Question

a solid oblique pyramid has a square base with an edge length of 2 cm. angle bac measures 45° and ac measures 3.6cm. what is the volume of the pyramid? 2.4 cm³ 3.6 cm³ 4.8 cm³ 7.2 cm³

Explanation:

Step1: Find pyramid height

In right triangle $BAC$, $\angle BAC=45^\circ$, $AC=3.6$ cm.
$\sin(45^\circ)=\frac{\text{height } h}{AC}$
$h = AC \cdot \sin(45^\circ) = 3.6 \cdot \frac{\sqrt{2}}{\sqrt{2}} = 3.6 \cdot \frac{1}{\sqrt{2}} \cdot \sqrt{2}$? No, simpler: $\tan(45^\circ)=\frac{h}{\text{adjacent}}$, but since $\angle BAC=45^\circ$, triangle is isosceles, so $h = AC \cdot \sin(45^\circ)$? Wait no, $\angle ACB$ is right angle, so $\sin(45^\circ)=\frac{BC}{AB}$, $\tan(45^\circ)=\frac{BC}{AC}=1$, so $BC=AC=3.6$? No, wait $\angle BAC=45^\circ$, right angle at $C$, so $BC=AC \cdot \tan(45^\circ)=3.6 \cdot 1=3.6$ cm. This $BC$ is the height of the pyramid.

Step2: Calculate base area

Square base edge = 2 cm.
$\text{Base Area} = s^2 = 2^2 = 4$ cm²

Step3: Compute pyramid volume

Volume formula: $V=\frac{1}{3} \times \text{Base Area} \times \text{Height}$
$V = \frac{1}{3} \times 4 \times 3.6$

Answer:

$4.8$ cm³ (Option C: $4.8\ \text{cm}^3$)