Question

which value is closest to the area of the following figure, in square inches? 50 in² 123 in² 100 in² 89 in²

Explanation:

Step1: Recall the formula for the area of a sector

The formula for the area of a sector of a circle is \(A=\frac{1}{2}r^{2}\theta\). But when we are given the arc - like figure (assuming it's a quarter - circle, since the right - angle is implied by the perpendicular lines). The formula for the area of a quarter - circle is \(A = \frac{1}{4}\pi r^{2}\). Here, we can also use the formula for the area of a sector \(A=\frac{1}{2}lr\), where \(l\) is the arc length and \(r\) is the radius. However, if we assume it's a quarter - circle (a sector with \(\theta = 90^{\circ}=\frac{\pi}{2}\) radians), and \(r = 9\) (approximate value for calculation purposes as the radius - like measure).
The formula for the area of a sector \(A=\frac{\theta}{2\pi}\times\pi r^{2}=\frac{\theta}{2}r^{2}\). For \(\theta = 90^{\circ}=\frac{\pi}{2}\) radians (in terms of proportion, \(\frac{90^{\circ}}{360^{\circ}}=\frac{1}{4}\)), \(A=\frac{1}{4}\pi r^{2}\). Substituting \(r = 9\) (approximate value for calculation), \(A=\frac{1}{4}\times3.14\times9^{2}=\frac{1}{4}\times3.14\times81 = 63.585\). But if we use the trapezoid - like wrong approach (which is incorrect, but another way - assume it's a combination of a triangle and a sector, no, it's a sector). Wait, another way: if we consider the formula \(A=\frac{1}{2}(a + b)h\) (wrong, no). Wait, correct formula for a sector (quarter - circle) \(A=\frac{1}{4}\pi r^{2}\). But if we use the approximation:
We know that the area of a circle is \(A=\pi r^{2}\). For a quarter - circle with \(r = 9\), \(A=\frac{1}{4}\times3.14\times81=63.585\). But if we use the formula \(A=\frac{1}{2}lr\) (where \(l\) is the arc length and \(r\) is the radius). For a quarter - circle, \(l=\frac{1}{4}(2\pi r)=\frac{\pi r}{2}\). Then \(A=\frac{1}{2}\times\frac{\pi r}{2}\times r=\frac{1}{4}\pi r^{2}\).
Alternatively, if we use the formula \(A=\frac{1}{2}bh\) (wrong for a sector). But if we assume it's a right - angled sector (quarter - circle) with \(r = 9\) (approximate, since \(9.8\) is close to \(10\) and \(4\) is a distractor? No, wait, no - the figure is a sector. The formula \(A=\frac{1}{4}\pi r^{2}\) (assuming \(r = 9\) for calculation, as \(9\) is the base - like measure related to the sector's radius). \(A=\frac{1}{4}\times3.14\times81 = 63.585\). But if we use \(r = 10\) (approximate \(9.8\approx10\)), \(A=\frac{1}{4}\times3.14\times100 = 78.5\). But another approach: the formula for the area of a sector \(A=\frac{1}{2}r^{2}\theta\) (in radians). If we assume \(\theta=\frac{\pi}{2}\) (quarter - circle) and \(r = 9\), \(A=\frac{1}{2}\times81\times\frac{\pi}{2}\approx63.6\). But wait, no - the figure is a sector. Wait, the formula \(A=\frac{1}{2}lr\). If we assume \(l\) (arc length) is approximately \(\frac{1}{4}(2\pi r)\) (for a quarter - circle). But if we use the wrong approach (sum of a triangle and a sector? No, it's a single sector). Wait, no - the formula for the area of a sector \(A=\frac{\theta}{360^{\circ}}\times\pi r^{2}\). If we assume \(\theta = 90^{\circ}\) (quarter - circle) and \(r\approx9\) (since \(9.8\) is close to \(10\) but \(9\) is given as a measure). \(A=\frac{90^{\circ}}{360^{\circ}}\times3.14\times9^{2}=\frac{1}{4}\times3.14\times81 = 63.585\). But if we use \(r = 10\) (approximate \(9.8\) as \(10\)), \(A=\frac{1}{4}\times3.14\times100=78.5\). But wait, another way: the formula \(A = \frac{1}{2}(a + b)h\) is for trapezoid. If we wrongly use \(a = 4\), \(b=9.8\), \(h = 9\) (wrong, but let's calculate \(A=\frac{1}{2}(4 + 9.8)\times9=\frac{1}{2}\times13.8\times9=62.1\). But the closest value to \(63.585\) (using \(r = 9\)) or \(78.5\) (using \(r = 10\)) among the options:
If we calculate \(A=\frac{1}{4}\times3.14\times9^{2}=63.585\), \(A=\frac{1}{4}\times3.14\times(9.8)^{2}=\frac{1}{4}\times3.14\times96.04 = 75.4\).
Another approach: the formula \(A=\frac{1}{2}r^{2}\theta\) (in radians). If \(\theta=\frac{\pi}{2}\) and \(r = 9.8\), \(A=\frac{1}{2}\times(9.8)^{2}\times\frac{\pi}{2}=\frac{1}{4}\times3.14\times96.04=75.4\).
If we consider the options:
- Option 1: \(50\)
- Option 2: \(123\)
- Option 3: \(100\)
- Option 4: \(89\)

\(75.4\) is closest to \(89\) (maybe there was a mis - take in radius assumption. If we assume the figure is a sector with \(r = 10\) (approximate \(9.8\) as \(10\)) and \(\theta = 90^{\circ}\), \(A=\frac{1}{4}\times3.14\times100 = 78.5\) (still closer to \(89\) than to \(50\) or \(100\) or \(123\))

Answer:

\(89\ in^{2}\)