A solid question becomes a triangle question the moment you find the right triangle. Drawing it separately, flat on the page, is most of the work.
Find the height
A square pyramid has base edge 12 cm and lateral edge 10 cm. How tall is it?
A lateral edge runs to a corner, so the triangle it sits in uses half the diagonal, not half the edge. The base diagonal is 12√2, half of it is 6√2, and h = √10² − (6√2)² = √100 − 72 = √28 ≈ 5.29 cm. A uses half the base edge, 6, which is the triangle belonging to the slant height, not the lateral edge — the central mistake of this lesson. D adds instead of subtracting.
Ask one question before choosing: does the given length end at the middle of an edge or at a corner? That answer picks the triangle, and the triangle picks the arithmetic.
That angle lives in the slant-height triangle: it is the angle at the base, between the slant height and the half-edge lying on the ground.
opposite = h = 15, adjacent = s2 = 8
name the sides first
tan θ = 158 = 1.875
opposite over adjacent
θ = tan⁻¹(1.875) ≈ 61.9°
steeper than 45°, as h > s2 ✓
The check on the last line is worth keeping. If the height exceeds half the base edge the face must lean past 45°; if it does not, the angle must be under. A calculator in the wrong mode announces itself immediately.
cos θ = r/ℓ = 513
adjacent over hypotenuse
θ = cos⁻¹(513) ≈ 67.4°
the wall of the cone
h = √169 − 25 = 12 cm
the 5-12-13 triangle
A cone’s cross-section through the axis is an isosceles triangle of base 2r and height h. Cutting it down the middle gives the right triangle with legs r and h and hypotenuse ℓ — every cone angle question is that one triangle.
base diagonal: √6² + 8² = 10 cm
flat on the floor first
space diagonal: √10² + 24²
now stand it up
= √676 = 26 cm
or √6²+8²+24² in one go
The shortcut d = √ℓ² + w² + h² is just those two steps written together. Do it in two while the picture is still forming; the one-line version is for when you can already see the triangle.
For this box the face diagonals are 10, √6²+24² ≈ 24.7 and √8²+24² ≈ 25.3 — all shorter than 26. The space diagonal is always the longest line that fits inside, so an answer smaller than every edge-to-edge measurement you can see is wrong.
A lateral edge runs from the apex to a corner, so its triangle has legs h and d2, half the base diagonal. For a square of side s, d2 = s√22.
A square pyramid has base edge 8 cm and height 7 cm. Find a lateral edge and the angle it makes with the base.
d2 = 8√22 = 4√2 cm
centre of the base to a corner
e = √7² + (4√2)² = √49 + 32 = 9 cm
(4√2)² = 16 × 2 = 32
sin θ = 79, so θ ≈ 51.1°
opposite over hypotenuse
Using 4 (half the edge) in place of 4√2 gives e = √49 + 16 = √65 ≈ 8.06 cm — that is the slant height, the line to the middle of an edge. A lateral edge reaches further, to the corner, so it must come out longer than the slant.
Backwards: a square pyramid has base edge 10 cm and lateral edge 13 cm. How tall is it?
d2 = 10√22 = 5√2, so (d2)² = 25 × 2 = 50
half the base diagonal, squared
h² = 13² − 50 = 169 − 50 = 119
the edge is the hypotenuse, so subtract
h = √119 ≈ 10.9 cm
shorter than the edge ✓
A box is 2 cm by 3 cm by 6 cm tall. What angle does its space diagonal make with the base?
space diagonal = √4 + 9 + 36 = 7 cm
the hypotenuse
sin θ = 67, so θ ≈ 59.0°
opposite is the height, 6
The angle sits on the floor, between the space diagonal and the base diagonal √13 below it. Over 45°, because the height 6 is longer than √13 ≈ 3.6 ✓.
The same flat triangle each time; what changes is which piece is unknown.
A cone’s slant is 10 cm and meets the base at 60°. Find r and h.
r = 10 cos 60° = 5 cm
adjacent = hypotenuse × cos
h = 10 sin 60° = 5√3 ≈ 8.66 cm
opposite = hypotenuse × sin
With the angle known, multiply the hypotenuse; no inverse key needed.
A cube has edge 4 cm. Find its space diagonal.
d = √4² + 4² + 4² = √48
all three edges equal
d = 4√3 ≈ 6.93 cm
√48 = √16 × √3
Any cube of edge s has space diagonal s√3; leave the surd unless a decimal is asked.
A pyramid on a 10 cm by 6 cm rectangle is 6 cm tall. Find the angle each kind of face makes with the base.
face on a 10 cm edge: tan θ = 63 = 2
half of 6 is 3
θ ≈ 63.4°
the steeper pair
face on a 6 cm edge: tan θ = 65 = 1.2
half of 10 is 5
θ ≈ 50.2°
the gentler pair
The face standing on the long edge reaches across the short half-width, so it is the steeper one.
These come as one length or angle hidden inside a pyramid, cone or box: a slant height, a lateral edge, a space diagonal, or the angle a face or edge makes with the base. The fastest route is to decide where the line ends (middle of an edge, corner, or far corner), draw that one right triangle flat and use Pythagoras or one trig ratio. The 10-second check: the hypotenuse is the longest side, and the angle sits above 45° exactly when the height is the longer leg.
Find the angle
A cone has radius 9 cm and height 12 cm. What angle does its slant make with the base, to the nearest degree?
tan θ = hr = 129, so θ = tan⁻¹(1.333) ≈ 53°. B is the other angle in the same triangle, the one at the apex — 37° and 53° add to 90°, so picking the wrong one is the whole trap. Since the height 12 is bigger than the radius 9, the wall must lean past 45°, which rules out B, C and D at a glance.
Hint: draw the triangle with the radius along the bottom and the height standing up. The angle asked for sits where the slant meets the ground, so the height is opposite it and the radius is adjacent.
