Geometry
Real geometry questions asked by students, solved step by step Every question below was submitted by a real student and answered step by step.
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.
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.
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 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.
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 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.
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.
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 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 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.
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.
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.
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.
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).
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.
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.
Answer: vertex (1,2), focus (1,4), axis x = 1, directrix y = 0. Comparing with (x-h)^2 = 4p(y-k) gives 4p = 8, so p = 2 and the parabola opens upward.
Answer: vertex (1,-3), focus (-2,-3), axis y = -3, directrix x = 4. The left side is the perfect square (y+3)^2, so the equation becomes (y+3)^2 = -12(x-1) and opens left.
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 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 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.
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.
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.
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.
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.