Question

1. choose the correct answer.

a = 3 cm, b = 7 cm, c = 7.4 cm
the triangle is a(n) ____ triangle.

• right
• acute
• obtuse

Explanation:

Step1: Recall the Pythagorean theorem and its converse for triangle types.

For a triangle with sides \(a\), \(b\), and hypotenuse \(c\) (the longest side), the Pythagorean theorem is \(a^{2}+b^{2}=c^{2}\) for a right triangle. For an acute triangle, \(a^{2}+b^{2}>c^{2}\), and for an obtuse triangle, \(a^{2}+b^{2}<c^{2}\). Here, \(c = 7.4\) cm (the longest side), \(a = 3\) cm, \(b = 7\) cm.

Step2: Calculate \(a^{2}+b^{2}\) and \(c^{2}\).

First, calculate \(a^{2}+b^{2}\):
\(a^{2}=3^{2} = 9\), \(b^{2}=7^{2}=49\), so \(a^{2}+b^{2}=9 + 49=58\).

Then, calculate \(c^{2}\):
\(c^{2}=(7.4)^{2}=54.76\).

Step3: Compare \(a^{2}+b^{2}\) and \(c^{2}\).

We have \(a^{2}+b^{2}=58\) and \(c^{2}=54.76\). Since \(58>54.76\) (i.e., \(a^{2}+b^{2}>c^{2}\)), the triangle is acute? Wait, no, wait. Wait, maybe I mixed up. Wait, the longest side is \(c = 7.4\), so we should check \(a^{2}+b^{2}\) vs \(c^{2}\). Wait, \(a = 3\), \(b = 7\), \(c = 7.4\). Wait, \(3^{2}+7^{2}=9 + 49 = 58\), \(7.4^{2}=54.76\). Wait, but if \(c\) is the longest side, then for a right triangle, \(a^{2}+b^{2}=c^{2}\). For acute, \(a^{2}+b^{2}>c^{2}\) (since the square of the longest side is less than the sum of squares of the other two, so the angle opposite \(c\) is acute, so all angles are acute). For obtuse, \(a^{2}+b^{2}<c^{2}\). Wait, but wait, maybe I made a mistake. Wait, let's re - check the side lengths. Wait, the triangle is a right - angled triangle? Wait, no, wait the original figure is a right - angled triangle? Wait, no, the problem is to determine the type. Wait, maybe I miscalculated \(c^{2}\). Wait, \(7.4\times7.4\): \(7\times7 = 49\), \(7\times0.4=2.8\), \(0.4\times7 = 2.8\), \(0.4\times0.4 = 0.16\), so \((7 + 0.4)^{2}=7^{2}+2\times7\times0.4+0.4^{2}=49 + 5.6+0.16 = 54.76\). And \(3^{2}+7^{2}=9 + 49 = 58\). Since \(58>54.76\), the square of the longest side is less than the sum of the squares of the other two sides. So the angle opposite the longest side (angle \(C\)) is acute. Since the longest side's opposite angle is acute, all angles are acute? Wait, no, wait maybe the original figure is a right triangle, but the given lengths are approximate. Wait, wait \(3^{2}+7^{2}=58\), and \(7.4^{2}=54.76\). Wait, but if it's a right triangle, \(3^{2}+7^{2}\) should be equal to \(c^{2}\). But \(58\) is not equal to \(54.76\). Wait, maybe there is a typo, but according to the calculation, \(a^{2}+b^{2}>c^{2}\), so the triangle is acute? Wait, no, wait the options are right, acute, obtuse. Wait, maybe I messed up the sides. Wait, maybe \(a\) and \(b\) are the legs, and \(c\) is the hypotenuse. Wait, \(3^{2}+7^{2}=9 + 49 = 58\), and \(7.4^{2}=54.76\). So \(58>54.76\), which means that \(c^{2}<a^{2}+b^{2}\), so the angle opposite \(c\) is acute. Since \(c\) is the longest side, all angles are acute? But that contradicts the figure which looks like a right triangle. Wait, maybe the given \(c\) is an approximation. Wait, let's check \(3 - 7 - \sqrt{3^{2}+7^{2}}=\sqrt{58}\approx7.62\). Oh! Wait, maybe there is a mistake in the given \(c\) value. Wait, the user provided \(c = 7.4\) cm, but \(\sqrt{3^{2}+7^{2}}=\sqrt{58}\approx7.62\) cm. So if \(c = 7.4\) cm, which is less than \(7.62\) cm, then \(a^{2}+b^{2}>c^{2}\), so the triangle is acute? But that seems odd. Wait, no, maybe I got the sides wrong. Wait, maybe \(a = 3\), \(b = 7\), and \(c\) is the hypotenuse. Wait, no, in a right triangle, \(c=\sqrt{3^{2}+7^{2}}\approx7.62\). So if \(c = 7.4\), which is less than \(7.62\), then the triangle is acute? But the options include right. Wait, maybe the problem has a typo, but according to the given values:

\(a = 3\), \(b = 7\), \(c = 7.4\).

Calculate \(a^{2}+b^{2}=9 + 49 = 58\)

\(c^{2}=54.76\)

Since \(a^{2}+b^{2}>c^{2}\), the triangle is acute? But that seems wrong. Wait, no, wait the formula for triangle types:

For a triangle with sides \(a\leq b\leq c\),

- If \(a^{2}+b^{2}=c^{2}\), right triangle.

- If \(a^{2}+b^{2}>c^{2}\), acute triangle (because the angle opposite \(c\) is acute, and since \(c\) is the longest side, all angles are acute).

- If \(a^{2}+b^{2}<c^{2}\), obtuse triangle (angle opposite \(c\) is obtuse).

So in this case, \(a^{2}+b^{2}=58\), \(c^{2}=54.76\), so \(a^{2}+b^{2}>c^{2}\), so the triangle is acute. But wait, the figure looks like a right triangle. Maybe the given \(c\) is incorrect. But according to the given numbers, we have to go with the calculation.

Wait, wait, maybe I made a mistake in calculating \(c^{2}\). Let's recalculate \(7.4\times7.4\):

\(7\times7 = 49\), \(7\times0.4 = 2.8\), \(0.4\times7 = 2.8\), \(0.4\times0.4 = 0.16\)

So \((7 + 0.4)^{2}=7^{2}+2\times7\times0.4+0.4^{2}=49 + 5.6+0.16 = 54.76\). That's correct.

And \(3^{2}+7^{2}=9 + 49 = 58\). So \(58>54.76\), so \(a^{2}+b^{2}>c^{2}\), so the triangle is acute. But the options have right, acute, obtuse. So according to the calculation, the answer should be acute. But that seems conflicting with the figure. Maybe the problem has a typo, but based on the given values, we proceed.

Answer:

acute (the option corresponding to "acute")