple to find.

3. drop a perpendicular line from line ED passing through F, that surely HELPS.

4. warriors837, PLEASE find fault with this method.

KLAMKIN AREA METHOD:

(i) Area of BDC + Area of BDE + Area of BAE = (1/2)(Base x Height)

x + y + x = (1/2)(2)(2) = 2, then

2x + y = 2 - - - eq'n 1

(ii) Area of BDE + Area of BAE = Area of BAD

x + y = (1/2)(r^2)(theta) = (1/2)(2^2)(pi/4) = pi/2, then

x + y = pi/2,

y = (pi/2 - x) - - - eq'n 2

substitute eq'n 1 in eq'n 2.

2x + (pi/2 - x) = 2, then

2x - x = 2 - pi/2

x = 2 - pi/2

But we are looking for the unshaded portion of the figure,

thus, Area of unshaded portion = Area = 2x = 2(2 - pi/2) = 4 - pi = 0.8584 square units.

i hope it is correct . . . .

5. Since the triangle is isosceles right, you know that the angles at A and C are 45 degrees. This means that their arcs are 1/8 of a circle. And you know the circle's radius.

Draw the altitude line from B to AC, and find the height of the triangle. Also, find the length of the base. From this, find the area of the whole triangle. Also, find the area of half of the triangle (that is, find the area of one of the smaller two triangles formed by the altitude line).

Find the area of one of the sectors, and subtract out the area of a small triangle. This gives you half the area of the curvy middle portion. Multiply by 2, and subtract from the area of the original triangle.

7. Find the area of one of the sectors, and subtract out the area of a small triangle.

How do I go about that?

The white part's area is 2-(pi-2). Correct?

9. Originally Posted by pacman
warriors837, PLEASE find fault with this method.

KLAMKIN AREA METHOD:

(i) Area of BDC + Area of BDE + Area of BAE = (1/2)(Base x Height)

x + y + x = (1/2)(2)(2) = 2, then

2x + y = 2 - - - eq'n 1

(ii) Area of BDE + Area of BAE = Area of BAD

x + y = (1/2)(r^2)(theta) = (1/2)(2^2)(pi/4) = pi/2, then

x + y = pi/2,

y = (pi/2 - x) - - - eq'n 2

substitute eq'n 1 in eq'n 2.

2x + (pi/2 - x) = 2, then

2x - x = 2 - pi/2

x = 2 - pi/2

But we are looking for the unshaded portion of the figure,

thus, Area of unshaded portion = Area = 2x = 2(2 - pi/2) = 4 - pi = 0.8584 square units.

i hope it is correct . . . .

It was...

10. Originally Posted by warriors837
Find the area of one of the sectors, and subtract out the area of a small triangle.
How do I go about that?
Apply the area-of-a-sector formula you've memorized.

11. triangle minus area inside both sectors = answer to your question
2 sectors minus triangle = area inside both sectors
therefore, triangle minus (2 sectors minus triangle) = answer to question

note that each arc is 45 degrees, or 1/8 of a circle
2 arcs would then be 1/4 of a circle

$\frac{1}{2} b h - (\frac{1}{4} \pi r^2 - \frac{1}{2} b h)$
$bh-\frac{1}{4}\pi r^2$
$2*2-\frac{1}{4} \pi 2^2$
And so you are correct. The answer to your question is $4-\pi$