QUESTION IMAGE
Question
- x = ??? (show your work below!!)
(image of a right triangle with legs 3, 3, angles 45°, 45°, right angle, and hypotenuse x)
- x = ??? (show your work below!!)
(image of a right triangle with legs 4, 4, angles 45°, 45°, right angle, and hypotenuse x)
- x = ??? (show your work below!!)
(image of a right triangle with legs 2, 2, angles 45°, 45°, right angle, and hypotenuse x)
Problem 2:
Step1: Identify Triangle Type
The triangle is a right - isosceles triangle (two angles \(45^{\circ}\), one right angle). In a right - isosceles triangle, the legs are equal, and we can use the Pythagorean theorem \(c^{2}=a^{2}+b^{2}\), where \(a = b = 3\) and \(c=x\).
Step2: Apply Pythagorean Theorem
Substitute \(a = 3\) and \(b = 3\) into the formula: \(x^{2}=3^{2}+3^{2}=9 + 9=18\). Then \(x=\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\).
Step1: Identify Triangle Type
The triangle is a right - isosceles triangle (two angles \(45^{\circ}\), one right angle). The legs \(a = b = 4\), and we use the Pythagorean theorem \(c^{2}=a^{2}+b^{2}\) with \(c = x\).
Step2: Apply Pythagorean Theorem
Substitute \(a = 4\) and \(b = 4\) into the formula: \(x^{2}=4^{2}+4^{2}=16 + 16 = 32\). Then \(x=\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}\).
Step1: Identify Triangle Type
The triangle is a right - isosceles triangle (two angles \(45^{\circ}\), one right angle). The legs \(a = b = 2\), and we use the Pythagorean theorem \(c^{2}=a^{2}+b^{2}\) with \(c=x\).
Step2: Apply Pythagorean Theorem
Substitute \(a = 2\) and \(b = 2\) into the formula: \(x^{2}=2^{2}+2^{2}=4 + 4=8\). Then \(x=\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\).
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\(3\sqrt{2}\)