Geometry
Drag & drop or click to add images
Real geometry questions asked by students, solved step by step Every question below was submitted by a real student and answered step by step.
Angle ADB = 42°. Doubling the bisected half gives angle A = 56°, then the angle sum gives B = 110°, and triangle ABD closes at 180° − 28° − 110°.
The altitude is 2*sqrt(3) cm and the angle AGM is 60 degrees. SAS congruence of triangles ABN and BCM makes AN = BM and fixes the angle at their crossing point.
BD = 20 cm and CD = 30 cm from the bisector ratio 30:45, and both parallels give DE = DF = 18 cm - the equality is what makes ADEF a rhombus.
Draw the line through P parallel to both. Co-interior angles give 70 and 40 degrees on the two sides of it, so the zig-zag angle at P is 110 degrees.
Angle A = 83 degrees. When AB = BC and AD = CD the quadrilateral is a kite, so the other two angles are equal: A = C = (360 - 58 - 136) / 2 = 83 degrees.
Answer: 36, 72, 108 and 144 degrees. Set the common value to x, so the angles are x, 2x, 3x, 4x; their sum is 360 degrees, giving 10x = 360 and x = 36.
Write all four angles in terms of M, use the 360 degree angle sum to get 4x + 60 = 360, so x = 75 and the angles step up by 10 degrees each.
With angle B = 36 and angle C = 76, angle A = 68, so angle BAD = 34 and angle BAF = 54. Then angle DAF = 20, angle AFD = 90 and angle ADF = 70.
The statement is false. Equal angles force the same shape but not the same size, so AAA gives similarity only; a 3-4-5 and a 6-8-10 triangle are the standard counterexample.
Perimeter is 8 + 6r and area is √3·r(r + 4). The overhang on each side is r, so the leg 2r, the offset r and the height form a 30-60-90 triangle with height r√3.
The area is 156 m². Multiply the two side lengths, and note that the unit becomes squared metres because two lengths measured in metres are multiplied together.
The area is 168 square units. Multiply 14 by 12; the perimeter, by contrast, is 2(14 + 12) = 52 units — a different formula for a different quantity.
The area is 12 m². The rectangle covers a 4-by-3 grid of one-metre squares, which is exactly what multiplying the two side lengths counts up.
Answer: 3500 cm^2, which is 0.35 m^2. Area is base times height, and the units multiply too, so centimetres times centimetres gives square centimetres.
With A = pi r^2 and pi = 22/7, cancel 35^2 = 1225 against the 7 to get 22 x 175 = 3850 square inches. The exact value is 1225 pi = 3848.45 in^2.
The area is 962.5 square cm. Solving 2 pi r = 110 with pi = 22/7 gives exactly r = 17.5, and squaring that radius keeps the arithmetic in whole fractions.
Answer: 282 square metres. When the two straight sides are radii of 15 m and the curved edge is 37.6 m, the sector area is A = rL/2 = 15 x 37.6 / 2.
The larger hexagon has area 225 square metres. Areas of similar figures scale by the square of the side ratio, so multiply 81 by (5/3)^2 = 25/9, not by 5/3.
The rectangle has area 54 cm^2. Divide the 12 cm perimeter by 4 to get a 3 cm side, square it for 9 cm^2 per square, then multiply by the six squares.
The area is 238.524 square units. Splitting 9.174 into 9 + 0.174 gives 234 + 4.524, which is faster and easier to check than multiplying the decimals directly.
Answer: 40 square units. The area of a rectangle is length times width, so A = 8 x 5 = 40, and the same product also counts the unit squares that tile it.
The ring has area 423.5 square inches. Factoring R^2 - r^2 as (R - r)(R + r) turns 441 - 306.25 into the much easier product 3.5 times 38.5.
The area is 128 cm². Each triangle has area ½ × 8 × 8 = 32 cm², and because the copies do not overlap the arrangement never affects the total.
The altitude to the base bisects it, giving a 5-12-13 right triangle, so the height is 5 cm and the area is (1/2)(24)(5) = 60 cm^2.
Four side lengths alone do not determine a quadrilateral's area — here is why, and what extra measurement (a diagonal, an angle) you need to finish the calculation.
The smaller polygon has area 180 square metres. Scaling down from 320 uses the square of the ratio, (3/4)^2 = 9/16, not the ratio 3/4 itself.
The area is 42. Slide one diagonal to build a triangle with sides 15, 7 and 20 and use Heron's formula, or find the height 4.2 directly and multiply by 10.
The area is y = 24 - 3x for 0 < x < 8. Take PB = 8 - x as the base; because angle C is right, leg AC = 6 is the height for every position of P.
The area ratio is 4:9. Similar polygons scale areas by the square of the similarity ratio, so a side ratio of 2:3 becomes 2 squared to 3 squared, or 4:9.
The shared area is exactly 2 pi, about 6.283. The centre lies on the hypotenuse and the circle is tangent to both legs, so the overlap is precisely half the disc.
Because Bm is parallel to Ay, angle xBm equals angle xAy, so their bisectors make equal corresponding angles with Ax and are parallel; and the bisectors of the supplementary pair xBm and ABm meet at 90 degrees.
Match against (x-h)^2 + (y-k)^2 = r^2. Because y + 3 is y - (-3), the y-coordinate of the centre is -3, and r is the square root of 9, not 9 itself.
Using C = 2 pi r with pi = 22/7, the 7 in the denominator cancels the 28 exactly: 2 x 22 x 4 = 176 cm. With the true pi the answer is 175.93 cm.
AB is root 29 (about 5.39) and CD is root 5 (about 2.24), so AB is longer. Squaring the coordinate differences makes the order of subtraction irrelevant.
The Cartesian form is (√3, 1, 5) ≈ (1.732, 1, 5). Apply x = r cos θ and y = r sin θ with the exact values for 30°, while the height z passes through unchanged.
Answer: AB = 17. A is two-thirds of the way from (1,10) to (-8,4), giving A(-5,6); B is the midpoint of (0,-7) and (6,-11), giving B(3,-9); then AB = sqrt(64+225).
The distance is √11562 ≈ 107.53. The coordinate differences are 83, −7 and −68, and 11562 = 2·3·41·47 has no square factor, so the radical cannot be simplified.
The distance is sqrt(22), about 4.6904. The 3D formula adds a third squared difference to the Pythagorean theorem, and squaring removes every sign.
The distance to the line is 11/√10 ≈ 3.4785, reached at a = 0.7. The distance to the point with parameter a is √(10a² − 14a + 17), a parabola minimised at its vertex.
The distance is √(227/14) ≈ 4.0267. The cross product of the point vector with the direction (3, 1, 2) gives (9, −5, −11), and dividing its length by √14 is the distance.
The pair is (λ, μ) = (1/2, 1/4). Place every point as a position vector, use D = (B+C)/2 and M = (A+D)/2, then subtract B and match coefficients against BA and BC.
Angle ACB = 44 degrees. Let angle C = x, write angle A as 120 - x from the triangle sum, halve it because AL bisects A, then solve the small triangle ALC.
The answer is AC = 9 cm. The bisector ratio AD/DC = AB/BC gives BC = 14, and then BO bisecting angle B inside triangle ABE gives OA/OE = AB/BE.
QR = 2 1/2 metres. Because PQ is the longer segment, R lies between P and Q, so QR = PQ − PR = 35/6 − 20/6; the other order would give an impossible negative length.
The other endpoint is S(5, 0). Reverse the midpoint formula with x2 = 2xM - x1, so the step from R to M gets repeated once more to land exactly on S.
The segment is 17 - 5 = 12 long, two thirds of that is 8, and the 8 units must be measured from X, giving 5 + 8 = 13 - not 17 - 8.
The triangle inequality needs the sum of the two shorter sides to be strictly greater than the longest. Here 4 + 7 = 11, so the figure is degenerate and zero triangles exist.
Exactly one triangle exists. All three side lengths are fixed, so SSS makes the triangle unique, and the triangle inequality holds since 4 + 8 = 12 is greater than 11.
The hypotenuse is c = 5. Squaring the legs gives 9 + 16 = 25, a perfect square, which is why 3-4-5 is the best known Pythagorean triple.
The hypotenuse is c = sqrt(61) = 7.81. Since 61 is prime it has no square factor, so the radical cannot be simplified and the exact answer stays in surd form.
Negating both coordinates sends each point to the far side of the origin at the same distance, which is a half-turn - not a reflection, which negates only one coordinate.
The exterior angles are 24, 48, 72, 96 and 120 degrees, so the interior angles are 156, 132, 108, 84 and 60 degrees — a ratio of 13:11:9:7:5 that sums to 540.
The two intersection points are (0.8800, 0.5081) and (0.3821, 0.2206). Substituting the line into the circle gives a quadratic with discriminant 0.44074, so two crossings.
Two altitudes of an isosceles triangle cut off equal segments: BK = CH follows from congruent right triangles, and KH is parallel to BC by the converse of Thales.
If AB = AC and D is the midpoint of BC, then AD is perpendicular to BC. Proof by SSS congruence plus the linear pair, with a coordinate proof as a second check.
The maximum area is 121/4 = 30.25, reached by a rectangle 5.5 tall and 5.5 wide. The trapezoid's width at height x is 11 - x, so the area is x(11 - x).
AB is 5 root 2, about 7.07, and the midpoint is (10, 0). Distance squares the coordinate differences; the midpoint averages them, so signs matter there.
The midpoint is (3.5, 2.5). Average the x-coordinates and the y-coordinates separately - subtracting them instead is what produces the common wrong answers.
The midpoint is (1843.5, 135.5, −4558). Average the three coordinate pairs independently; two of the three averages land on a half, which is expected from odd sums.
The closest point is (21/10, 7/10) = (2.1, 0.7), the orthogonal projection of (1, 4) onto (3, 1). The residual (−1.1, 3.3) is perpendicular to the direction vector.
Opposite angles of a quadrilateral inscribed in a circle are supplementary, so angle ADC = 180 - 136 = 44 degrees.
The overlap square has side 6 cm. Each sheet has area d^2/2 = 98, two sheets total 196, and subtracting the covered area 160 leaves an overlap of 36 square cm.
The overlap square has side 4 cm. Each sheet has area 200 square cm, two sheets total 400, and the covered area of 384 leaves an overlap of just 16 square cm.
The overlap square has side 8 cm. Each sheet covers 392 square cm, so two sheets give 784, and the stated covered area of 720 leaves 64 square cm counted twice.
Standard form is (x + 4)² = −4(y − 2), so the vertex is (−4, 2), the focus is (−4, 1), the axis of symmetry is x = −4 and the directrix is the line y = 3.
The track is 88 m. Each corner swaps 14 m of straight edge for a quarter arc of 3.5π ≈ 11 m, so the 100 m perimeter shrinks by about 3 m four times over.
The diagonal splits the rectangle into two equal triangles, so the rectangle's area is 16. Dividing by 4 sqrt 2 gives the other side 2 sqrt 2, and the perimeter is 12 sqrt 2.
The track is 44 + 14pi = 88 m. Each corner trades 7 + 7 m of straight edge for a quarter-circle arc, so the perimeter drops by 3 m per corner.
The area formula gives height 4 m; the base overhang is (18-12)/2 = 3 m, so each leg is a 3-4-5 hypotenuse of 5 m and the perimeter is 12 + 18 + 5 + 5 = 40 m.
The third side must equal 5 or 6, giving triangles 5-5-6 and 6-6-5. Both satisfy the triangle inequality, so the perimeter is not unique: 16 or 17.
A polygon with 14 diagonals is a heptagon (n = 7), and the only polygon whose interior angle sum equals its exterior sum is the quadrilateral, since (n - 2)180 = 360.
Both proofs follow from coordinates B(0,0), A(a,0), C(0,a), H(t,0), D(0,-t). The dot products of DH with AC and of CH with AD are each exactly zero.
The ratio is 4:3. For similar figures the perimeter ratio equals the ratio of corresponding sides, so 8 cm to 6 cm reduces to 4:3 with no extra work.
The larger room costs $384. Cost follows area, and area scales as the square of the side ratio, so multiply $216 by (4/3) squared = 16/9 to get $384.
Two consecutive vertices of a square are (2, −1) and (−1, 3). Rotating the side vector 90° gives two valid squares: (3, 6) and (6, 2), or (−5, 0) and (−2, −4).
CD = 7. Solving 7 + (x + 5) = 7x gives x = 2, so CD = x + 5 = 7, and the whole segment BD = 14 turns out to split into two equal halves of 7 each.
DF = 4. Here the whole segment also contains x, so the postulate gives 2 + 2x = x + 3 and hence x = 1; substituting back makes DF = 4 with two equal parts of 2.
FH = 20. Solving 4 + 8x = 10x gives x = 2, so FH = 10x = 20, made up of FG = 4 and GH = 16 — the constant term is what pins the variable down here.
LM = 17. The postulate gives (x + 10) + 4 = 3x, so x = 7; substituting back into LM = x + 10 is the step that answers the question actually asked.
LN = 51, from x = 19. Because M is between L and N, LM + MN = LN gives one linear equation in x — then substitute into LN rather than reporting x.
MN = 12. The equation 6 + (x + 9) = 6x has solution x = 3, so MN = x + 9 = 12 and the whole LN = 18 equals 6 + 12, confirming the split is consistent.
MN = 15. From 9 + (x + 7) = 3x we get x = 8, so MN = x + 7 = 15, and the whole segment LN = 24 equals 9 + 15 exactly as the postulate requires.
QR = 33, from x = 3. R lies between Q and S, so QR + RS = QS; the 11x and 19x terms combine to 8x = 24, and substituting back gives the length QR, not x.
RT = 18. From 6x + 6 = 9x we get x = 2, so RT = 9x = 18, split into RS = 12 and ST = 6 — and the 6 : 9 coefficient ratio predicts that two-thirds split.
TU = 40, from x = 7. Set ST + TU = SU because T lies between S and U, solve the linear equation, then substitute back into the expression for TU only.
The square's side is 7.8 cm. The equilateral triangle's perimeter is 3 x 10.4 = 31.2 cm, and dividing that shared perimeter by 4 sides gives 7.8 cm.
The answer is root 78, about 8.83 cm. Each square has area d^2/2 = 162, so the overlap is 162 + 162 - 246 = 78 square cm and its side is its square root.
The answer is 8 root 3, about 13.86 cm. Each square has area 288, so the overlap is 576 - 384 = 192 square cm and root 192 simplifies to 8 root 3.
The formula is r = sqrt(A/pi) and the radius is 5 cm. Divide by pi before taking the root, and keep only the positive root since a radius cannot be negative.
The result is h = S/(2 pi r) - r. Subtract the two circular ends first, then divide the whole remaining side by 2 pi r - dividing before subtracting is the usual error.
The far arc is 140 degrees. An external angle equals half the difference of the two intercepted arcs, so 38 = (far - 64)/2 gives far = 76 + 64.
The power of the point P satisfies PA^2 = PD x PC, so PC = 100/8 = 12.5 and the chord CD = PC - PD = 4.5.
The total surface area is 486 cm². Take the cube root of 729 to get an edge of 9 cm, then apply S = 6a² to the six identical square faces: 6 × 81 = 486 cm².
The surface area is 486 square cm. Take the cube root of 729 to get an edge of 9, square it for one face area of 81, then multiply by the cube's six faces.
A trapezoid with leg BC = 7 cm whose opposite leg has midpoint 4 cm from BC has area 28 cm². The trick: triangle IBC is always exactly half the trapezoid.
Answer: 556.5 square units. A shape like a triangle with a flat top is a trapezoid, so average the parallel sides (45+8)/2 = 26.5 and multiply by the height 21.
The perimeter is 12, not “10 or 12”. The quadratic gives sides 2 and 4, but 2 fails the triangle inequality with sides 3 and 5, so only the root 4 is usable.
A straight row of five 1 cm squares has perimeter 12 cm and a 3-plus-2 block has perimeter 10 cm. The rule is perimeter = 20 - 2 x (number of shared edges).
Vertex (1, 2), focus (1, 4), axis x = 1, directrix y = 0. Match the equation to (x - h)^2 = 4p(y - k), read off h, k and p, then place the focus.
Vertex (1, -3), focus (-2, -3), axis y = -3, directrix x = 4. Spot the perfect square, factor to (y + 3)^2 = -12(x - 1), then read 4p = -12.
V = 1.331 sqrt(3) pi / 3 = 2.41 cubic metres. The 60 degree apex angle halves to a 30 degree half-angle, so the height is r/tan30 = 1.1 sqrt 3 = 1.9052 m.
The volume is 36 pi, about 113.1 cubic inches, matching choice 113.04. Halve the diameter to get r = 3 first; using d = 6 instead inflates the answer eightfold.
The volume is about 19.313 m³. The 60° apex gives a half-angle of 30°, so the height is r√3 = 3.8105 m before applying V = πr²h/3.
The volume is 729 cubic cm. Divide 486 by the cube's six faces to get a face area of 81, take the square root for an edge of 9, then cube it.
Answer: 90*pi, about 282.74 cubic units. The volume of a cylinder is the base area pi*r^2 times the height, so pi*3^2*10 = 90*pi.
The volume is 2,958,816 cubic units. Multiply length x width x height, splitting 444 x 476 into partial products so the three-factor arithmetic stays checkable by hand.
The volume is 384 cubic inches. Square the base edge for 144, multiply by the height 8 to get 1152, then divide by 3 because a pyramid is a third of its prism.
The volume is 600 cubic metres. Square the base edge to get 225, multiply by the height 8, then take one third; the one third is what separates a pyramid from a prism.
Halve the diameter to get r = 6 mm, then V = (4/3)pi r^3 = 288 pi. That is 904.78 mm^3 with the true pi, or the 904.32 mm^3 that appears in answer keys built on pi = 3.14.
The volume is 5827.5 m^3. Find the right-triangle cross-section area (1/2 x 21 x 15 = 157.5 m^2), then multiply by the 37 m prism length.
An oblique cylinder has a characteristic cross-section that is a rhombus with side 2 cm and a 60° angle. Find the volume using the rhombus area to recover radius and height.
The plane through the midpoints of AD and AB parallel to TA splits the pyramid 9 : 23. The section is a pentagon, and slicing by height gives the exact 3/32.
The two pieces measure 2a^2*t/45 and 13a^2*t/45. Use coordinates to find where plane PQK meets the four lateral edges, then apply the tetrahedron ratio rule.