Constructions and Loci · applications

Applications: Constructions and Loci

10 question types · Secondary 2 · each worked step by step with a figure that follows the steps

GCSE Higher

01

Two Lines of Paving Laid 2 m from the Walls of a Slanting Corner, and the Lamp Post Where They Cross

methodKeep the Wall's Gradient for the Parallel Line, Step 2 m Along a Perpendicular to Find One Point of It, Then Solve the Two Lines of Paving Together

A courtyard lies between two straight walls that meet at a corner O. On a plan marked in meters, O is at (0, 0), one wall runs along the x-axis, the other runs along the line y = 34x, and the courtyard is the space between them for x > 0. A gardener lays two straight lines of paving stones inside the courtyard, each parallel to one wall and 2 m from it. (a) Find the equation of the line of paving that is parallel to the slanting wall. (b) A lamp post stands where the two lines of paving cross. Find its position, and its distance from the corner O.

246810246Oy = 22 mwallwallpaving beside the x-axis: y = 2
The paving beside the x-axis is the line y = 2.
The paving beside the x-axis is parallel to it and 2 m inside the courtyard, so it is the line y = 2.
step 1 of 5

A line parallel to a wall has the same gradient as the wall, and its distance from the wall is measured at right angles to the wall, not straight up or straight across. So a point 2 m from the slanting wall is found by stepping 2 m along a perpendicular, and the parallel line is drawn through that point. Where the two lines of paving cross is a point 2 m from both walls.

  1. The paving beside the x-axis is parallel to it and 2 m inside the courtyard, so it is the line y = 2.
  2. A perpendicular to the slanting wall has gradient −43, because 34 × (−43) = −1. Going 3 m across and 4 m down is a step of √32 + 42 = 5 m, so a step of 2 m is 25 of it: 1.2 m across and 1.6 m down. From the point (4, 3) on the wall, this reaches (5.2, 1.4), which is inside the courtyard and 2 m from the wall.
  3. (a) The paving has the wall's gradient, 34, and passes through (5.2, 1.4): y − 1.4 = 34(x − 5.2), so y = 34x − 3.9 + 1.4 = 34x − 2.5. Multiplied by 4 and rearranged, this is 3x − 4y = 10.
  4. The lamp post is on both lines of paving, so substitute y = 2: 2 = 34x − 2.5, so 34x = 4.5 and x = 6. The lamp post is at (6, 2).
  5. (b) Its distance from O is √62 + 22 = √40 = 2√10 ≈ 6.32 m. Check: stepping 2 m from the lamp post back toward the slanting wall, 1.2 m to the left and 1.6 m up, reaches (4.8, 3.6), and 34 × 4.8 = 3.6, so that point is on the wall.

answer(a) y = 34x − 2.5, that is 3x − 4y = 10; (b) at (6, 2), 2√10 ≈ 6.32 m from O

techniqueDrawing Parallel and Perpendicular Lines · Parallel and Perpendicular Lines

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Moving the wall's line 2 m straight down, to y = 34x − 2. That shift is measured vertically, not at right angles to the wall, and the lines y = 34x and y = 34x − 2 are only 2 × 45 = 1.6 m apart.
  • Stepping 2 m from the slanting wall on the wrong side, which gives 3x − 4y = −10. That line lies above the slanting wall, outside the courtyard, and it meets y = 2 at x = −23, which is not in the courtyard at all.
02

A Drain Laid at Right Angles to a Road, and Where It Reaches a Stream

methodFind the Perpendicular Gradient from Gradients That Multiply to −1, Write the Drain's Line Through the Gully, Solve It with the Stream's Line, Then Scale the Length

On a plan of a housing estate, 1 unit represents 10 m. A straight road follows the line y = 12x + 2, and a straight stream follows the line y = x − 7. A drain is to be laid from a gully at P(6, 5), at the side of the road, at right angles to the road, until it reaches the stream. (a) Find the equation of the line of the drain. (b) Find the point where the drain reaches the stream, and the length of the drain in meters.

246810122468Proadstreamgradient 1/2road: gradient 1/2; drain: gradient −2
The road has gradient 12, so the drain has gradient −2, because 12 × (−2) = −1.
The gully is on the road, since 12 × 6 + 2 = 5. The road's gradient is 12, so the drain's gradient m satisfies 12 × m = −1, and m = −2.
step 1 of 5

Two lines at right angles have gradients that multiply to −1. The drain's gradient comes from the road's gradient, its equation from that gradient and the gully, and the end of the drain from solving its equation together with the stream's. The length is measured on the plan with Pythagoras' theorem and then multiplied by the scale.

  1. The gully is on the road, since 12 × 6 + 2 = 5. The road's gradient is 12, so the drain's gradient m satisfies 12 × m = −1, and m = −2.
  2. (a) The drain passes through (6, 5) with gradient −2: y − 5 = −2(x − 6), so y = −2x + 17.
  3. The drain reaches the stream where both equations hold: −2x + 17 = x − 7, so 3x = 24 and x = 8. Then y = 8 − 7 = 1, and the drain ends at (8, 1).
  4. On the plan the drain goes 2 units across and 4 units down, so its length is √22 + 42 = √20 = 2√5 units.
  5. (b) The drain reaches the stream at (8, 1), and it is 2√5 × 10 = 20√5 ≈ 44.7 m long. Check: the drain falls 4 units for 2 across, a gradient of −2, and (8, 1) is on the stream because 8 − 7 = 1.

answer(a) y = −2x + 17; (b) at (8, 1), and the drain is 20√5 ≈ 44.7 m long

techniqueDrawing Parallel and Perpendicular Lines · Parallel and Perpendicular Lines

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Taking the drain's gradient as −12 or as 2, which makes only one of the two changes. The road's gradient must be turned upside down AND given the opposite sign, so that the two gradients multiply to −1: 12 × (−2) = −1.
  • Giving the length as √20 ≈ 4.47 m. That is the length in plan units; each unit represents 10 m, so the drain is ten times as long.
03

A Phone Mast the Same Distance from Two Villages, Placed as Close as Possible to a Town

methodBuild the Perpendicular Bisector of the Two Villages from Their Midpoint and the Perpendicular Gradient, Then Drop a Perpendicular to It from the Town

On a map marked in kilometers, two villages stand at A(1, 1) and B(7, 5), and a town stands at C(5, 8). A phone company will build one mast that is the same distance from both villages, and it wants the mast as close to the town as that allows. (a) Find the equation of the line on which the mast must stand. (b) Find the position of the mast, its distance from each village, and its distance from the town.

24682468ABC(4, 3)midpoint (4, 3); gradient of AB = 4/6 = 2/3
The midpoint of AB is (4, 3), and AB has gradient 23.
The midpoint of AB is (1 + 72, 1 + 52) = (4, 3), and the gradient of AB is 5 − 17 − 1 = 23.
step 1 of 5

Every point the same distance from A and B lies on the perpendicular bisector of AB: the line through the midpoint of AB at right angles to it. The point of that line nearest to C is the foot of the perpendicular from C, so a second perpendicular, this time drawn from the town, finds the mast.

  1. The midpoint of AB is (1 + 72, 1 + 52) = (4, 3), and the gradient of AB is 5 − 17 − 1 = 23.
  2. (a) The perpendicular bisector has gradient −32, since 23 × (−32) = −1, and it passes through (4, 3): y − 3 = −32(x − 4), so y = −32x + 9, which is 3x + 2y = 18.
  3. The nearest point of this line to C is where the perpendicular from C meets it. That perpendicular has gradient 23 and passes through (5, 8): y − 8 = 23(x − 5), so y = 23x + 143.
  4. Solve the two equations together: −32x + 9 = 23x + 143. Multiply every term by 6: −9x + 54 = 4x + 28, so 13x = 26 and x = 2. Then y = −3 + 9 = 6, and the mast is at (2, 6).
  5. (b) The mast at (2, 6) is √12 + 52 = √26 ≈ 5.10 km from A and √52 + 12 = √26 ≈ 5.10 km from B, equal as required, and √32 + 22 = √13 ≈ 3.61 km from the town.

answer(a) y = −32x + 9, that is 3x + 2y = 18; (b) at (2, 6), √26 ≈ 5.10 km from each village and √13 ≈ 3.61 km from the town

techniqueConstructing a Perpendicular Bisector · Constructing a Perpendicular from a Point

examsO-Level · GCSE Higher

Common pitfalls

  • Putting the mast at the midpoint (4, 3) because it is the same distance from both villages. Every point of the perpendicular bisector is; the midpoint is √12 + 52 = √26 ≈ 5.10 km from the town, farther than the foot of the perpendicular.
  • Going from C straight down to the bisector. The shortest distance from a point to a line is along the perpendicular; straight down from C reaches (5, 1.5), which is 6.5 km from the town.
04

The Shortest Pipe from a Water Trough to a Channel, Found with a Tape Measure

methodA Point the Same Distance from Two Points of a Line Lies on Their Perpendicular Bisector: Take the Midpoint for the Foot, Then Use Pythagoras for the Length

On a plan of a field marked in meters, a water trough stands at T(7, 9), and the straight edge of an irrigation channel runs below it. The farmer holds one end of a 10 m tape at the trough, pulls it tight and swings it round: the tape just reaches the channel's edge at A(1, 1) and at B(15, 3). (a) Find the point of the channel's edge nearest to the trough. (b) Find the length of the shortest pipe from the trough to the channel's edge.

481216246810TABchannel10 m10 mTA = TB = 10 m: T is on the bisector of AB
The tape of 10 m reaches the edge at A and B, so T is on the perpendicular bisector of AB.
TA = TB = 10 m, so T is on the perpendicular bisector of AB. That line crosses AB at right angles at its midpoint, so the foot of the perpendicular from T to the edge is the midpoint of AB.
step 1 of 4

This is the construction of a perpendicular from a point: an arc centered at the point cuts the line twice, and the perpendicular from the point bisects the piece of line between the two cuts. The trough is 10 m from A and from B, so it lies on the perpendicular bisector of AB, and that bisector meets the channel's edge at right angles, at the midpoint of AB.

  1. TA = TB = 10 m, so T is on the perpendicular bisector of AB. That line crosses AB at right angles at its midpoint, so the foot of the perpendicular from T to the edge is the midpoint of AB.
  2. (a) The midpoint of AB is M = (1 + 152, 1 + 32) = (8, 2), the point of the edge nearest the trough.
  3. Check the right angle: the edge has gradient 3 − 115 − 1 = 17, TM has gradient 2 − 98 − 7 = −7, and 17 × (−7) = −1.
  4. (b) TM = √12 + 72 = √50 = 5√2 ≈ 7.07 m. Check with Pythagoras' theorem in triangle TMA: AM = √72 + 12 = √50, and 50 + 50 = 100 = 102.

answer(a) (8, 2); (b) 5√2 ≈ 7.07 m

techniqueConstructing a Perpendicular from a Point · Constructing a Perpendicular Bisector

examsPSLE · O-Level · GCSE Higher

Common pitfalls

  • Laying the pipe straight down from the trough, along x = 7. The edge is not level, so straight down is not at right angles to it: that pipe meets the edge at (7, 137) and is about 7.14 m long, longer than 7.07 m.
  • Taking the shortest pipe as 10 m, the length of the tape. The tape reaches the edge only at A and B, the two ends of the piece it cuts off; every point of the edge between them is nearer, and the nearest is the midpoint.
05

A Goat Tied to the Corner of a Shed, and the Grass It Can Reach

methodSplit the Region the Rope Sweeps into Sectors, a New Smaller One at Each Corner the Rope Bends Round, and Add Their Areas

A goat is tied by a rope 8 m long to a post at one outside corner of a shed. The shed is a rectangle 6 m by 4 m standing in a large field of grass. The goat cannot go into the shed, and the rope bends round the shed's corners. (a) Find the area of grass the goat can reach, in terms of π. (b) The farmer moves the post to the middle of one 6 m wall, on the outside, and keeps the same rope. Find the new area in terms of π, and say which post gives the goat more grass.

post8 mshed6 m4 m270/360 × pi × 82= 48 pi
From the corner the rope sweeps 270° of a circle of radius 8 m: 48π m².
From the corner post the rope swings freely through 360° − 90° = 270°, because the shed fills a right angle. That sector has area 270360 × π × 82 = 48π m².
step 1 of 6

The goat can reach every point within 8 m of the post that the rope can reach without passing through the shed: a locus made of sectors of circles. Where the rope runs along a wall and bends round a corner, a new, smaller sector begins at that corner, with a radius equal to the length of rope that is left.

  1. From the corner post the rope swings freely through 360° − 90° = 270°, because the shed fills a right angle. That sector has area 270360 × π × 82 = 48π m².
  2. Along the 6 m wall the rope bends round the next corner with 8 − 6 = 2 m left, and it sweeps a quarter circle: 14 × π × 22 = π m².
  3. Along the 4 m wall it bends round the other corner with 8 − 4 = 4 m left: 14 × π × 42 = 4π m². These two quarter circles lie on different sides of the shed, so they do not overlap.
  4. (a) The goat can reach 48π + π + 4π = 53π ≈ 166.5 m² of grass.
  5. From the middle of a 6 m wall the rope sweeps a semicircle, 12 × π × 82 = 32π m². It reaches each end of that wall with 8 − 3 = 5 m left, which sweeps two quarter circles of 254π m² each, and then each far corner with 5 − 4 = 1 m left, which sweeps two quarter circles of 14π m² each.
  6. (b) The new area is 32π + 252π + 12π = 45π ≈ 141.4 m². The post at the corner gives the goat 53π − 45π = 8π ≈ 25.1 m² more grass.

answer(a) 53π ≈ 166.5 m²; (b) 45π ≈ 141.4 m², so the post at the corner gives the goat 8π ≈ 25.1 m² more

techniqueLoci

examsGCSE Higher

Common pitfalls

  • Counting a full circle, 64π m², or the 270° sector alone, 48π m². The shed blocks a right angle of the full circle, and the rope that bends round a corner reaches grass that the 270° sector misses.
  • Using the whole rope, 8 m, as the radius at the next corner. The part of the rope lying along the wall is used up: only 8 − 6 = 2 m is left to swing round the far end of the 6 m wall.
06

A Guard Dog on a Running Wire Along a Wall: the Ground It Covers, and What It Can Reach

methodSplit the Region into a Rectangle Beside the Wire and a Quarter Circle at Each End, and Measure Each Point to the Nearest Point of the Wire: Straight Across Beside It, to the End Beyond It

A guard dog's lead, 4 m long, is clipped to a ring that slides along a straight wire. The wire is fixed along the foot of a long straight wall, and the dog stays in the yard on one side of the wall. On a plan in meters, the wall is the x-axis, the yard is the region y ≥ 0, and the wire runs from (6, 0) to (16, 0). (a) Find the area of the yard the dog can reach, in terms of π. (b) The owner wants the dog kept away from a trash can at P(19, 2) and a bird table at Q(11, 5). Find how far each of P and Q is from the nearest point of the wire, and say which of them the dog can reach.

6111619PQ4 mwallwirebeside the wire: 10 × 4 = 40 m2
Beside the wire the dog reaches a rectangle 10 m by 4 m: 40 m².
Beside the wire, the points within 4 m of it form a rectangle 10 m long and 4 m deep: 10 × 4 = 40 m².
step 1 of 6

The dog can reach every point of the yard within 4 m of some point of the wire. Beside the wire that is a band 4 m deep; beyond each end it is part of a circle about that end. The distance from a point to the wire is measured to the NEAREST point of the wire: straight across when the point is beside the wire, and to the end when the point is beyond it.

  1. Beside the wire, the points within 4 m of it form a rectangle 10 m long and 4 m deep: 10 × 4 = 40 m².
  2. Beyond each end of the wire, the points within 4 m of that end, on the yard side of the wall, form a quarter circle of radius 4 m. Together the two quarter circles make 2 × 14 × π × 42 = 8π m².
  3. (a) The dog can reach 40 + 8π ≈ 65.1 m² of the yard.
  4. Q(11, 5) is beside the wire, since 11 is between 6 and 16, so its nearest point of the wire is straight below it, (11, 0), which is 5 m away.
  5. P(19, 2) is beyond the end of the wire at (16, 0), since 19 > 16, so its nearest point of the wire is that end: √32 + 22 = √13 ≈ 3.61 m away.
  6. (b) √13 ≈ 3.61 m is less than 4 m, so the dog can reach the trash can; 5 m is more than 4 m, so it cannot reach the bird table.

answer(a) 40 + 8π ≈ 65.1 m²; (b) the trash can is √13 ≈ 3.61 m from the wire, so the dog can reach it; the bird table is 5 m from the wire, so the dog cannot

techniqueLoci

examsGCSE Higher

Common pitfalls

  • Giving the distance from P as 2 m, its distance from the wall. The point (19, 0) straight below P is not on the wire, which stops at x = 16; the nearest point of the wire is its end.
  • Adding a semicircle at each end, 16π m² in all, as if the dog could go behind the wall. The wall cuts each end circle in half again, leaving a quarter circle on the yard side.
07

A Bicycle Rack Near the South Wall of a School Yard and Nearer the Main Gate

methodFind the Perpendicular Bisector of the Two Gates, Test One Point to See Which Side Is Nearer the Main Gate, Then Cut That Side with the 3 m Strip and Find Its Area

A school yard is a rectangle 20 m by 12 m. On a plan in meters its corners are (0, 0), (20, 0), (20, 12) and (0, 12), and its south wall runs from (0, 0) to (20, 0). The main gate is at A(4, 12) and the side gate is at B(20, 4). A bicycle rack must stand no more than 3 m from the south wall, and nearer the main gate than the side gate. (a) Find the equation of the line of points that are the same distance from both gates, and the point where it meets the south wall. (b) Find the area of the part of the yard where the rack may stand.

89.5122034812ABsouth wall(12, 8)midpoint (12, 8); gradient of AB = −8/16 = −1/2
The midpoint of the gates is (12, 8), and AB has gradient −12.
The midpoint of AB is (4 + 202, 12 + 42) = (12, 8), and the gradient of AB is 4 − 1220 − 4 = −12, so the perpendicular bisector has gradient 2.
step 1 of 5

The rack's region is where two loci overlap. The points no more than 3 m from the south wall form a strip 3 m wide along it. The points nearer A than B lie on A's side of the perpendicular bisector of AB. The region is the part of the strip on that side of the bisector.

  1. The midpoint of AB is (4 + 202, 12 + 42) = (12, 8), and the gradient of AB is 4 − 1220 − 4 = −12, so the perpendicular bisector has gradient 2.
  2. (a) The bisector is y − 8 = 2(x − 12), that is y = 2x − 16. It meets the south wall, y = 0, where 2x = 16: at (8, 0).
  3. To see which side is nearer A, test the corner (0, 0): its squared distance to A is 42 + 122 = 160 and to B is 202 + 42 = 416. So the part of the yard west of the bisector, where y > 2x − 16, is nearer the main gate.
  4. The strip runs from y = 0 to y = 3. At y = 3 the bisector has 3 = 2x − 16, so x = 9.5. The region is a trapezoid with corners (0, 0), (8, 0), (9.5, 3) and (0, 3): its parallel sides are 8 m and 9.5 m, and they are 3 m apart. Check: (9.5, 3) is the same distance from both gates, since 5.52 + 92 = 10.52 + 12 = 111.25.
  5. (b) The area is 12 × (8 + 9.5) × 3 = 26.25 m².

answer(a) y = 2x − 16, meeting the south wall at (8, 0); (b) 26.25 m²

techniqueLoci · Constructing a Perpendicular Bisector

examsGCSE Higher

Common pitfalls

  • Using the side of the bisector that contains B. Testing one point, such as a corner of the yard, settles which side is nearer the main gate; the part of the strip east of the bisector is where the rack must NOT stand.
  • Taking the region as a rectangle 8 m by 3 m, 24 m². The bisector is slanted, so the region is 9.5 m wide at 3 m from the wall, not 8 m.
08

Four Sprinklers at the Corners of a Lawn: the Dry Ground, and the Reach That Waters All of It

methodTake a Quarter Circle at Each Corner from the Rectangle; for the Whole Lawn, Find the Point Farthest from Its Nearest Corner

A rectangular lawn is 12 m long and 8 m wide. A sprinkler at each of its four corners waters every point of the lawn within 4 m of it. (a) Find the area of the lawn that no sprinkler waters, in terms of π. (b) The four sprinklers are replaced by stronger ones, all with the same reach. Find the least reach that waters the whole lawn, and the point of the lawn that is reached last.

4 m12 m8 m1/4 × pi × 42= 4 pi each
Each sprinkler waters a quarter circle of radius 4 m: 4π m².
The lawn's corners are right angles, so each sprinkler waters a quarter circle of radius 4 m on the lawn: 14 × π × 42 = 4π m².
step 1 of 5

Each sprinkler waters the locus of points within its reach, a circle, and the lawn holds the quarter of it at its corner. The dry area is what the quarters leave. For the whole lawn, every point must be within reach of its NEAREST sprinkler, so the reach needed is the greatest distance from any point of the lawn to its nearest corner.

  1. The lawn's corners are right angles, so each sprinkler waters a quarter circle of radius 4 m on the lawn: 14 × π × 42 = 4π m².
  2. Along the 8 m sides two sprinklers are 8 m apart and 4 + 4 = 8, so their quarter circles only touch; along the 12 m sides they are farther apart. No point is watered twice, and the four sprinklers water 4 × 4π = 16π m².
  3. (a) The dry area is 12 × 8 − 16π = 96 − 16π ≈ 45.7 m².
  4. Split the lawn into four rectangles 6 m by 4 m, one at each corner. A point in one of them is nearest that corner, and the point of the rectangle farthest from the corner is the rectangle's opposite corner, which is the center of the lawn.
  5. (b) The center is √62 + 42 = √52 = 2√13 ≈ 7.21 m from every sprinkler, so the least reach is 2√13 ≈ 7.21 m, and the center of the lawn is the last point reached.

answer(a) 96 − 16π ≈ 45.7 m²; (b) 2√13 ≈ 7.21 m, and the last point reached is the center of the lawn

techniqueLoci

examsGCSE Higher

Common pitfalls

  • Giving the length of the diagonal, √122 + 82 ≈ 14.4 m, as the least reach. That is what ONE sprinkler at a corner would need; with four, each point only has to be within reach of its nearest sprinkler.
  • Taking four whole circles, 4 × 16π = 64π ≈ 201 m², from the lawn, which leaves a negative area. Only a quarter of each circle lies on the lawn.
09

The Inside Edge of a Running Track, and the Extra Distance in the Next Lane Out

methodSplit the Locus into Two Straights Parallel to the Segment and Two Semicircles About Its Ends, Which Together Make One Circle

The inside edge of a school's running track is the locus of points exactly 30 m from a straight line segment 80 m long, drawn down the middle of the field. (a) Find the length of the inside edge, and the area of the field inside it, in terms of π. (b) In the next lane out a runner keeps to the path whose every point is 31.2 m from the same segment. How much farther does she run in one lap than a runner on the inside edge?

80 m30 m80 mstraights: 2 × 80 = 160 m
Beside the segment the edge is two straights of 80 m, 30 m on each side.
Beside the segment the edge is two straight lines parallel to it, 30 m on each side, each as long as the segment: 2 × 80 = 160 m.
step 1 of 5

Beside the segment, the points at a fixed distance from it lie on two lines parallel to it. Beyond each end, they lie on a semicircle about that end. So the locus is a stadium shape: two straights as long as the segment, and two semicircles that together make one circle.

  1. Beside the segment the edge is two straight lines parallel to it, 30 m on each side, each as long as the segment: 2 × 80 = 160 m.
  2. Beyond each end the edge is a semicircle of radius 30 m about that end. The two semicircles make one circle, whose circumference is 2 × π × 30 = 60π m.
  3. (a) The inside edge is 160 + 60π ≈ 348.5 m long. The field inside it is a rectangle 80 m by 60 m and a circle of radius 30 m: 80 × 60 + π × 302 = 4800 + 900π ≈ 7627 m².
  4. The outside path has the same two straights, and its two semicircles make a circle of radius 31.2 m, so its length is 160 + 2 × π × 31.2 = 160 + 62.4π m.
  5. (b) The difference is 62.4π − 60π = 2.4π ≈ 7.54 m in one lap, all of it on the bends. Check: 2 × π × 1.2 = 2.4π, the circumference of a circle whose radius is the 1.2 m between the two paths.

answer(a) the inside edge is 160 + 60π ≈ 348.5 m long and encloses 4800 + 900π ≈ 7627 m²; (b) 2.4π ≈ 7.54 m farther

techniqueLoci · Drawing Parallel and Perpendicular Lines

examsGCSE Higher

Common pitfalls

  • Adding 2 × 1.2 = 2.4 m, as if the lap grew only by the width between the paths on each side. The bends are circles, and a circle's circumference grows by 2π times the increase in its radius.
  • Making the straights longer as well. Every point of a straight is measured to the segment at right angles, so the outside straights are still 80 m long; only the semicircles grow.
10

Lamps Beside a Straight Footpath: the Stretch One Lamp Lights, and How Many Are Needed

methodDrop a Perpendicular from the Pole to the Path and Use Pythagoras for Half the Lit Stretch, Then Cover the Path with Stretches That Overlap

A straight footpath across a park is 40 m long. Lamps are to stand on poles in a line parallel to the path and 6 m from it, and each lamp lights all the ground within 10 m of the foot of its pole. (a) Find the length of the path that one lamp lights. (b) Find the fewest lamps that light the whole path. When that many lamps are spaced equally, and the first and last light the path exactly to its ends, how far apart are neighboring poles?

F6 mpolepathpole to path at right angles: 6 m
The perpendicular from the pole meets the path at F, 6 m away.
The perpendicular from a pole meets the path at a point F, 6 m from the pole. The lit stretch of the path is centered on F, and its two ends are exactly 10 m from the pole.
step 1 of 5

The ground a lamp lights is the locus of points within 10 m of its pole: a circle. The path crosses the circle in a chord, and the perpendicular from the pole to the path cuts that chord in half. Pythagoras' theorem in the right-angled triangle gives half the chord.

  1. The perpendicular from a pole meets the path at a point F, 6 m from the pole. The lit stretch of the path is centered on F, and its two ends are exactly 10 m from the pole.
  2. In the right-angled triangle with hypotenuse 10 m and one side 6 m, the third side is √102 − 62 = √64 = 8 m, so the lit stretch reaches 8 m on each side of F.
  3. (a) One lamp lights 2 × 8 = 16 m of the path.
  4. Two lamps light at most 2 × 16 = 32 m, less than 40 m, so at least 3 are needed. With 3, the first stands opposite the point 8 m from one end of the path, the last opposite the point 40 − 8 = 32 m from that end, and the middle one halfway between them.
  5. (b) 3 lamps are needed, and neighboring poles are 32 − 82 = 12 m apart. Check: the lamps light the path from 0 to 16 m, from 12 to 28 m and from 24 to 40 m, so every point of the path is lit.

answer(a) 16 m; (b) 3 lamps, with neighboring poles 12 m apart

techniqueLoci · Constructing a Perpendicular from a Point

examsGCSE Higher

Common pitfalls

  • Taking one lamp's stretch as 20 m, the diameter of its circle. The path is 6 m from the pole, so it crosses the circle in a chord shorter than the diameter: 16 m.
  • Dividing 40 by 16 and rounding down to 2 lamps. Two lamps leave 8 m of the path dark; the number of lamps must be rounded UP.
Mr. Chalk Read the guide