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I was sent this problem by a friend, they say it is a year 6 question. Using trigonometry I got the shaded area as $4 - \frac{\pi}{3}$.

Now Year six, I doubt if they do trigonometry. Is there another approach out there?

enter image description here

Anubhab
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  • Are you sure about your answer? It looks a little large to me. I would say that using the symmetry of the diagram you should be expecting something like $8-2\pi$. – postmortes Dec 09 '18 at 08:55
  • Can you provide details about what you'd expect in sixth year? Visitors here come from a wide variety of contries and educational systems. I assume they wouldn't be doing integrals either? – MvG Dec 09 '18 at 11:51
  • @MvG, it doesn't matter now as the actual answer is $\frac{32}{5}-4\tan^{-1}(\frac{3}{4})$, and so you can't do it without trigonometry. – Anubhab Dec 09 '18 at 13:11
  • This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|\text{region}PAT|$. – Blue Dec 09 '18 at 14:24

3 Answers3

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Diagram

Label the vertices as shown in the diagram. $\tan\alpha=\frac{1}{2}\implies \tan 2\alpha=\frac{4}{3}\implies 2\alpha=\tan^{-1}(\frac{4}{3})$

$\therefore$ Area of circular sector$ BAG=8\tan^{-1}(\frac{4}{3})$

Area of $\triangle AGC=16\cos\alpha\sin\alpha=8\sin2\alpha=\frac{32}{5}$

Thus, area of curved figure $BEG=\frac{48}{5}-8\tan^{-1}(\frac{4}{3})$

Therefore, Area of shaded part$=\frac{1}{2}$ (Area of $BEFC$)$-\frac{1}{2}$ (Area of semicircle)$-$Area of curved figure $BEG$.

This gives $\frac{32}{5}-4\pi+8\tan^{-1}(\frac{4}{3})=\frac{32}{5}-4\tan^{-1}(\frac{3}{4})\approx1.252$

Anubhab
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Flip the image by vertically.

It is not as simple as you think. Year six is probably high school when they can do geometry and calculus. I have done it using calculus and the area is as shown in the figure enter image description here

Satish Ramanathan
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enter image description here

$\color{blue}{\text{Triangle Area }}$(the larger blue triangle) = $\color{blue}{\text{Sector Area }}$(of the circle) + one $\color{blue}{\text{Corner Area }}$+ $\color{red}{\text{x Area }}$

If radius (r) = 1

Steps Formulas Values
1. $A_{\color{green}{\bigcirc}} = \pi r^2$ $\pi (1)^2 = \pi \times 1 = \pi \approx 3.1416$
2. $A_{\color{brown}{\square}} = \text{Side}^2$ $2 \times 2 = 4$
3. $A_{\color{blue}{\triangle}} = \frac{1}{2} \times \text{Base} \times \text{Height}$ $\frac{1}{2} \times 2 \times 1 = 1$
4. $A_{\text{corner}} = \frac{A_{\color{brown}{\square}} - A_{\color{green}{\bigcirc}}}{4}$ $\frac{4 - 3.1415}{4} = 0.2146$

$A_{\text{corner}}$ is the area outside the circle but inside the square for a single corner only.

Step 5 – The BLUE Triangle | Find the $\alpha$

Using trigonometry, given that Opposite side (a) = 1 and Adjacent side (b) = 2

$tan(\alpha)$ = 1/2 => $\alpha$ = $tan^{-1} \left( \frac{1}{2} \right)$ => $\alpha$ = 26.5650°

$\color{lightgray}{\boxed{\text{Hypotenuse } (c) \text{ is: } a^2 + b^2 = c^2 \Rightarrow 1^2 + 2^2 = c^2 \Rightarrow 1 + 4 = c^2 \Rightarrow \sqrt{5} = c \approx 2.236}}$

Step 6,7 – The ORANGE Triangle | Find the $\beta$ and Area $A_{\color{orange}{\triangle}}$

Intuitively analyze it, the Orange Triangle is Isosceles, and one of its angles is an alternate interior angle of $\alpha$.

∴ both of its Base Angles are 26.5650$^\circ$

∴ the Vertex Angle ($\beta$) = 180$^\circ$ - 26.565$^\circ$ - 26.565$^\circ$ $\approx$ 126.8700$^\circ$

Splitting the orange isosceles triangle from its center creates two equal right-angled triangles. By drawing a perpendicular from the vertex angle, or equivalently from the center of the circle to the base of the orange isosceles triangle, divide it into two congruent (or equal) right-angled triangles.

Now, consider The Orange Triangle as two right-angled triangles with the Hypotenuse (c) is equal to the radius (r) = 1 and $\alpha$ = 26.5650$^\circ$

Using Trigonometry

Opposite side (a) = c $\cdot sin$(26.5650$^\circ$) $\approx$ 0.4472
Adjacent side (b) = c $\cdot cos$(26.5650$^\circ$) $\approx$ 0.8944

Since, the area of The Orange Triangle is twice the sum of both halves

Area $A_{\color{orange}{\triangle}}$ = 2 $\times \frac{1}{2} \times b \times a$ $\Rightarrow$ 0.4472 $\times$ 0.8944 $\approx$ 0.4

Alternatively, a more common approach is to duplicate the orange isosceles triangle along its non radii side to form a parallelogram. The area of this parallelogram is given by
$A = r \times h$, where the height $h$ is $r \cdot \sin(2\alpha)$. The factor $2$ appears because the angle is also duplicated at this corner, angle $\beta$ doesn't have this issue. Taking half of this area gives the area of the orange triangle.

Step 8,9 – The Non-GREEN region of the Circle | Find Sector & Segment

Area of the minor sector of the circle using Vertex Angle ($\beta$)
$A_{\text{sector}} = \frac{A_{\color{green}{\bigcirc}}}{360} \times \beta$ $\Rightarrow$ $A_{\text{sector}} = \frac{3.1416}{360} \times$ 126.8700 $\approx$ 1.1072

By removing the orange triangle from this sector left with the area of the minor segment (or blue Segment)

$A_{\color{blue}{segment}}$ = $A_{\text{sector}}$ - $A_{\color{orange}{\triangle}}$ $\Rightarrow$ 1.1072 - 0.4 $\approx$ 0.7072


Step 10 – Find area X

$A_{\color{blue}{\triangle}}$ = $A_{\color{blue}{segment}}$ + $A_{\color{blue}{corner}}$ + $\color{red}{A}_{\color{red}{x}}$
$\Rightarrow$ $\color{red}{A}_{\color{red}{x}}$ = $A_{\color{blue}{\triangle}}$ - $A_{\color{blue}{segment}}$ - $A_{\color{blue}{corner}}$
$\Rightarrow$ $\color{red}{A}_{\color{red}{x}}$ = 1 - 0.7072 - 0.2146
$\Rightarrow$ $\color{red}{A}_{\color{red}{x}}$ = 0.0782

Note: ∵ r = 1, Ax = 0.0782. For r = 4 (diameter = 8), see the table below.


Summary

1.  - Area of the circle
2.  - Area of the circumscribed square
3.  - Area of the blue right-angle triangle (quarter of the square)
4.  - Area of the one Corner (quarter of the area of Square minus Circle)
5.  - Angle α of the blue triangle using tan(α) where a=radius and b=diameter
6.  - Angle β of the Orange Triangle, an isosceles triangle with two angles as α
7.  - Area of the Orange Triangle calculated using two right-angled triangles
8.  - Area of the minor sector using β
9.  - Area of the minor segment (sector minus Orange Triangle)
10. - AreaX = area of Blue triangle - area of segment - area of Corner

JavaScript

const diameters = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]; // Example diameters
const calculations = [];

function calculateProperties(diameter) { const radius = diameter / 2; const areaCircle = Math.PI * Math.pow(radius, 2); const areaSquare = Math.pow(diameter, 2); const areaBlueTriangle = areaSquare / 4; const areaCorner = (areaSquare - areaCircle) / 4;

const angleAlpha = Math.atan(radius / diameter) * (180 / Math.PI); // In degrees
const angleBeta = (180 - 2 * angleAlpha);

// Step 7: Calculate area of the Orange Isosceles Triangle using right-angled triangles
const x = radius * Math.cos(angleAlpha * Math.PI / 180);
const h = radius * Math.sin(angleAlpha * Math.PI / 180);
const areaHalfOrangeTriangle = 0.5 * x * h;
const areaOrangeTriangle = 2 * areaHalfOrangeTriangle;

// Step 8: Area of the minor sector (using Beta)
const areaMinorSector = (angleBeta / 360) * areaCircle;

// Step 9: Area of the minor segment (sector minus Orange Triangle)
const areaMinorSegment = areaMinorSector - areaOrangeTriangle;

// Step 10: AreaX = area of Blue triangle - area of segment - area of Corner
const areaX = areaBlueTriangle - areaMinorSegment - areaCorner;

return { 
    radius, 
    areaCircle, 
    areaSquare, 
    areaBlueTriangle, 
    angleAlpha, 
    angleBeta, 
    areaOrangeTriangle, 
    areaCorner, 
    areaMinorSector, 
    areaMinorSegment, 
    areaX
};

}

function printResults(results) { console.log("Calculated Values:"); results.forEach((res, index) => { console.log(Input Diameter: ${diameters[index]}); console.log(0. Radius: ${res.radius.toFixed(4)}); console.log(1. Area of Circle: ${res.areaCircle.toFixed(4)}); console.log(2. Area of Square: ${res.areaSquare.toFixed(4)}); console.log(3. Area of Blue Triangle: ${res.areaBlueTriangle.toFixed(4)}); console.log(4. Area of Corner: ${res.areaCorner.toFixed(4)}); console.log(5. Angle α: ${res.angleAlpha.toFixed(4)}°); console.log(6. Angle β (Orange Triangle): ${res.angleBeta.toFixed(4)}°); console.log(7. Area of Orange Triangle: ${res.areaOrangeTriangle.toFixed(4)}); console.log(8. Area of Minor Sector: ${res.areaMinorSector.toFixed(4)}); console.log(9. Area of Minor Segment: ${res.areaMinorSegment.toFixed(4)}); console.log(10. AreaX: ${res.areaX.toFixed(4)}); console.log("------------------------------"); }); }

diameters.forEach(d => calculations.push(calculateProperties(d))); printResults(calculations);

Output:

Property Diameter=1 Diameter=2 Diameter=3 Diameter=4 Diameter=5 Diameter=6 Diameter=7 Diameter=8 Diameter=9 Diameter=10 Diameter=11 Diameter=12 Diameter=13 Diameter=14 Diameter=15 Diameter=16 Diameter=17 Diameter=18 Diameter=19 Diameter=20
Radius 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000 4.5000 5.0000 5.5000 6.0000 6.5000 7.0000 7.5000 8.0000 8.5000 9.0000 9.5000 10.0000
Area of Circle 0.7854 3.1416 7.0686 12.5664 19.6350 28.2743 38.4845 50.2655 63.6173 78.5398 95.0332 113.0973 132.7323 153.9380 176.7146 201.0619 226.9801 254.4690 283.5287 314.1593
Area of Square 1.0000 4.0000 9.0000 16.0000 25.0000 36.0000 49.0000 64.0000 81.0000 100.0000 121.0000 144.0000 169.0000 196.0000 225.0000 256.0000 289.0000 324.0000 361.0000 400.0000
Area of Blue Triangle 0.2500 1.0000 2.2500 4.0000 6.2500 9.0000 12.2500 16.0000 20.2500 25.0000 30.2500 36.0000 42.2500 49.0000 56.2500 64.0000 72.2500 81.0000 90.2500 100.0000
Area of Corner 0.0537 0.2146 0.4829 0.8584 1.3413 1.9314 2.6289 3.4336 4.3457 5.3650 6.4917 7.7257 9.0669 10.5155 12.0714 13.7345 15.5050 17.3827 19.3678 21.4602
Angle α 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651° 26.5651°
Angle β (Orange Triangle) 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699° 126.8699°
Area of Orange Triangle 0.1000 0.4000 0.9000 1.6000 2.5000 3.6000 4.9000 6.4000 8.1000 10.0000 12.1000 14.4000 16.9000 19.6000 22.5000 25.6000 28.9000 32.4000 36.1000 40.0000
Area of Minor Sector 0.2768 1.1071 2.4911 4.4286 6.9197 9.9643 13.5626 17.7144 22.4198 27.6787 33.4912 39.8574 46.7770 54.2503 62.2771 70.8575 79.9915 89.6790 99.9202 110.7149
Area of Minor Segment 0.1768 0.7071 1.5911 2.8286 4.4197 6.3643 8.6626 11.3144 14.3198 17.6787 21.3912 25.4574 29.8770 34.6503 39.7771 45.2575 51.0915 57.2790 63.8202 70.7149
AreaX 0.0196 0.0782 0.1761 0.3130 0.4891 0.7042 0.9586 1.2520 * 1.5846 1.9562 2.3670 2.8170 3.3060 3.8342 4.4015 5.0080 5.6535 6.3382 7.0620 7.8249
Waqar
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