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The total surface area of a square-based pyramid is 528 cm². If its slant height is 16 cm, find the length of a side of its base.

Sagot :

Sure, let's determine the length of a side of the base for a square-based pyramid given its total surface area and slant height.

Given:
- Total surface area (A) = 528 cm²
- Slant height (l) = 16 cm

We need to find the length of a side (s) of the square base.

The formula for the surface area of a square-based pyramid is:
[tex]\[ A = \text{Base Area} + \text{Lateral Surface Area} \][/tex]

The Base Area for a square is:
[tex]\[ \text{Base Area} = s^2 \][/tex]

The Lateral Surface Area is the sum of the areas of the four triangular faces. Each triangle has a base s and a height (the slant height l). The area of each triangle is:
[tex]\[ \text{Area of one triangle} = \frac{1}{2} \cdot s \cdot l \][/tex]
Therefore, the total Lateral Surface Area is:
[tex]\[ \text{Lateral Surface Area} = 4 \cdot \left(\frac{1}{2} \cdot s \cdot l\right) = 2 \cdot s \cdot l \][/tex]

Now, the total surface area formula can be written as:
[tex]\[ 528 = s^2 + 2 \cdot s \cdot 16 \][/tex]
Simplifying, we get:
[tex]\[ 528 = s^2 + 32s \][/tex]

Rearranging the equation to form a standard quadratic equation:
[tex]\[ s^2 + 32s - 528 = 0 \][/tex]

We solve this quadratic equation using the quadratic formula:
[tex]\[ s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 32 \)[/tex], and [tex]\( c = -528 \)[/tex].

First, we calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = 32^2 - 4 \cdot 1 \cdot (-528) = 1024 + 2112 = 3136 \][/tex]

Next, we find the roots:
[tex]\[ s = \frac{-32 \pm \sqrt{3136}}{2 \cdot 1} = \frac{-32 \pm 56}{2} \][/tex]

This gives us two potential solutions:
[tex]\[ s = \frac{-32 + 56}{2} = \frac{24}{2} = 12 \][/tex]
[tex]\[ s = \frac{-32 - 56}{2} = \frac{-88}{2} = -44 \][/tex]

Since the length of a side cannot be negative, we discard [tex]\( s = -44 \)[/tex].

Hence, the length of a side of the base of the square-based pyramid is:
[tex]\[ s = 12 \, \text{cm} \][/tex]