KCSE 2025 Maths P1 Q18 — Volume & Surface Area of a Pyramid on a Block

KCSE 2025 Form 4 Measurement

Published 9 July 2026

The Question

“A solid consists of a right pyramid mounted on a rectangular block. The pyramid and the block share an identical square base of side 6 cm, the block is 10 cm tall, and each slant edge of the pyramid from the apex to a base corner is 5 cm. (a) Calculate the volume of the whole solid. (b) The solid is cut into two identical halves along the vertical plane of symmetry that passes through the apex and a diagonal of the base. Calculate the surface area of one of the pieces.”

Find the pyramid's vertical height

The apex sits directly above the centre of the square base, so the perpendicular height, half of a base diagonal, and a slant edge form a right-angled triangle. Half the diagonal of a 6 by 6 square is three root two. Use Pythagoras with the slant edge of 5 as the hypotenuse to get the vertical height.

\text{half diagonal} = \frac{1}{2}\sqrt{6^{2}+6^{2}} = 3\sqrt{2}

H = \sqrt{5^{2} - \left(3\sqrt{2}\right)^{2}} = \sqrt{25-18} = \sqrt{7}

Add the block volume and the pyramid volume

Work out each solid separately. The block is a cuboid, so multiply its three dimensions. The pyramid uses one third of the base area times the vertical height. Adding them gives the total volume of the solid.

V_{\text{block}} = 6 \times 6 \times 10 = 360

V_{\text{pyramid}} = \frac{1}{3} \times 36 \times \sqrt{7} = 12\sqrt{7} \approx 31.75

V_{\text{total}} = 360 + 12\sqrt{7} \approx 391.75 \ \text{cm}^{3}

Find the slant height of a triangular face

For the surface area you need the slant height of each triangular face, measured from the apex to the midpoint of a base edge. This is different from the slant edge. It forms a right triangle with the vertical height root seven and half of a base side, which is 3.

l = \sqrt{3^{2} + \left(\sqrt{7}\right)^{2}} = \sqrt{9+7} = \sqrt{16} = 4

Find the total outside surface area, then halve it

Add up the outer surfaces of the whole solid before cutting: the square bottom, the four rectangular sides of the block, and the four triangular faces of the pyramid. The top of the block is fully covered by the pyramid, so it is not counted. Cutting along the plane of symmetry divides this outer surface equally between the two pieces.

A_{\text{bottom}} = 6 \times 6 = 36

A_{\text{block sides}} = 4 \times (6 \times 10) = 240

A_{\text{triangles}} = 4 \times \left(\tfrac{1}{2} \times 6 \times 4\right) = 48

A_{\text{outside}} = 36 + 240 + 48 = 324 \Rightarrow \tfrac{1}{2} \times 324 = 162

Add the freshly exposed cut face

The cut creates a brand new flat face that only one piece keeps its own copy of, so it must be added on top of the 162. This face is a pentagon: a rectangle through the block plus a triangle through the pyramid. The rectangle is the base diagonal, six root two, wide and the block height, 10, tall. The triangle has that same diagonal as its base and the pyramid's vertical height root seven as its height.

A_{\text{cut rectangle}} = 6\sqrt{2} \times 10 = 60\sqrt{2}

A_{\text{cut triangle}} = \tfrac{1}{2} \times 6\sqrt{2} \times \sqrt{7} = 3\sqrt{14}

A_{\text{piece}} = 162 + 60\sqrt{2} + 3\sqrt{14} \approx 258.1 \ \text{cm}^{2}

The diagonal is used because the cut runs across the square base from corner to corner, not along a side.

Final Result

The volume of the whole solid is 360 + 12 root 7, about 391.75 cubic centimetres. The surface area of one cut piece is 162 + 60 root 2 + 3 root 14, about 258.1 square centimetres.

Why this method works

The method works because it separates each surface into shapes whose areas we already know how to find, and treats the cut carefully. The slant edge (5 cm) and the slant height (4 cm) are genuinely different lengths: the slant edge runs to a corner while the slant height runs to the midpoint of an edge, so each needs its own right-angled triangle built from the pyramid's vertical height. When the solid is sliced through its plane of symmetry, the original outer skin splits into two equal shares, but slicing also reveals an interior face that was not part of the outside before. That new face belongs to each piece separately, so it is added once, not halved.

✓

Check the piece area: 60 root 2 is about 84.85 and 3 root 14 is about 11.22, so 162 + 84.85 + 11.22 is about 258.1 square centimetres, matching the stated answer.