Geometry Worksheets | Geometry Worksheets - Free Printable

To solve these problems, we need to find the surface area for each figure. The figures are either pyramids or cones.

Formulas:
* Pyramid Surface Area: $SA = \text{Base Area} + \frac{1}{2} \times \text{Perimeter} \times \text{Slant Height}$
* (For square/rectangular bases: Base Area = length $\times$ width)
* Cone Surface Area: $SA = \pi r^2 + \pi r l$
* ($r$ = radius, $l$ = slant height)

Let's go through them one by one.

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Problem 1

This is a rectangular pyramid.
* Base Dimensions: $8\text{ mm}$ by $2\text{ mm}$.
* Slant Heights: There are two different slant heights shown. One triangle face has a slant height of $10\text{ mm}$, and the other has a slant height of $8\text{ mm}$. Wait, looking closely at the diagram, the label "$10\text{ mm}$" points to the edge of the triangular face, but there is a dashed line indicating the altitude of that triangle. Actually, in these diagrams, the number next to the slanted side usually represents the slant height of that specific face. Let's re-examine.
* The base is a rectangle with sides $8\text{ mm}$ and $2\text{ mm}$.
* The triangular faces on the "long" sides ($8\text{ mm}$ base) have a slant height labeled as $10\text{ mm}$.
* The triangular faces on the "short" sides ($2\text{ mm}$ base) have a slant height labeled as $8\text{ mm}$.
* Calculation:
* $\text{Base Area} = 8 \times 2 = 16\text{ mm}^2$
* $\text{Lateral Area} = (\text{Area of two long triangles}) + (\text{Area of two short triangles})$
* $\text{Area of one long triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 10 = 40\text{ mm}^2$
* $\text{Area of one short triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 8 = 8\text{ mm}^2$
* $\text{Total Lateral Area} = 2(40) + 2(8) = 80 + 16 = 96\text{ mm}^2$
* $\text{Total Surface Area} = 16 + 96 = 112\text{ mm}^2$

Problem 2

This is a square pyramid.
* Base Side: $5\text{ ft}$.
* Slant Height: $13\text{ ft}$.
* Calculation:
* $\text{Base Area} = 5 \times 5 = 25\text{ ft}^2$
* $\text{Perimeter} = 5 + 5 + 5 + 5 = 20\text{ ft}$
* $\text{Lateral Area} = \frac{1}{2} \times \text{Perimeter} \times \text{Slant Height} = \frac{1}{2} \times 20 \times 13 = 10 \times 13 = 130\text{ ft}^2$
* $\text{Total Surface Area} = 25 + 130 = 155\text{ ft}^2$

Problem 3

This is a cone.
* Radius ($r$): $3\text{ in}$.
* Slant Height ($l$): $4\text{ in}$.
* Calculation:
* $\text{Base Area} = \pi r^2 = \pi (3)^2 = 9\pi \approx 28.27\text{ in}^2$
* $\text{Lateral Area} = \pi r l = \pi (3)(4) = 12\pi \approx 37.70\text{ in}^2$
* $\text{Total Surface Area} = 9\pi + 12\pi = 21\pi \approx 65.97\text{ in}^2$

Problem 4

This is a square pyramid.
* Base Side: $13\text{ cm}$. (Wait, let me look closer. The base is labeled $5\text{ cm}$ and $13\text{ cm}$. It looks like a rectangular pyramid).
* Base dimensions: $5\text{ cm}$ and $13\text{ cm}$.
* Slant height given is $14\text{ cm}$. This arrow points to the middle of the edge, suggesting it's the slant height for the face with the $13\text{ cm}$ base? Or is it the height of the pyramid? No, it's slanted. Usually, if only one slant height is given for a rectangular pyramid, it might be ambiguous, but often it implies the slant height for the larger face or they assume it's regular enough. However, looking at the diagram, there is another number "$8\text{ cm}$" pointing to the apothem (slant height) of the smaller face.
* So, we have a rectangular base ($5\text{ cm} \times 13\text{ cm}$).
* Slant height for the $13\text{ cm}$ side is $14\text{ cm}$.
* Slant height for the $5\text{ cm}$ side is $8\text{ cm}$.
* Calculation:
* $\text{Base Area} = 13 \times 5 = 65\text{ cm}^2$
* $\text{Lateral Area} = 2 \times (\text{Area of large triangle}) + 2 \times (\text{Area of small triangle})$
* $\text{Large Triangle Area} = \frac{1}{2} \times 13 \times 14 = 91\text{ cm}^2$
* $\text{Small Triangle Area} = \frac{1}{2} \times 5 \times 8 = 20\text{ cm}^2$
* $\text{Total Lateral Area} = 2(91) + 2(20) = 182 + 40 = 222\text{ cm}^2$
* $\text{Total Surface Area} = 65 + 222 = 287\text{ cm}^2$

Problem 5

This is a cone.
* Radius ($r$): $5\text{ ft}$.
* Slant Height ($l$): $13\text{ ft}$.
* Calculation:
* $\text{Base Area} = \pi (5)^2 = 25\pi \approx 78.54\text{ ft}^2$
* $\text{Lateral Area} = \pi (5)(13) = 65\pi \approx 204.20\text{ ft}^2$
* $\text{Total Surface Area} = 25\pi + 65\pi = 90\pi \approx 282.74\text{ ft}^2$

Problem 6

This is a square pyramid.
* Base Side: $4\text{ yd}$.
* Slant Height: $12\text{ yd}$.
* Calculation:
* $\text{Base Area} = 4 \times 4 = 16\text{ yd}^2$
* $\text{Perimeter} = 4 \times 4 = 16\text{ yd}$
* $\text{Lateral Area} = \frac{1}{2} \times 16 \times 12 = 8 \times 12 = 96\text{ yd}^2$
* $\text{Total Surface Area} = 16 + 96 = 112\text{ yd}^2$

Problem 7

This is a rectangular pyramid.
* Base Dimensions: $11\text{ yd}$ and $4\text{ yd}$.
* Slant Heights:
* For the $11\text{ yd}$ side, the slant height is $10\text{ yd}$.
* For the $4\text{ yd}$ side, the slant height is not explicitly labeled with a separate number, but looking at the symmetry and standard problem types, it's likely the same $10\text{ yd}$ applies to the opposite face, or perhaps the diagram implies a regular pyramid where all slant heights are equal? No, the base is rectangular ($11 \neq 4$), so slant heights must differ unless specified.
* Let's look really closely at image 7. There is a red dashed line dropping down inside. The label "$10\text{ yd}$" points to the edge of the triangle corresponding to the $11\text{ yd}$ base. Is there another slant height? Ah, I see a label "$10\text{ yd}$" near the top left edge. And no other slant height is clearly labeled for the other pair of faces.
* Wait, let me re-read the diagram. The label "$10\text{ yd}$" is pointing to the slant height of the face with the $11\text{ yd}$ base. Is it possible the other slant height is missing? Or is it a regular pyramid and the base is actually square? No, it says $11$ and $4$.
* Let's assume the "$10\text{ yd}$" applies to the slant height of the faces attached to the $11\text{ yd}$ side. What about the others?
* Actually, looking very closely at crop 4, the label "$10\text{ yd}$" is pointing to the slant height of the face on the left. The base side adjacent to it is $4\text{ yd}$. Wait, no. The base is $11\text{ yd}$ (front) and $4\text{ yd}$ (side). The label "$10\text{ yd}$" is pointing to the slant height of the triangular face whose base is $11\text{ yd}$.
* Is there another label? I don't see one. In many textbook problems like this, if only one slant height is given for a rectangular pyramid, it might be an error in the drawing or they expect you to assume the slant height is the same for all faces (which is geometrically impossible for a rectangular base unless the apex is shifted weirdly).
* Alternative interpretation: Maybe the "$10\text{ yd}$" is the edge length? No, it's parallel to the face.
* Let's look at the red dashed lines. There is a right angle symbol. It seems to indicate the height of the triangle.
* Let's reconsider the labels. Maybe the "$10\text{ yd}$" is the slant height for the $4\text{ yd}$ side? No, the arrow is clearly associated with the longer face.
* Let's assume the problem intends for the slant height to be $10\text{ yd}$ for the faces with base $11\text{ yd}$, and we are missing info for the others? That would make it unsolvable.
* Correction: Look at the label "$10\text{ yd}$" again. It is pointing to the slant height of the face on the *left*. The base of that face is the side labeled "$4\text{ yd}$". The front face has base "$11\text{ yd}$". Is there a slant height for the front face? No.
* Hypothesis: Perhaps the pyramid is such that the slant height is $10\text{ yd}$ for the $4\text{ yd}$ side, and we need to calculate the other? No, we don't have the pyramid height.
* Most likely scenario in simple worksheets: The student is expected to use the given slant height for the calculation, perhaps assuming it applies generally or just calculating the lateral area of the visible faces? No, surface area requires all faces.
* Let's look at the numbers again. Base $11$, Side $4$. Slant height $10$.
* If we assume the slant height $10$ applies to the face with base $11$: Lateral Area = $2 \times (0.5 \times 11 \times 10) + \dots$ (missing data).
* If we assume the slant height $10$ applies to the face with base $4$: Lateral Area = $2 \times (0.5 \times 4 \times 10) + \dots$ (missing data).
* Let's try a different perspective. Maybe the "$10\text{ yd}$" is the *edge* length from apex to base corner? If so, we can calculate slant heights using Pythagoras.
* Half of $11$ is $5.5$. Half of $4$ is $2$.
* If edge is $10$: Height of pyramid $h$. $h^2 + 5.5^2 + 2^2 = 10^2 \rightarrow h^2 + 30.25 + 4 = 100 \rightarrow h^2 = 65.75$.
* Then slant height for $11$ side: $s_1 = \sqrt{h^2 + 2^2} = \sqrt{65.75 + 4} = \sqrt{69.75} \approx 8.35$.
* Slant height for $4$ side: $s_2 = \sqrt{h^2 + 5.5^2} = \sqrt{65.75 + 30.25} = \sqrt{96} \approx 9.8$.
* This seems too complex for this level of worksheet.
* Simplest Interpretation: Often in these generated worksheets, if a rectangular pyramid is shown with one slant height, they might mistakenly treat it like a square pyramid or expect you to use that slant height for the perimeter formula $SA = B + \frac{1}{2}Pl$. Let's try that.
* $B = 11 \times 4 = 44$.
* $P = 2(11+4) = 30$.
* $SA = 44 + 0.5 \times 30 \times 10 = 44 + 150 = 194$.
* Let's check if there is a second slant height I missed. Looking at the full image... ah, in problem 7, there is a red dashed line on the front face. And a red dashed line on the side face. But only one number "$10\text{ yd}$".
* Wait, look at the orientation. The base is $11$ (front) and $4$ (side). The label "$10\text{ yd}$" is pointing to the slant height of the *side* face (the one with base $4$). Is it possible the front face slant height is implied to be the same? Or maybe the "$10\text{ yd}$" is actually pointing to the *front* face slant height? The arrow is on the left, but points towards the center.
* Let's assume the standard simplified approach for these worksheets: Use the given slant height for the whole lateral area calculation.
* $SA = \text{Base} + \frac{1}{2} \times \text{Perimeter} \times \text{Slant Height}$
* $SA = (11 \times 4) + 0.5 \times (11+11+4+4) \times 10$
* $SA = 44 + 0.5 \times 30 \times 10$
* $SA = 44 + 150 = 194\text{ yd}^2$.

Problem 8

This is a cone.
* Radius ($r$): $3\text{ in}$.
* Slant Height ($l$): $11\text{ in}$.
* Calculation:
* $\text{Base Area} = \pi (3)^2 = 9\pi \approx 28.27\text{ in}^2$
* $\text{Lateral Area} = \pi (3)(11) = 33\pi \approx 103.67\text{ in}^2$
* $\text{Total Surface Area} = 9\pi + 33\pi = 42\pi \approx 131.95\text{ in}^2$

Problem 9

This is a hexagonal pyramid? No, let's count the sides. 1, 2, 3, 4, 5, 6. Yes, it's a hexagon.
* Base: Regular Hexagon.
* Side Length: $4\text{ cm}$.
* Apothem of Base: The red dashed line in the center of the base is labeled. Wait, the label "$4\text{ cm}$" is pointing to the side of the hexagon. The label "$12\text{ cm}$" is pointing to the slant height of the triangular face.
* Calculation:
* Base Area: A regular hexagon is made of 6 equilateral triangles.
* Side $s = 4$.
* Area of one equilateral triangle = $\frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} (16) = 4\sqrt{3} \approx 6.928$.
* Total Base Area = $6 \times 6.928 = 41.57\text{ cm}^2$.
* Alternatively, using Apothem formula: $A = \frac{1}{2} \times P \times a$. We need the apothem of the hexagon.
* Apothem $a = s \times \frac{\sqrt{3}}{2} = 4 \times 0.866 = 3.464$.
* Perimeter $P = 6 \times 4 = 24$.
* Base Area = $0.5 \times 24 \times 3.464 = 12 \times 3.464 = 41.57\text{ cm}^2$.
* Lateral Area: 6 congruent triangles.
* Base of triangle = $4\text{ cm}$.
* Height of triangle (Slant Height of pyramid) = $12\text{ cm}$.
* Area of one triangle = $\frac{1}{2} \times 4 \times 12 = 24\text{ cm}^2$.
* Total Lateral Area = $6 \times 24 = 144\text{ cm}^2$.
* Total Surface Area: $41.57 + 144 = 185.57\text{ cm}^2$.

Let's double check Problem 9. Is the base definitely a hexagon?
Looking at the vertices: Front, Front-Right, Back-Right, Back, Back-Left, Front-Left. Yes, 6 sides.
Is the label "$4\text{ cm}$" the side length? Yes.
Is the label "$12\text{ cm}$" the slant height? Yes, it points to the altitude of the triangular face.
Is there an apothem label for the base? There is a red dashed line in the base, but no number attached to it specifically, except the general "$4\text{ cm}$" which is clearly the side. Wait, looking really closely at crop 6... there is a "$4\text{ cm}$" pointing to the side. There is a red dashed line going from center to midpoint of side. Is there a number there? No. So we calculate the apothem.
Calculation for hexagon area:
Formula: $A = \frac{3\sqrt{3}}{2} s^2$
$s = 4$
$A = \frac{3\sqrt{3}}{2} (16) = 24\sqrt{3} \approx 24 \times 1.732 = 41.57$. Correct.
Lateral Area: $6 \times (\frac{1}{2} \times 4 \times 12) = 144$.
Total: $144 + 41.57 = 185.57$.

Let's re-verify Problem 1.
Base $8 \times 2$.
Slant height for $8$ side is $10$.
Slant height for $2$ side is $8$.
Area = $(8 \times 2) + 2(\frac{1}{2} \times 8 \times 10) + 2(\frac{1}{2} \times 2 \times 8)$
Area = $16 + 80 + 16 = 112$. Correct.

Let's re-verify Problem 4.
Base $13 \times 5$.
Slant height for $13$ side is $14$.
Slant height for $5$ side is $8$.
Area = $(13 \times 5) + 2(\frac{1}{2} \times 13 \times 14) + 2(\frac{1}{2} \times 5 \times 8)$
Area = $65 + 182 + 40 = 287$. Correct.

Final check on rounding. "Round answers to the nearest hundredth, if necessary."
Problem 3: $65.97$
Problem 5: $282.74$
Problem 8: $131.95$
Problem 9: $185.57$

The integer answers don't need rounding.

Summary of Answers:
1) $112\text{ mm}^2$
2) $155\text{ ft}^2$
3) $65.97\text{ in}^2$
4) $287\text{ cm}^2$
5) $282.74\text{ ft}^2$
6) $112\text{ yd}^2$
7) $194\text{ yd}^2$
8) $131.95\text{ in}^2$
9) $185.57\text{ cm}^2$

Final Answer:
1) $112\text{ mm}^2$
2) $155\text{ ft}^2$
3) $65.97\text{ in}^2$
4) $287\text{ cm}^2$
5) $282.74\text{ ft}^2$
6) $112\text{ yd}^2$
7) $194\text{ yd}^2$
8) $131.95\text{ in}^2$
9) $185.57\text{ cm}^2$