QUESTION IMAGE
Question
calculating the volume of an oblique pyramid
the area of the base of the oblique pentagonal pyramid is 50 cm² and the distance from the apex to the center of the pentagon is 6√2 cm. the measure of ∠acb is 45°
the height, ab, is cm.
the volume of the pyramid is cm³.
Step1: Find the height \( AB \)
We have a right triangle \( ABC \) with \( \angle ACB = 45^\circ \), \( \angle ABC = 90^\circ \), and \( AC = 6\sqrt{2} \) cm. In a right triangle, if one of the non - right angles is \( 45^\circ \), then it is an isosceles right triangle, and \( AB = BC \). We can also use the sine function: \( \sin(45^\circ)=\frac{AB}{AC} \). Since \( \sin(45^\circ)=\frac{\sqrt{2}}{2} \) and \( AC = 6\sqrt{2} \), we have \( AB=AC\times\sin(45^\circ)=6\sqrt{2}\times\frac{\sqrt{2}}{2} \).
Simplify the expression: \( 6\sqrt{2}\times\frac{\sqrt{2}}{2}=6\times\frac{2}{2}=6 \) cm.
Step2: Calculate the volume of the pyramid
The formula for the volume \( V \) of a pyramid is \( V=\frac{1}{3}Bh \), where \( B \) is the area of the base and \( h \) is the height of the pyramid. We know that \( B = 50\space cm^2 \) and \( h = AB=6\space cm \).
Substitute the values into the formula: \( V=\frac{1}{3}\times50\times6 \).
Simplify the expression: \( \frac{1}{3}\times50\times6 = 50\times2=100 \space cm^3 \).
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The height \( AB \) is \( \boldsymbol{6} \) cm. The volume of the pyramid is \( \boldsymbol{100} \) \( cm^3 \).