Mensuration and Coordinate Geometry

Graph of absolute value function

The absolute value of a real number a is |a| = a when a \ge 0 and |a| = -a when a < 0. Geometrically, it is the distance from a to 0, so |a| is never negative. The graph of y = |x| is a V with its vertex at the origin and slopes +1 and -1. Replacing x by x - a moves the vertex to x = a, and adding a constant c outside the bars shifts the whole graph vertically by c.

When one absolute value wraps another, as in f(x) = \bigl|\,|x+2| - |x-1|\,\bigr|, split the real line at the points where each inner expression changes sign, and handle each piece separately. A general fact is useful here: for a < b, the expression |x-a| - |x-b| equals -(b-a) for x \le a, equals 2x - a - b on [a,b], and equals b-a for x \ge b. The inner part is therefore a flat, steep, flat ramp that crosses zero at the midpoint x = \frac{a+b}{2}.

The outer absolute value then reflects the negative flat piece upward to height b-a, and folds the middle line into a V whose minimum of 0 sits at the midpoint. The resulting graph is flat, then V-shaped with its zero at the midpoint, then flat again, with both flat levels equal to b-a. Since the outer bars make every output nonnegative, a candidate graph that dips below the horizontal axis can be eliminated at once, and the zeros of f are exactly the zeros of the inner expression.

As a fresh example, h(x) = \bigl|\,|x-1| - 2\,\bigr| has an inner V with vertex (1,-2). Reflecting its negative part gives zeros at x=-1 and x=3, and a peak of height 2 at x=1, where the graph turns downward.

The expression \frac{|x|}{x} equals +1 for x>0 and -1 for x<0, and it is undefined at x=0; \frac{x}{|x|} is the same function. Hence -\frac{|x|}{x} equals -1 on the positive half-line and +1 on the negative half-line. Its graph consists of two horizontal half-lines with open ends at the origin, not a hyperbola like \frac{1}{x} and not a set of vertical segments. To check a plot, substitute one positive and one negative value of x and read off the sign directly.

The max function returns the larger of its two arguments, so \max(S-K, 0) is 0 while S \le K and equals S-K afterwards. This hinge function H_K(S) has slope 0 and then slope 1. The identities \max(x,0) = \frac{x+|x|}{2} and |x| = 2\max(x,0) - x connect hinges directly to absolute values. If K_1 < K_2, then H_{K_1}(S) - H_{K_2}(S) is 0 for S \le K_1, rises with slope 1 on [K_1,K_2], and stays constant at K_2 - K_1 for S \ge K_2. A piecewise linear graph that is flat, rises with slope 1, and flattens again is therefore a difference of two hinges, with breakpoints at the two values of K. To match such a plot, evaluate the candidate expression at one point left of both breakpoints, one between them, and one to the right; these three values quickly separate the options.

Mutually tangent circles with centre triangle

Two circles touch when they have exactly one common point. If they touch externally, the distance between their centres equals the sum of the radii, d = r_1 + r_2. If one circle lies inside the other and they touch internally, the distance equals the difference of the radii, d = |r_1 - r_2|. In either case the point of tangency lies on the segment joining the two centres.

When three circles touch one another pairwise, their centres form a triangle whose sides are sums of radii. For centres O_1, O_2, O_3 we have O_1O_2 = r_1 + r_2, O_1O_3 = r_1 + r_3, and O_2O_3 = r_2 + r_3. The unknown radius therefore appears linearly in two sides, and any extra condition on the triangle, such as a right angle at one centre, becomes an equation in r_3. If the angle at O_3 is 90^\circ, Pythagoras' theorem gives O_1O_2^2 = O_1O_3^2 + O_2O_3^2. For a general angle, use the cosine rule c^2 = a^2 + b^2 - 2ab\cos C.

Expanding the squared sides gives a quadratic r^2 + pr + q = 0, whose roots are r = \frac{-p \pm \sqrt{p^2 - 4q}}{2}. For a fresh example, let r_1 = 3, r_2 = 1, and let the angle at O_3 be right. Since O_1O_2 = 4, the equation (3+r)^2 + (1+r)^2 = 16 simplifies to r^2 + 4r - 3 = 0, so r = \sqrt{7} - 2 \approx 0.65. Discard negative roots and any root that makes the circle too large for its region. When the options are given as surds, comparing decimal approximations is the fastest check.

After finding r_3, confirm that it is positive and that it reproduces all three centre distances. Many distractors are the same roots with a wrong sign inside the square root, so match the signs carefully when simplifying.

In a square tangency problem, a circle touching two perpendicular sides has its centre on the diagonal through the corner, at distance r from each side. With the corner at the origin, the centre is (r, r); for a circle in the opposite corner of a square of side s, the centre is (s-r, s-r). If two such circles touch, the distance between their centres is \sqrt{2}\,(s - r_1 - r_2), so the tangency condition is \sqrt{2}\,(s - r_1 - r_2) = r_1 + r_2.

An equation of the form \sqrt{2}\,(a - r) = b + r gives r = \frac{a\sqrt{2} - b}{\sqrt{2} + 1}, and multiplying numerator and denominator by \sqrt{2} - 1 rationalises the denominator. Squaring instead gives a quadratic in r whose two roots include one extraneous value; reject the root that exceeds the side length of the square, and keep the root that lies inside it.

Graph of quadratic function

A graph-matching item shows a picture and asks which formula produces it. Do not begin by substituting many points. Read the qualitative features of the curve first and compare them with the formula. Four features decide most items: symmetry, sign, behaviour as |x|\to\infty, and the location of turning points.

• even function — f(-x)=f(x) for all x, so the graph is a mirror image across the y-axis. 예) g(x)=x^2\,3^{-|x|} is even.
• odd function — f(-x)=-f(x), so the graph is unchanged by a half-turn about the origin, and f(0)=0 when f is defined at 0. 예) h(x)=x\,3^{-|x|} is odd.
• decay at infinity — the behaviour of f(x) as x\to\pm\infty. 예) For x<0, 2^{-x}=2^{|x|} grows without bound, while 2^{-|x|} decays on both sides.

Symmetry is the first check. Replacing x by -x shows whether a formula is even or odd. Multiplying by x alone makes a product odd, while x^2 and |x| are even. Therefore x^2\cdot 2^{-|x|} is even and x\cdot 2^{-|x|} is odd, because 2^{-|x|} is even in both cases.

Sign is the second check. A factor x^2 or 2^{-|x|} is never negative, so x^2\cdot 2^{-|x|}\ge 0 everywhere, and it vanishes only at x=0. An odd function changes sign at the origin, so the graph must be negative on one side and positive on the other.

The tails matter as well. A graph that approaches zero on both sides needs a decaying factor on both sides, so an expression with 2^{-x} alone fails because it blows up as x\to-\infty. The expression x\cdot 2^{-|x|} approaches zero at both ends.

Turning points can be located by calculus. For x>0, f(x)=x\,2^{-x} has f'(x)=2^{-x}(1-x\ln 2), which vanishes at x=1/\ln 2\approx 1.44, and the peak value is e^{-1}/\ln 2\approx 0.53. Oddness then gives the matching trough on the negative side.

For quadratic curves, the number of points where two graphs meet is found by equating the expressions and counting real roots. Writing ax^2+bx+c=0, the discriminant \Delta=b^2-4ac decides the count: \Delta>0 gives two distinct real intersection points, \Delta=0 gives one point of tangency, and \Delta<0 gives none. Example: y=2x^2+1 and y=-x^2 give 3x^2+1=0 with \Delta=-12<0, so the parabolas do not meet. A quick check is to see whether one expression is always nonnegative and the other always nonpositive. Then they can meet only where both are zero, and one must check whether those zeros coincide.

Shaded region area with circular arcs

Area problems on circular figures are solved by splitting the shaded region into pieces whose areas are standard, then adding or subtracting them. First identify each arc's centre and radius. A semicircle whose diameter is a side of a square has its centre at the midpoint of that side and radius equal to half the side.

• disk — area \pi r^2. Semicircle — area \frac{\pi r^2}{2}. Quarter circle — area \frac{\pi r^2}{4}. 예) A quarter circle of radius 3 has area \frac{9\pi}{4}.
• sector — area \frac{\theta}{2}r^2, with \theta in radians.
• segment — the region between a chord and its arc, equal to a sector minus a triangle, with area \frac{r^2}{2}(\theta-\sin\theta).

The complement method is often the shortest route. If the total figure has a known area, the shaded area equals the total minus the white area. Decide which pieces are white before adding anything. Quarter circles centred at the corners of a square, together with a semicircle, often combine to give a whole number.

When two curved regions overlap, the shaded set may be the union minus the overlap, which is the symmetric difference. Its area is |A|+|B|-2|A\cap B|, because the overlap is counted in both pieces and must be removed twice. Two semicircles built on adjacent sides of a square often overlap in a lens-shaped region. For two circles of radius r whose centres are d apart, the lens area is 2r^2\cos^{-1}\!\left(\frac{d}{2r}\right)-\frac{d}{2}\sqrt{4r^2-d^2}. Example: for r=2 and d=2, the lens area is \frac{8\pi}{3}-2\sqrt{3}.

Two triangles that share a parallel side are similar. If a line parallel to one side of a triangle cuts the other two sides in the ratio m:n, every length in the smaller triangle scales by that ratio, and areas scale by its square. A semicircle on a segment is similar in scale to one on a proportional segment, so the areas of semicircles scale by the square of the diameter ratio.

An inscribed angle is half the central angle that subtends the same arc. So an inscribed angle of 45^\circ corresponds to a central angle of 90^\circ on the same arc. If the centre lies inside a triangle whose vertices are on the circle, that triangle splits into three triangles that share the centre as a vertex. Each piece has area \frac{1}{2}ab\sin C for two sides a,b and included angle C.

Concentric rings are regions between two circles with the same centre. The area between radii r_1<r_2 is \pi(r_2^2-r_1^2), so ring area depends on the squares of the radii, not on widths alone. For a ratio of land to water, add the areas of each kind separately, keep \pi symbolic until the end, and then round as the question specifies.

Trigonometric ratio from angle measurement

In a right triangle with an acute angle \theta, the three basic ratios are \sin\theta = \text{opposite}/\text{hypotenuse}, \cos\theta = \text{adjacent}/\text{hypotenuse}, and \tan\theta = \text{opposite}/\text{adjacent}. Their reciprocals are \sec\theta = 1/\cos\theta, \csc\theta = 1/\sin\theta, and \cot\theta = 1/\tan\theta. The words opposite and adjacent are defined relative to the angle being used, not to a fixed position in the figure, so the first step in any such problem is to name which leg is adjacent to the given angle and which side is the hypotenuse. \n\nWhen the angle is close to 90^\circ, the adjacent leg is very short compared with the hypotenuse, so \cos\theta is very small and \sec\theta is very large. Complementary angles satisfy \cos(90^\circ-\delta)=\sin\delta, so for a small deviation \delta we have \sec(90^\circ-\delta)=1/\sin\delta. A tiny change in the measured angle therefore produces a large change in the ratio, and the answer depends heavily on the last digits of the angle. \n\nFor small \delta measured in degrees, use \sin\delta \approx \delta_{\text{rad}}, where \delta_{\text{rad}} = \delta \cdot \pi/180. For example, 0.1^\circ = 1.745\times10^{-3} rad, so \sec(89.9^\circ) \approx 1/(1.745\times10^{-3}) \approx 573. Always convert degrees to radians before applying \sin x \approx x; using the degree value directly gives a wrong answer. \n\nIn an isosceles triangle, the altitude from the apex bisects the base and creates two congruent right triangles. If the base has length b and each base angle is \alpha, the altitude is h = (b/2)\tan\alpha = (b/2)\cot(90^\circ-\alpha). When \alpha is close to 90^\circ, the complement form is the more reliable one to evaluate, because the small residual angle carries the precision. As a check, the law of sines gives the side opposite the base angle, and the Pythagorean theorem then gives the altitude; both routes should agree.

Cylinder formed from rolled rectangle

A rectangular sheet can be rolled into a cylinder by bringing two opposite edges together. The joined edges run parallel to the axis, so their length becomes the height h of the cylinder. The other pair of edges wraps around the circular cross-section, so its length is the circumference 2\pi r, which gives r = L/(2\pi) where L is the wrapping side. Which pair is joined therefore decides which side becomes the height and which becomes the circumference. \n\nA sheet of dimensions 6\times1 rolled with its short edges joined has height 1 and circumference 6, so r = 3/\pi. A sheet of dimensions 54\times4 rolled with its long edges joined has height 54 and circumference 4, so r = 2/\pi. Each case must be read from the joining rule, not from the sheet's size alone. \n\nFor a closed cylinder, the volume is V = \pi r^2 h, because the two circular ends enclose the solid. A worked example: a 40\times3 sheet rolled with the 40 cm edges joined has h = 40 and r = 3/(2\pi), so V = \pi \cdot \dfrac{9}{4\pi^2}\cdot 40 = \dfrac{90}{\pi}. The surface area of a closed cylinder is 2\pi r^2 + 2\pi r h, but the volume formula does not change with the surface. \n\nA cube of edge a has surface area 6a^2 and volume a^3. When a cube's surface area is set equal to a sheet's area S, its edge is a = \sqrt{S/6}; no cutting is implied in that case. When the problem asks for the largest cube formed from equal square patches, the patches must tile the sheet exactly, and a cube of edge a needs six square faces of side a. For the 6\times1 sheet, the total area is 6 square units, so the largest cube has edge 1 unit and volume 1 cubic unit. \n\nFor a 54\times4 sheet, the area is 216 square units, so the cube edge satisfies 6a^2 = 216, giving a = 6 and volume 216. The ratio of the cylinder volume 216/\pi to the cube volume 216 is 1/\pi. Since both volumes must be computed in the same units, the ratio is dimensionless and any unit error cancels only if both volumes are converted consistently.

Coordinate displacement and distance

A route described as a sequence of compass moves is easiest to analyse by placing the starting point at the origin of a coordinate plane, with East as the positive x-direction and North as the positive y-direction. Each leg then changes the position by a displacement. A move of d units East adds (d, 0), a move of d units South adds (0, -d), and a diagonal leg adds equal components along both axes, so a leg of length d in the North-East direction adds (d/\sqrt{2}, d/\sqrt{2}). Displacement is the net change in position, whereas the distance travelled is the total length of all the legs. Only the displacement determines the final point.

When a leg is specified by the coordinate where it ends rather than by its length, the unknown component must be found from that coordinate. A later leg stated as a pure North or South move changes only the y-coordinate. Two legs whose east-west components are equal and opposite cancel in the x-direction, so their combined effect is a pure north-south displacement.

The distance of a point from the origin is the hypotenuse of the right triangle formed by its coordinates, \sqrt{x^2+y^2}, by Pythagoras. The distance between two points (x_1,y_1) and (x_2,y_2) is \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. A frequent error is to report a single coordinate as the distance, which ignores the other component.

An intermediate point reached partway along a route is not the endpoint, and its coordinates should not be taken as the answer. A distance that equals a coordinate of an intermediate point is a common distractor.

The same reasoning applies when a path is specified by turning angles instead of compass directions. Track the heading, where an anticlockwise turn adds a positive angle and a clockwise turn subtracts it. If equal moves and equal turns repeat, the path closes when the total turning is a multiple of 360^\circ. A regular star polygon \{n/k\} is traced when each turn is k \times 360^\circ / n. The pentagon uses a turn of 72^\circ, while the pentagram \{5/2\} uses 144^\circ, and its sides cross one another. An initial turn only rotates the whole figure and does not change its shape.

When two bodies leave the same point along perpendicular directions, each one's distance from the start equals its own displacement along its axis. If a body stops for a time t within a total travel time T, its moving time is T-t and its distance is speed times moving time. Equating two such distances gives a linear equation in t, and converting minutes to hours before solving avoids errors.

Points with integer coordinates form a lattice. Each lattice point has six nearest neighbours at unit distance along the coordinate axes, while the next nearest neighbours lie at \sqrt{2} along face diagonals, so the ratio of the next-nearest distance to the nearest distance is \sqrt{2}.

Parallelogram side from perpendicular distances

In a parallelogram the opposite sides are parallel and equal in length, so each side has a partner of the same length and the figure can be read from either pair.

The area of a parallelogram equals any side taken as the base multiplied by the perpendicular height to that base. The height is the perpendicular distance between the base line and the parallel line through the opposite side. It is not the length of an adjacent slanted side, and the foot of the perpendicular may fall outside the segment that forms the base.

Since the area is the same whichever side is chosen as base, two expressions for the area can be equated. If side a has height h_a and side b has height h_b, then a\,h_a = b\,h_b. For example, if one side is 8 and its height is 3, the other side with height 4 has length (8 \times 3)/4 = 6.

Once the area is known from one base and height, the unknown side is found by dividing the area by the given perpendicular height. Fractional results such as 28/5 should be kept exact until the end of the calculation.

A perpendicular drawn from a vertex is sometimes mistaken for an adjacent side. The two coincide only in a rectangle, where the adjacent sides are perpendicular to each other.

A useful check is to recompute the area with the other base and height. The two products must be equal, and any mismatch shows a misidentified height.

• altitude (height) — the perpendicular distance from a vertex, or between two parallel lines, used as the height for a chosen base. 예) In a parallelogram with base 10 and height 4, the area is 40 square units.
• slant side — a side that is not perpendicular to the base, so its length is not the height. 예) In a parallelogram with base 10 and slant side 5, the height is not necessarily 5.

Triangle formation from segment lengths

Three segments can form a triangle only if the length of each segment is strictly less than the sum of the other two. This is the triangle inequality. Equality in any of the three inequalities gives a degenerate figure in which the points are collinear, and that figure is not a triangle.

When the three pieces come from one stick of total length L, their lengths sum to L, so the three inequalities reduce to a single requirement. Each piece must be strictly shorter than L/2. The reduction holds because a piece shorter than L/2 is automatically shorter than the sum of the other two, which exceeds L/2.

A stick broken at two points gives three pieces. If the stick has length 1 and the break points are at positions b_1 < b_2 measured from one end, the pieces are b_1, b_2 - b_1 and 1 - b_2. Writing each piece condition in terms of b_1 and b_2 gives the constraints on the break points.

A necessary condition is one that must hold in every case where a triangle is formed. To show that a stated condition is not necessary, it is enough to find one valid triangle in which that condition fails. For instance, a stick of length 10 broken into pieces 4, 4 and 2 forms a triangle, since 4+2>4, even though the sum of the two break positions, 4+8=12, exceeds the total length 10.

Conditions that involve sums of break positions rather than individual piece lengths are not triangle inequalities, so they must be tested against a concrete example before being accepted as necessary.

• triangle inequality — the rule that each side is less than the sum of the other two. 예) Segments of lengths 3, 4 and 8 cannot form a triangle, because 3+4<8.
• degenerate triangle — the case where one length equals the sum of the other two, so the three points are collinear. 예) Segments of lengths 2, 3 and 5 give a degenerate figure.

Equilateral triangle side and area

An equilateral triangle has three equal sides and three interior angles of 60^\circ. If its side is s, its altitude is h=\frac{\sqrt{3}}{2}s and its area is A=\frac{\sqrt{3}}{4}s^2. Because the area carries the irrational factor \sqrt{3}, an equilateral triangle with an integer side always has an irrational area, so a plain rational answer can be rejected at once when the side is an integer.

Key terms:

• subdivide into equal parts — cut a side into n congruent pieces, each of length s/n. Example: a side of 12 cm cut into 3 equal parts gives pieces of 4 cm.
• common multiple — a number divisible by each of the given divisors. The least such number is the least common multiple (lcm). Example: \operatorname{lcm}(3,5)=15.

When every side is cut into a given number of equal parts and each part must have integer length, the side s must be a common multiple of all those divisors, so the minimum side is their lcm. For example, if the sides are cut into 3 and 5 equal parts, the minimum side is 15 cm and the area is \frac{\sqrt{3}}{4}\times 15^2=\frac{225\sqrt{3}}{4} cm^2.

Cutting equal corners from an equilateral triangle: if each corner is cut off at the points that divide the sides into three equal parts, each removed corner is an equilateral triangle of side s/3. The remaining figure is a regular hexagon of side s/3. Each corner has area \frac{1}{9} of the whole triangle, so the three corners remove \frac{1}{3} of the area and the hexagon keeps \frac{2}{3}. The hexagon-to-triangle area ratio is therefore 2:3.

Inscribed and circumscribed circles: for an equilateral triangle of side a, the circumradius is R=\frac{a}{\sqrt{3}} and the inradius is r=\frac{a}{2\sqrt{3}}, so R=2r. Both circles share the centroid as centre. The ratio of their areas is \left(\frac{r}{R}\right)^2=\frac{1}{4}, not \frac{1}{2}, because area scales with the square of the radius.

Sphere inscribed in cube

A sphere of radius r is enclosed by the smallest cube when every face of the cube just touches the sphere. In that case the edge of the cube equals the diameter 2r, because the sphere touches two opposite faces and the distance between them is the diameter.

Key terms:

• enclosing — containing the whole solid inside a boundary, so the sphere must not cross any face.
• inscribed — drawn inside a figure and touching its faces from within.

Since the edge is a=2r, the volume of the minimum cube is V=a^3=(2r)^3=8r^3.

A cube whose edge is smaller than the diameter cannot contain the sphere, however its volume is expressed. Writing the volume as r^3 treats the radius as the edge, and writing 2r^3 forgets to cube the factor 2.

For comparison, the sphere's own volume is \frac{4}{3}\pi r^3, so the sphere occupies \frac{\pi}{6}\approx 0.52 of the volume of its enclosing cube. Worked check: a sphere of radius 3 cm needs an edge of 6 cm, so the cube has volume 6^3=216 cm^3.

Square areas in tangent circle diagram

A tangent to a circle is a line that touches the circle at exactly one point, called the point of contact. The radius drawn to the point of contact is perpendicular to the tangent, so the radius and the tangent segment meet at a right angle.

Key terms:

• tangent — a line touching a circle at one point.
• point of contact — the single common point of the line and the circle.

Because of this right angle, the triangle formed by the centre, the point of contact and an external vertex is right-angled, and Pythagoras' theorem applies. The square constructed on the hypotenuse has area equal to the sum of the squares constructed on the two legs. Hence the square on one leg equals the square on the hypotenuse minus the square on the other leg.

In a tangent diagram, one leg is a radius, so the square on it has the area of the radius squared, while the other leg is a tangent segment. The largest square is always the one on the hypotenuse, so the relation to remember is: square on hypotenuse = square on leg 1 + square on leg 2, rearranged as square on tangent segment = square on hypotenuse minus square on radius.

Circles touching sides of a square: a circle of radius r that touches two perpendicular sides of a square, and lies inside it, has its centre at distance r from each side. If the square's corner is at the origin, the centre is at (r,r). Two circles touching each other externally have centres a distance r_1+r_2 apart. If one circle sits in a corner at (r_1,r_1) and the other in the opposite corner at (a-r_2,a-r_2), the centre distance is \sqrt{2}\,(a-r_1-r_2). Setting this equal to r_1+r_2 gives r_2=\frac{\sqrt{2}\,a-(1+\sqrt{2})r_1}{1+\sqrt{2}}.

Rationalising with \frac{1}{1+\sqrt{2}}=\sqrt{2}-1 simplifies the result. Worked check: for a square of side 6 cm and r_1=1 cm, r_2=(5\sqrt{2}-1)(\sqrt{2}-1)=11-6\sqrt{2}\approx 2.52 cm, which is small enough to fit inside the square.

Geometric growth of iterated curve length

In an iterative curve construction, every straight segment at one stage is replaced by a fixed pattern of smaller segments at the next stage. If the pattern has k segments, each scaled by the factor \frac{1}{m} of the segment it replaces, then the number of segments is multiplied by k and the segment length is divided by m at every iteration.

Key terms:

• iteration — one application of the replacement rule to every segment.
• common ratio — the factor by which the total length is multiplied at each step, here \frac{k}{m}.

The total length after n iterations is therefore a geometric progression, L_n=L_0\left(\frac{k}{m}\right)^n, where L_0 is the initial length. After n iterations there are k^n segments, each of length m^{-n}.

The exponent is n, not 2n or \frac{n}{2}, because the ratio is applied once per iteration. A useful check is to compute the totals for iterations 1 and 2 and confirm that the formula reproduces both. Example: if each segment is replaced by 4 segments of half length, the ratio is \frac{4}{2}=2, so the length after n iterations is 2^n.

A common error is to count only the length of one piece or to square the ratio. Another is to forget that the ratio is the product of the segment count and the scaling factor, not their sum. Even when the length grows without bound, the curve is still made of a finite number of segments at each stage.

Circle area from right triangle

The area of a circle is \pi r^2, so when a triangle is given, the radius must be found first. If M and N lie on a circle and O is the centre, then triangle MON has its right angle at O. Both OM and ON are radii, so they have the same length r and act as the two legs of the right triangle. The area of a right triangle is half the product of its legs, so the triangle's area is \frac{1}{2}r^2.

Setting the given triangle area equal to \frac{1}{2}r^2 gives r^2 directly, and the circle's area follows from \pi r^2. A common error is to treat the triangle area as r^2 and skip the factor \frac{1}{2}, which makes the radius squared too small by half.

The pattern is: if \frac{1}{2}r^2 = A, then r^2 = 2A and the circle's area is 2\pi A. For A = 50, this gives 100\pi. Wrong options in such questions usually come from ignoring the factor 2 (giving A\pi), using the triangle area A itself as the answer, or squaring the radius incorrectly. Each option can be checked by substituting back into \frac{1}{2}r^2 = A.

Cone formed by revolving triangle

When a plane figure is revolved about one of its sides, that side becomes the axis of the solid. For a right triangle revolved about one of its legs, the fixed leg is the height h, and the other leg, which is perpendicular to the axis, sweeps out a circle of radius r at the base. The resulting solid is a right circular cone with volume V = \frac{1}{3}\pi r^2 h.

The key step is identifying which side is the radius and which is the height. The hypotenuse is never the radius or height when the triangle is revolved about a leg, even though it is the longest side. For example, cutting a unit square along its diagonal gives a right isosceles triangle with legs of 1 unit and hypotenuse \sqrt{2}. Revolving this triangle about a leg of length 1 gives r = h = 1, so V = \frac{\pi}{3}.

A quick check is that a cone occupies one third of the cylinder with the same radius and height, whose volume is \pi r^2 h. Any answer larger than this cylinder volume is impossible, so values such as \frac{3\pi}{2} or 3\pi can be rejected when r = h = 1. The typical errors are forgetting the factor \frac{1}{3} or using a radius or height that is not a leg.

Solids made of two cones are analysed by an axial cross-section. If two identical right circular cones share a common circular base and their apexes lie on the common axis, a plane through both apexes contains that axis. Each cone's axial section is an isosceles triangle, and the outer boundary of the combined section has four vertices: the two apexes and the two ends of the common diameter. Each of its four sides is the slant height of one cone, so all four sides are equal and the boundary is a rhombus.

Trapezium height from side lengths

The height of a trapezium is the perpendicular distance between its two parallel sides. It is not the length of either slanted leg. When only the side lengths are given, the height is found by forming a right triangle. Draw a line through one vertex parallel to a leg, which creates a parallelogram and a triangle whose sides are the two legs and the difference of the parallel sides.

In such a triangle, the difference of the parallel sides is the base, and the legs are the other two sides. If the sides are 3, 4 and 5, then 3^2 + 4^2 = 5^2, so the triangle is right-angled with area \frac{1}{2}\cdot 3 \cdot 4 = 6. Taking the base as 5, the altitude h satisfies \frac{1}{2}\cdot 5 \cdot h = 6, so h = \frac{12}{5}. In general, the height equals twice the triangle's area divided by the base.

An algebraic approach gives the same result. Let x and y be the horizontal offsets of the two legs, so that x + y equals the difference of the parallel sides. The legs satisfy x^2 + h^2 = (\text{leg})^2 and y^2 + h^2 = (\text{other leg})^2. Solving these equations simultaneously recovers the height.

The height can never exceed either slanted leg, because each leg is the hypotenuse of a right triangle whose other side is the height. Any option larger than the shorter leg can therefore be eliminated immediately.

Quadrilateral area from coordinates

The area of a quadrilateral given by coordinates is often easiest to find from the geometry of its vertices rather than by a long computation. When two vertices share an x-coordinate, the segment joining them is vertical, and its length is the difference of their y-coordinates. Similarly, a horizontal segment has length equal to the difference of the x-coordinates.

For a quadrilateral whose diagonals are perpendicular, the area is \frac{1}{2}d_1 d_2, where d_1 and d_2 are the diagonal lengths. The formula applies to any convex quadrilateral with perpendicular diagonals, including rhombi and kites. The vertices must be joined in an order so that the segments joining opposite vertices are the diagonals.

If both diagonals measure 4 units and are perpendicular, the area is \frac{1}{2}(4)(4) = 8 square units. The same answer can be checked with the shoelace formula or by splitting the quadrilateral into two triangles along one diagonal, so one method can verify another.

Common errors include taking only one diagonal, which gives half the area, or mistaking a side length for a diagonal. A side such as 2\sqrt{2} introduces an irrational factor that does not belong to the formula.

Cube wire frame cutting

A wire frame of a cube can be treated as a graph whose vertices are the corners and whose edges are the twelve edges of the cube. Each corner meets three edges, so every vertex has odd degree 3. A wire passing through a vertex uses two of its edges, so an odd-degree vertex must be an endpoint of at least one wire piece.

Each wire piece has only two ends, so eight odd vertices require at least 8/2 = 4 separate pieces. Soldering lets pieces be joined at the corners, but it does not reduce the number of pieces that the edge structure forces. If k pieces are cut from one wire, exactly k-1 cuts are needed, so the minimum is 4 - 1 = 3 cuts.

This minimum is achievable. The twelve edges can be split into four edge-disjoint trails of three edges each, which can be checked edge by edge. Since the wire length equals the total edge length when each edge is 1 m, no wire is wasted, and the pieces can be bent and soldered into the frame.

Four cuts would produce five pieces, and six or twelve cuts divide the wire far more finely than necessary. Such options are wrong because the question asks for the minimum.

Intersection of two circles

A circle is the set of all points at a fixed distance r (the radius) from a fixed centre (a,b). Its standard equation is (x-a)^2+(y-b)^2=r^2, and a point (x,y) lies on the circle exactly when its coordinates satisfy this equation. To test whether a given point lies on a circle, substitute its coordinates into the equation and check for equality. This substitution method is the standard approach when the candidate points are already listed as options.