A sector of a circle is inside a rectangle. The rectangle has the side lengths 25cm and 15cm. This also means that the radius of the sector is 25cm. What would be the area in the rectangle that is outside the sector?
The answer is 185cm^2 but i dont no how to get it. Please help!
+26387 A sector of a circle is inside a rectangle. The rectangle has the side lengths 25cm and 15cm. This also means that the radius of the sector is 25cm. What would be the area in the rectangle that is outside the sector?
The answer is 185cm^2 but i dont no how to get it. Please help!
\(\begin{array}{rcll} A_{\text{rectangle}} &=& 25\cdot 15 \ cm^2 \\ &=& 375 \ cm^2 \\\\ A_{\text{sector of a circle}} &=& \pi \cdot 25^2 \cdot \frac{\varphi}{360^\circ} \ cm^2 \\ A_{\text{Area in the rectangle and outside sector}} &=& A_{\text{rectangle}} - A_{\text{sector of a circle}}\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 25^2 \cdot \frac{\varphi}{360^\circ} \ cm^2\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 625 \cdot \frac{\varphi}{360^\circ} \ cm^2\\\\ \sin{(\frac{\varphi}{2})} &=& \frac{\frac{15}{2}}{25} \\ \sin{(\frac{\varphi}{2})} &=& \frac{7.5}{25} \\ \sin{(\frac{\varphi}{2})} &=& 0.3 \\ \frac{\varphi}{2} &=& \arcsin{(0.3)} \\ \frac{\varphi}{2} &=& 17.4576031237^\circ \\ \varphi &=& 2\cdot 17.4576031237^\circ \\ \varphi &=& 34.9152062474^\circ \\\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 625 \cdot \frac{34.9152062474^\circ}{360^\circ} \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - 190.432908760 \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &=& 184.567091240 \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &\approx& 185 \ cm^2\\\\ \end{array}\)
+8581 Well, You know that the formula is for the area of a rectangle. So, When you do all those steps and such, it kind of makes sense on how that answer came to be.
Did i make any sense to you?
It doesn't work!. Because it is a "sector of a circle is inside a rectangle", and NOT a circle?
+26387 A sector of a circle is inside a rectangle. The rectangle has the side lengths 25cm and 15cm. This also means that the radius of the sector is 25cm. What would be the area in the rectangle that is outside the sector?
The answer is 185cm^2 but i dont no how to get it. Please help!
\(\begin{array}{rcll} A_{\text{rectangle}} &=& 25\cdot 15 \ cm^2 \\ &=& 375 \ cm^2 \\\\ A_{\text{sector of a circle}} &=& \pi \cdot 25^2 \cdot \frac{\varphi}{360^\circ} \ cm^2 \\ A_{\text{Area in the rectangle and outside sector}} &=& A_{\text{rectangle}} - A_{\text{sector of a circle}}\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 25^2 \cdot \frac{\varphi}{360^\circ} \ cm^2\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 625 \cdot \frac{\varphi}{360^\circ} \ cm^2\\\\ \sin{(\frac{\varphi}{2})} &=& \frac{\frac{15}{2}}{25} \\ \sin{(\frac{\varphi}{2})} &=& \frac{7.5}{25} \\ \sin{(\frac{\varphi}{2})} &=& 0.3 \\ \frac{\varphi}{2} &=& \arcsin{(0.3)} \\ \frac{\varphi}{2} &=& 17.4576031237^\circ \\ \varphi &=& 2\cdot 17.4576031237^\circ \\ \varphi &=& 34.9152062474^\circ \\\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - \pi \cdot 625 \cdot \frac{34.9152062474^\circ}{360^\circ} \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &=& 375 \ cm^2 - 190.432908760 \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &=& 184.567091240 \ cm^2\\\\ A_{\text{Area in the rectangle and outside sector}} &\approx& 185 \ cm^2\\\\ \end{array}\)
+129847 Here's a pic of the situation :
The intersection point of the side of the rectangle and the circle at F = (sqrt(25^2 -7.5^2), 7.5)
And, by symmetry.....the central angle of the sector is given by 2* tan-1(7.5/ sqrt(25^2 -7.5^2)) = about 34.915206247444°
So.....the area of the rectangle outside the sector = 375 - pi*(25)^2* [ 34.915206247444 / 360) = [375 - 190.432] cm^2 ≈ 185 cm^2
