3 a student finds that a regular polygon has no interior angles greater than or equal to 90°. what is the largest number of sides the polygon could have? record your answer and fill in the bubbles in the grid below. be sure to use the correct place value.

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# Explanation: ## Step1: Recall the formula for the interior angle of a regular polygon. The formula for the measure of each interior angle $ I $ of a regular polygon with $ n $ sides is $ I=\frac{(n - 2)\times180^{\circ}}{n} $. ## Step2: Set up the inequality based on the problem. We know that $ I<90^{\circ} $ (since no interior angles are greater than or equal to $ 90^{\circ} $). So we set up the inequality: $$ \frac{(n - 2)\times180}{n}<90 $$ ## Step3: Solve the inequality. First, multiply both sides of the inequality by $ n $ (assuming $ n>0 $, which it is for a polygon): $$ (n - 2)\times180<90n $$ Expand the left - hand side: $$ 180n-360 < 90n $$ Subtract $ 90n $ from both sides: $$ 180n-90n-360<0 $$ $$ 90n - 360<0 $$ Add 360 to both sides: $$ 90n<360 $$ Divide both sides by 90: $$ n < 4 $$ Wait, this seems wrong. Wait, we made a mistake. The interior angle of a regular polygon: for a regular polygon, the interior angle $ I=\frac{(n - 2)\times180}{n} $, and we want $ I\leq90^{\circ} $? No, the problem says "no interior angles greater than or equal to $ 90^{\circ} $", so $ I < 90^{\circ} $. Wait, but for a triangle ($ n = 3 $), $ I=60^{\circ} $, for a quadrilateral ($ n = 4 $), $ I = 90^{\circ} $ (square). Wait, maybe we misread the problem. Wait, the problem says "no interior angles greater than or equal to $ 90^{\circ} $", so $ I<90^{\circ} $. But let's re - derive the formula. Wait, the formula for the interior angle of a regular polygon is $ I=\frac{(n - 2)\times180}{n} $. Let's solve $ \frac{(n - 2)\times180}{n}<90 $ Multiply both sides by $ n $ ( $ n>0 $): $ (n - 2)\times180<90n $ $ 180n-360 < 90n $ $ 180n-90n<360 $ $ 90n<360 $ $ n < 4 $. But that would mean $ n = 3 $, but that's not right. Wait, maybe the problem is "no interior angles greater than $ 90^{\circ} $", i.e., $ I\leq90^{\circ} $. Let's try that. Set $ \frac{(n - 2)\times180}{n}\leq90 $ Multiply both sides by $ n $: $ (n - 2)\times180\leq90n $ $ 180n-360\leq90n $ $ 180n - 90n\leq360 $ $ 90n\leq360 $ $ n\leq4 $. But a square has interior angle $ 90^{\circ} $, the problem says "no interior angles greater than or equal to $ 90^{\circ} $", so $ n < 4 $, so $ n = 3 $. But this contradicts the initial thought. Wait, maybe we made a mistake in the formula. Wait, no, the exterior angle of a regular polygon is $ E = 180 - I $, and the sum of exterior angles of any polygon is $ 360^{\circ} $, so each exterior angle $ E=\frac{360}{n} $, and $ I = 180 - E $. We want $ I<90 $, so $ 180 - E<90 $, so $ E > 90 $. Since $ E=\frac{360}{n} $, then $ \frac{360}{n}>90 $ Solve $ \frac{360}{n}>90 $ Multiply both sides by $ n $ ( $ n>0 $): $ 360>90n $ Divide both sides by 90: $ 4>n $ So $ n < 4 $, so $ n = 3 $. But this seems too small. Wait, maybe the problem was "no interior angles less than or equal to $ 90^{\circ} $"? No, the original problem says "no interior angles greater than or equal to $ 90^{\circ} $". Wait, maybe there is a misinterpretation. Let's re - read the problem: "A student finds that a regular polygon has no interior angles greater than or equal to $ 90^{\circ} $. What is the largest number of sides the polygon could have?" Wait, if $ n = 3 $, interior angle is $ 60^{\circ} $; $ n = 4 $, interior angle is $ 90^{\circ} $ (which is equal to $ 90^{\circ} $, so it's not allowed); $ n = 5 $, interior angle is $ \frac{(5 - 2)\times180}{5}=\frac{540}{5} = 108^{\circ}>90^{\circ} $, which is not allowed. Wait, this can't be. There must be a mistake in the inequality direction. Wait, maybe the problem is "no interior angles less than or equal to $ 90^{\circ} $", i.e., all interior angles are greater than $ 90^{\circ} $. Let's try that. Then $ I>90^{\circ} $ $ \frac{(n - 2)\times180}{n}>90 $ Multiply both sides by $ n $ ( $ n>0 $): $ (n - 2)\times180>90n $ $ 180n-360>90n $ $ 180n - 90n>360 $ $ 90n>360 $ $ n > 4 $ Then the largest number of sides? No, the problem says "no interior angles greater than or equal to $ 90^{\circ} $", so $ I<90 $. But according to this, the only regular polygons with $ I<90 $ are triangles ($ n = 3 $). But this seems odd. Wait, maybe the problem was "no interior angles less than $ 90^{\circ} $"? Let's check the original problem again. Wait, the user's problem: "A student finds that a regular polygon has no interior angles greater than or equal to $ 90^{\circ} $. What is the largest number of sides the polygon could have?" Wait, maybe there is a typo, and it's "no interior angles less than or equal to $ 90^{\circ} $", i.e., all interior angles are greater than $ 90^{\circ} $. Let's solve that case. $ I>90 $ $ \frac{(n - 2)\times180}{n}>90 $ $ 180n-360>90n $ $ 90n>360 $ $ n > 4 $ To find the largest number of sides? No, that would be infinite, which is not possible. Wait, no, the sum of interior angles of a polygon is $ (n - 2)\times180 $, and each interior angle $ I=\frac{(n - 2)\times180}{n} $. Wait, maybe the problem is "no interior angles greater than $ 90^{\circ} $", i.e., $ I\leq90^{\circ} $. Then: $ \frac{(n - 2)\times180}{n}\leq90 $ $ 180n-360\leq90n $ $ 90n\leq360 $ $ n\leq4 $ Since $ n = 4 $ gives $ I = 90^{\circ} $ (square), which is equal to $ 90^{\circ} $, and the problem says "no interior angles greater than or equal to $ 90^{\circ} $", so $ n < 4 $, so $ n = 3 $. But this seems too small. There must be a misinterpretation. Wait, maybe the problem is "no interior angles less than $ 90^{\circ} $", i.e., all interior angles are at least $ 90^{\circ} $. Then $ I\geq90^{\circ} $ $ \frac{(n - 2)\times180}{n}\geq90 $ $ 180n-360\geq90n $ $ 90n\geq360 $ $ n\geq4 $ To find the largest number of sides? No, as $ n $ increases, the interior angle increases (for $ n>4 $, interior angle of regular polygon is greater than $ 90^{\circ} $, and as $ n $ approaches infinity, the interior angle approaches $ 180^{\circ} $). So there is no largest number of sides in that case. Wait, going back to the original problem, maybe there was a mistake in the problem statement, or in our interpretation. Wait, another approach: maybe the problem is about exterior angles. Wait, no. Wait, let's calculate interior angles for different $ n $: - $ n = 3 $: $ I = 60^{\circ} $ - $ n = 4 $: $ I = 90^{\circ} $ - $ n = 5 $: $ I = 108^{\circ} $ - $ n = 6 $: $ I = 120^{\circ} $ -... The problem says "no interior angles greater than or equal to $ 90^{\circ} $", so $ I<90^{\circ} $. So only $ n = 3 $ satisfies this, because for $ n = 4 $, $ I = 90^{\circ} $ (which is equal, so not allowed), and for $ n>4 $, $ I>90^{\circ} $. But this seems to be the case. So the largest number of sides is 3. But this seems too small, maybe the problem was "no interior angles less than or equal to $ 90^{\circ} $", in which case we want $ I>90^{\circ} $, and as $ n $ increases, $ I $ increases, but there is no upper limit. But that can't be. Wait, maybe the original problem was "no interior angles less than $ 90^{\circ} $", i.e., all interior angles are $ \geq90^{\circ} $. Then we solve $ \frac{(n - 2)\times180}{n}\geq90 $, which gives $ n\geq4 $. But the question is "the largest number of sides", which would be infinite, which is not possible. So there must be a misinterpretation. Wait, maybe the problem is "no interior angles greater than $ 90^{\circ} $", i.e., $ I\leq90^{\circ} $. Then $ n = 3 $ ( $ I = 60^{\circ} $) and $ n = 4 $ ( $ I = 90^{\circ} $). Since the problem says "no interior angles greater than or equal to $ 90^{\circ} $", $ n = 4 $ is excluded (because $ I = 90^{\circ} $), so $ n = 3 $. But this seems to be the case. # Answer: 3

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Explanation:

Step1: Recall the formula for the interior angle of a regular polygon.

Step2: Set up the inequality based on the problem.

Step3: Solve the inequality.

Answer: