Question

determining necessary information for calculating the volume of an oblique pyramid

what lengths would allow you to calculate the volume of the oblique pyramid with a square base? check all that apply.

□ ab and ef

□ ac, cb, and ef

□ ac and ef

□ ad, db, and ef

□ ad, dc, and ef

Explanation:

The volume formula for a pyramid (including oblique pyramids) is \( V=\frac{1}{3}Bh \), where \( B \) is the area of the base and \( h \) is the height (perpendicular height, not the slant height). For a square base, \( B = s^2 \), where \( s \) is the side length of the square base.

Step 1: Analyze each option

- **Option 1: AB and EF**
\( AB \) is the height (perpendicular height, since \( \angle ABG \) is a right angle), and \( EF \) is the side length of the square base. So we can find \( B = EF^2 \) and \( h = AB \), then use \( V=\frac{1}{3}Bh \). So this works.
- **Option 2: AC, CB, and EF**
Using the Pythagorean theorem, \( AB=\sqrt{AC^{2}-CB^{2}} \) (wait, no—actually, since \( \angle ABC \) is a right angle, \( AB \) can be related to \( AC \) and \( CB \) via Pythagoras? Wait, no, the diagram shows \( \angle ABG \) is right, so \( AB \) is vertical. Wait, maybe \( AC \) and \( CB \) can be used to find \( AB \) (if \( \triangle ABC \) is right-angled, but actually, the right angle is at \( B \) for \( AB \) and \( BG \)). Wait, maybe I misread. Wait, the base is square \( EFGH \) (or \( EFDH \)? Wait, the base is a square, so \( EF \) is a side. Then, \( AC \), \( CB \): if \( AB \) is the height, and \( AC \) and \( CB \) form a right triangle with \( AB \), then \( AB \) can be found (but actually, \( AB \) is the height directly? Wait, no—maybe \( AC \) and \( CB \) are legs to find \( AB \) (but \( AB \) is vertical). Wait, maybe this is a mistake. Wait, no—let's re-express. The height is \( AB \) (perpendicular to the base). The base area is \( EF^2 \). If we can find \( AB \) from \( AC \) and \( CB \) (if \( \triangle ACB \) is right-angled, then \( AB=\sqrt{AC^{2}-CB^{2}} \)? No, Pythagoras is \( AC^{2}=AB^{2}+CB^{2} \) if \( \angle ABC \) is right. So \( AB=\sqrt{AC^{2}-CB^{2}} \). Then with \( EF \) (side of base), we can compute volume. So this works.
- **Option 3: AC and EF**
We need the height \( AB \), but \( AC \) alone doesn't give \( AB \) (we need another leg, like \( CB \)). So \( AC \) and \( EF \) are not enough.
- **Option 4: AD, DB, and EF**
\( AD \) and \( DB \): since the base is square, \( DB \) is a side? Wait, no—\( D \) is a vertex of the base, \( B \) is a point. Wait, the base is square \( EFDH \) (assuming), so \( EF \) is a side. \( AD \) and \( DB \): if \( \triangle ADB \) is right-angled, then \( AB=\sqrt{AD^{2}-DB^{2}} \) (if \( \angle ABD \) is right). Then \( EF \) is the side of the base. So we can find \( AB \) (height) and \( EF \) (side of base, so \( B = EF^2 \)), then volume. So this works.
- **Option 5: AD, DC, and EF**
\( AD \) and \( DC \): \( DC \) is a side of the base? Wait, the base is square, so \( DC \) would be a side? No, \( D \), \( C \), \( B \): \( C \) is the midpoint? Wait, no—if \( D \), \( C \), \( B \) are colinear, and the base is square, then \( DC \) and \( DB \) would be related, but \( AD \) and \( DC \): if \( \angle ACD \) is right, then \( AB \) (height) would be equal to... Wait, no—\( AB \) is the perpendicular height. \( AD \) and \( DC \) don't give the perpendicular height. So this doesn't work.

Wait, let's correct:

The volume of a pyramid is \( V = \frac{1}{3}Bh \), where \( B \) is the area of the base (square, so \( B = s^2 \), \( s \) is side length of base, e.g., \( EF \)), and \( h \) is the perpendicular height (the distance from the apex to the base, which is \( AB \) in the diagram, since \( AB \perp \) base? Wait, the diagram has a right angle at \( B \) between \( AB \) and \( BG \), so \( AB \) is the perpendicular height.

- **Option 1: AB (height) and EF (side of base)**
\( B = EF^2 \), \( h = AB \), so \( V = \frac{1}{3}(EF^2)(AB) \). Valid.

- **Option 2: AC, CB, EF**
Since \( \triangle ABC \) is right-angled (right angle at \( B \)), \( AB = \sqrt{AC^2 - CB^2} \) (by Pythagoras: \( AC^2 = AB^2 + CB^2 \)). Then \( B = EF^2 \), \( h = AB \), so volume can be calculated. Valid.

- **Option 3: AC and EF**
We need \( AB \) (height), but \( AC \) alone doesn't give \( AB \) (we need \( CB \) to use Pythagoras). So insufficient.

- **Option 4: AD, DB, EF**
\( \triangle ADB \) is right-angled (right angle at \( B \)), so \( AB = \sqrt{AD^2 - DB^2} \) (Pythagoras: \( AD^2 = AB^2 + DB^2 \)). Then \( B = EF^2 \), \( h = AB \), so volume can be calculated. Valid.

- **Option 5: AD, DC, EF**
\( AD \) and \( DC \): \( DC \) is not related to the height \( AB \) (since \( DC \) is in the base plane or not perpendicular). So we can't find \( AB \) from \( AD \) and \( DC \). Invalid.

So the valid options are AB and EF; AC, CB, and EF; AD, DB, and EF.

Answer:

• AB and EF
• AC, CB, and EF
• AD, DB, and EF