Question

1. the surface area of the triangular prism is 216 in². what is the height, x,?

Explanation:

Step1: Calculate the area of the triangular bases

The formula for the area of a triangle is \(A=\frac{1}{2}bh\). Here, \(b = 6\) cm and \(h = 4\) cm.
\(A_{triangle}=\frac{1}{2}\times6\times4= 12\) \(cm^{2}\). Since there are two triangular bases, \(A_{total - triangles}=2\times12 = 24\) \(cm^{2}\).

Step2: Calculate the area of the rectangular faces

The three rectangular faces have dimensions:

• One with dimensions \(6\times x\), area \(A_{1}=6x\)
• One with dimensions \(5\times x\), area \(A_{2}=5x\)
• One with dimensions \( \sqrt{4^{2}+6^{2}}\times x\) (but wait, no - using the fact that for a right - angled triangle (since \(3 - 4 - 5\) is a Pythagorean triple, but here base \(b = 6\), height \(h = 4\), hypotenuse \(c=\sqrt{4^{2}+6^{2}}=\sqrt{16 + 36}=\sqrt{52}\approx7.21\) is wrong. Wait, no - the prism has right - angled triangular bases (because \(3 - 4 - 5\) is wrong, but if we assume the base triangle has base \(b = 6\), height \(h = 4\), hypotenuse \(c=\sqrt{4^{2}+6^{2}}=\sqrt{52}\). But actually, using the formula for the surface area of a triangular prism \(SA=2A_{base}+(a + b + c)l\), where \(a,b,c\) are the sides of the base triangle and \(l\) is the length of the prism.

The sides of the base triangle: \(a = 6\), \(b = 5\), \(c=\sqrt{4^{2}+6^{2}}=\sqrt{16 + 36}=\sqrt{52}\) is wrong. Wait, no - if the base triangle has base \(b = 6\), height \(h = 4\), and one side is \(5\) (maybe a mis - draw). Let's use the formula \(SA = 2\times(\frac{1}{2}\times6\times4)+(6 + 5+ \sqrt{4^{2}+6^{2}})x\). But wait, if we assume the base triangle is a right - angled triangle with legs \(3\) and \(4\) (no, the base is \(6\)). Wait, another approach:
The surface area formula for a triangular prism \(SA=2\times(\frac{1}{2}bh)+(b + s_{1}+s_{2})x\). If the base triangle has \(b = 6\), height \(h = 4\), and the other two sides of the base triangle: using Pythagoras, if it's a right - angled triangle (assuming the height is perpendicular to the base), and one of the other sides is \(5\) (maybe a \(3 - 4 - 5\) triangle scaled wrong. Wait, no - let's re - express.
The surface area \(SA=2\times(\frac{1}{2}\times6\times4)+(6 + 5+ 5)x\) (assuming the non - base sides of the triangular base are equal. Wait, no - the correct formula \(SA = 2A_{base}+(perimeter_{base})x\). \(A_{base}=\frac{1}{2}\times6\times4 = 12\), \(perimeter_{base}=6 + 5+ 5\) (if it's an isosceles triangle with two sides of length \(5\)).
So \(216=2\times12+(6 + 5+ 5)x\)
\(216=24+(16)x\)

Step3: Solve for \(x\)

Subtract \(24\) from both sides: \(216-24=16x\)
\(192 = 16x\)
Divide both sides by \(16\): \(x=\frac{192}{16}=12\)

Answer:

\(x = 12\) \(cm\)