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 $BAC$, $\sin(45^\circ)=\frac{h}{AC}$.
$h = AC \times \sin(45^\circ) = 3.6 \times \frac{\sqrt{2}}{2} = 3.6 \times \frac{\sqrt{2}}{2}$? No, wait, actually for oblique pyramid, the height is $AC \times \sin(45^\circ)$? Wait no, $\sin(45^\circ)=\frac{h}{AB}$? No, angle at A is 45°, right angle at C, so $\tan(45^\circ)=\frac{h}{AC}$? No, $\tan(\theta)=\frac{opposite}{adjacent}$, angle BAC is 45°, so $\tan(45^\circ)=\frac{BC}{AC}$, and BC is the height h. Since $\tan(45^\circ)=1$, so $h = AC \times \tan(45^\circ)=3.6 \times 1=3.6$? No, wait no: angle BAC is 45°, right angle at C, so $\sin(45^\circ)=\frac{BC}{AB}$, $\cos(45^\circ)=\frac{AC}{AB}$, $\tan(45^\circ)=\frac{BC}{AC}$. Since $\tan(45^\circ)=1$, so $BC=AC=3.6$? No, that's wrong. Wait no, actually, the height of the oblique pyramid is the perpendicular distance from apex to base, which is $BC \times \sin(45^\circ)$? No, wait, angle BAC is 45°, AC is 3.6 cm, right angle at C, so triangle BAC is right-angled at C, so the height h of the pyramid is $AB \times \sin(45^\circ)$? No, no, let's correct: the height h of the pyramid is the length of the perpendicular from B to the base, which is $AC \times \sin(45^\circ)$? No, $\sin(45^\circ)=\frac{h}{AB}$, $\cos(45^\circ)=\frac{AC}{AB}$, so $AB=\frac{AC}{\cos(45^\circ)}$, then $h=AB \times \sin(45^\circ)=AC \times \tan(45^\circ)=3.6 \times 1=3.6$? Wait no, $\tan(45^\circ)=1$, so $h=3.6$ cm? No, wait, no, angle BAC is 45°, so the height h is $AC \times \sin(45^\circ)$? No, I'm mixing up. Let's do it properly:

In $\triangle BAC$, right-angled at C, $\angle BAC=45^\circ$, $AC=3.6$ cm.
$\sin(45^\circ)=\frac{BC}{AB}$, $\cos(45^\circ)=\frac{AC}{AB}$, $\tan(45^\circ)=\frac{BC}{AC}$
Since $\tan(45^\circ)=1$, $BC=AC \times 1=3.6$ cm. But BC is the slant edge, no, wait no: the height of the pyramid is the perpendicular distance from B to the base, which is $BC \times \sin(45^\circ)$? No, no, the base is the square, point A is on the base, AC is perpendicular to the base? Wait, the diagram shows AC is 3.6 cm, right angle at C, so AC is along the base, BC is perpendicular to AC, so BC is the height of the pyramid? No, angle BAC is 45°, so $\angle BAC=45^\circ$, so in triangle BAC, right-angled at C, $BC=AC \times \tan(45^\circ)=3.6 \times 1=3.6$ cm. That BC is the perpendicular height of the pyramid from B to the base.

Step2: Calculate base area

Base is square with edge 2 cm.
$A = s^2 = 2^2 = 4$ cm²

Step3: Compute pyramid volume

Volume formula: $V=\frac{1}{3} \times \text{base area} \times \text{height}$
First, find the correct height: wait, no, angle BAC is 45°, so the height h is $AC \times \sin(45^\circ)$? No, wait, $\sin(45^\circ)=\frac{h}{AB}$, $\cos(45^\circ)=\frac{AC}{AB}$, so $h=AC \times \tan(45^\circ)=3.6 \times 1=3.6$? No, that can't be, because if angle is 45°, then opposite and adjacent are equal, so BC=AC=3.6, which is the height.

Wait no, wait, the height of the oblique pyramid is the perpendicular distance from apex to base, which is $AB \times \sin(45^\circ)$? No, I'm making a mistake. Let's use $\sin(45^\circ)=\frac{h}{AC}$? No, angle at A is 45°, so the height h is $AC \times \sin(45^\circ)$? No, $\sin(\theta)=\frac{opposite}{hypotenuse}$, so if angle at A is 45°, opposite side is h (height), hypotenuse is AB, adjacent is AC. So $\sin(45^\circ)=\frac{h}{AB}$, $\cos(45^\circ)=\frac{AC}{AB}$, so dividing the two equations: $\tan(45^\circ)=\frac{h}{AC}$, so $h=AC \times \tan(45^\circ)=3.6 \times 1=3.6$ cm. Yes, that's correct, because $\tan(\theta)=\frac{opposite}{adjacent}$.

Now volume:
$V=\frac{1}{3} \times 4 \times 3.6$? No, that gives 4.8, which is an option. Wait, wait, no! Wait, no, the height is not 3.6. Wait, angle BAC is 45°, so the height h is $AC \times \sin(45^\circ)$? No, no, if angle BAC is 45°, then the height h is the length of the perpendicular from B to the base, which is $AB \times \sin(45^\circ)$, but $AB=\frac{AC}{\cos(45^\circ)}$, so $h=\frac{AC}{\cos(45^\circ)} \times \sin(45^\circ)=AC \times \tan(45^\circ)=3.6 \times 1=3.6$. But then volume is $\frac{1}{3} \times 2^2 \times 3.6 = \frac{1}{3} \times 4 \times 3.6 = 4.8$ cm³.

Wait, let's recheck:
Base area: $2 \times 2 = 4$ cm².
Height: from angle BAC=45°, right triangle, so height $h = AC \times \sin(45^\circ)$? No, $\sin(45^\circ)=\frac{h}{AB}$, $\cos(45^\circ)=\frac{AC}{AB}$, so $h=AC \times \tan(45^\circ)=3.6 \times 1=3.6$ cm. Correct, because $\tan(45^\circ)=1$.

Then volume: $\frac{1}{3} \times 4 \times 3.6 = 4.8$ cm³.

Answer:

4.8 cm³ (Option C)