Geometry

Geometry Solver with Step-by-Step Answers

Areas, volumes, angle chases, and coordinate proofs, with the theorem that licenses each step named on the line it is used.

Solve a geometry problem now

First solution free

Geometry marks are awarded for reasons, not numbers. A correct angle with no justification beside it scores a fraction of what the same angle scores when it is followed by alternate angles are equal. Every solution here is written in that form, so the working can be copied into the shape your teacher expects rather than translated first.

Photograph the diagram if there is one. If you are typing, list the given measurements and the relationships between the labelled points — a description such as ABCD is a cyclic quadrilateral, angle A is 70 degrees carries everything the solution needs.

What this solver handles

Perimeter, area, surface area, and volume for the standard plane figures and solids, including composite shapes.
Pythagoras and its converse, plus the special right triangles that appear constantly once you know their exact ratios.
Angle chasing with parallel lines, triangle and polygon angle sums, and exterior angles.
Similarity and congruence: proving triangles similar, then using the scale factor on lengths, areas, and volumes.
Circle theorems — inscribed angle, angles in a semicircle, tangent and radius, cyclic quadrilaterals, arc length, and sector area.
Coordinate geometry: distance, midpoint, gradient and line equations, and the conditions for parallel or perpendicular lines.
Problems where an unknown length is labelled and the geometry produces an equation to solve.
Transformations: translation, reflection, rotation, and enlargement.

When you type instead of photographing, describe the figure in the order someone would draw it, and be explicit about two things in particular: whether a circular measurement is a radius or a diameter, and whether a height is perpendicular or slant. Those two ambiguities cause more wrong geometry answers than any theorem does. Say too whether the question wants an exact surd or a rounded decimal, since both are correct and only one earns the mark.

Three problems, worked

Worked example

Points A (- 2, 5) and B (4, - 3)

1
The distance formula is Pythagoras applied to the horizontal and vertical gaps. The horizontal gap is 6, the vertical gap is −8, and squaring removes the sign.
$A B = (4 - (- 2))^{2} + (- 3 - 5)^{2} = 36 + 64$
2
One hundred is a perfect square, so the distance is exact.
$A B = 10$
3
The midpoint is the average of the coordinates, taken one axis at a time.
$M = (\frac{- 2 + 4}{2}, \frac{5 + ( - 3 )}{2}) = (1, 1)$
4
The gradient is rise over run, keeping the points in the same order top and bottom.
$m = \frac{- 3 - 5}{4 - ( - 2 )} = \frac{- 8}{6} = - \frac{4}{3}$
5
Use point-slope form with the midpoint, then rearrange. Substituting either original point returns the correct y-value, which is the check.
$y - 1 = - \frac{4}{3} (x - 1) ⟹ y = - \frac{4}{3} x + \frac{7}{3}$

Answer

A B = 10, M = (1, 1), y = - \frac{4}{3} x + \frac{7}{3}

Worked example

Triangle with sides 7, 8, 9

1
No angle is given and no side is obviously a height, so Heron's formula is the route. Start with the semi-perimeter.
$s = \frac{7 + 8 + 9}{2} = 12$
2
Substitute into Heron's formula. Each bracket is the semi-perimeter minus one side.
$A = 12 (12 - 7) (12 - 8) (12 - 9) = 12 \cdot 5 \cdot 4 \cdot 3$
3
The product is 720, and 720 is 144 times 5, so the root simplifies.
$A = 720 = 125 \approx 26.83$
4
The height on any side now follows from the standard area formula, because the area is already known.
$\frac{1}{2} (9) h = 125$
5
Solve for h. Taking the height on the longest side gives the shortest height, which is a useful sanity check.
$h = \frac{24 5}{9} = \frac{8 5}{3} \approx 5.96$

Answer

A = 125 \approx 26.83, h \approx 5.96 on the side of length 9

Worked example

A cylinder of radius 3 cm, height 10 cm, capped by a hemisphere

1
Split the solid into parts you have formulas for, and note which faces are internal. The circle where the hemisphere meets the cylinder is internal, so it belongs to no surface.
$V = V_{cylinder} + V_{hemisphere}$
2
The cylinder is a prism: base area times height.
$V_{cyl} = π r^{2} h = π (3)^{2} (10) = 90 π cm^{3}$
3
A hemisphere is half a sphere, so halve the four-thirds.
$V_{hem} = \frac{2}{3} π r^{3} = \frac{2}{3} π (27) = 18 π cm^{3}$
4
Add. Keeping pi symbolic until the end avoids compounding rounding error.
$V = 108 π \approx 339 cm^{3}$
5
For the surface area, count only the faces exposed to air: the flat base, the curved side of the cylinder, and the curved cap.
$S = π r^{2} + 2 π r h + 2 π r^{2} = 9 π + 60 π + 18 π$
6
Total the terms. Note the units: volume is in cubic centimetres, surface area in square centimetres.
$S = 87 π \approx 273 cm^{2}$

Answer

V = 108 π \approx 339 cm^{3}, S = 87 π \approx 273 cm^{2}

Estimate before computing, then compare the two. The triangle with sides 7, 8 and 9 came out at 26.83, and it had to fall below 28: the largest area two sides of 7 and 8 can enclose is half their product, reached when the angle between them is a right angle. Do not read more into the closeness than it supports. Sine is flat near 90 degrees, so an area within four per cent of that ceiling still leaves the angle between those two sides at $73. 4^{\circ}$ by the law of cosines — an ordinary acute triangle.

The composite solid supports a different check, on dimensions rather than size. Every term in the volume line is a length multiplied by a length multiplied by a length, and every term in the surface line is two lengths multiplied. If a term ever carries the wrong number of factors, a radius has gone missing or a height has been squared, and the mistake is visible before any arithmetic happens.

Work one through yourself

A ladder problem, because it hides a piece of geometry students consistently get wrong: the relationship between how far the top slides down and how far the foot slides out is not linear. Try each step before revealing it.

A 13 m ladder rests against a vertical wall, its foot 5 m from the base

0 of 4 steps

Try it yourself first — you'll remember it longer.

Where students go wrong

Scaling lengths, areas, and volumes by the same factor

If every length of a solid is doubled, its surface area quadruples and its volume goes up eightfold. Lengths scale by $k$ , areas by $k^{2}$ , volumes by $k^{3}$ . The exponent is the number of dimensions being multiplied together, which is also why a question that mixes $cm^{2}$ and $cm^{3}$ almost always has a scale factor hidden in it.

Trusting the diagram

Diagrams are usually not to scale, and an angle that looks like a right angle is only a right angle if it is marked or provable. Two sides that look equal are only equal if the tick marks say so. Assuming from appearance is the single fastest way to construct a proof that proves nothing.

Using the slant height as the perpendicular height

A cone has two heights: the perpendicular height $h$ from apex to base centre, and the slant height $ℓ$ along the surface. Volume needs $h$ , curved surface area needs $ℓ$ , and they are linked by $ℓ^{2} = h^{2} + r^{2}$ . Substituting the wrong one is the most common error in solid geometry.

Radius against diameter

A circle of diameter 10 has area $25 π$ , not $100 π$ . Questions give the diameter deliberately, because halving it is a step that can be forgotten. Write $r = 5$ on its own line before you touch a formula.

A cone volume with one wrong substitution

One of these lines is wrong. Click it.

None of those four is an arithmetic error. Each is the wrong quantity substituted into a correct formula, which is exactly why rechecking the numbers never finds them. Labelling every given value before you start — radius or diameter, slant or perpendicular, area or volume — costs about fifteen seconds and removes the whole category. The broader set of habits for catching your own errors is collected in the guide to verifying your own work.

Formulas worth knowing cold

Geometry reference

Tap any formula with a derivation to see where it comes from.

Plane figures

Solids

Coordinate geometry

Practice

Four to try

Answers are checked here — nothing is sent anywhere.

1
$Area of a triangle with base 12 cm and height 7 cm$
2
$Right triangle with legs 9 and 12. Hypotenuse = ?$
3
$Volume of a sphere of radius 3, in terms of π$
4
$Sum of the interior angles of a decagon$

Frequently asked questions

Is the geometry solver free?

Your first solution is free with no account needed. A free account then gives three solutions a day, and Gauth Plus removes the cap for $11.99 a month with a three-day free trial.

Does it show the steps or just the answer?

Each line names the theorem it leans on — angles on a straight line, alternate angles, the inscribed angle theorem — because a geometry answer with no reasons attached earns almost no method marks.

Can I photograph a diagram?

Yes, and for geometry a photograph is usually better than typing, since the labelled diagram carries information that is tedious to describe in words. Make sure every label and given measurement is legible in the shot.

Does it handle three-dimensional solids?

Volume and surface area for prisms, cylinders, pyramids, cones, spheres, and composite solids built from them, including the frustum left when a cone is cut.

Will it give an exact answer or a decimal?

Both. Exact forms such as twelve root five or 108 pi are given first, with a rounded decimal after, so you can match whichever your question asked for.

Can it do coordinate geometry proofs?

Yes. Showing a quadrilateral is a parallelogram from its vertices, or that a triangle is right-angled from its slopes, is worked as a chain of stated conditions rather than a single computation.

Keep going

The unit circle you actually needExact ratios for the special right triangles, derived not memorised.Turning words into equationsFor diagram questions that hand you a sentence instead of a length.Trigonometry solverOnce the diagram becomes a triangle with an angle in it.Slope calculator with stepsGradient, midpoint, and distance from two points in one pass.