Pythagoras and similar shapes · applications

Applications: Pythagoras and Similar Shapes

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

O-Level · SAT · GCSE Higher

01

A Ladder Against a Wall That Slips Down

methodThe Ladder Is the Hypotenuse: Subtract the Known Square from the Square of the Hypotenuse, Once for Each Position

A ladder 2.5 m long leans against a vertical wall on level ground. Its foot is 0.7 m from the wall. (a) How high up the wall does the ladder reach? (b) The ladder slips until its top is 2.0 m above the ground. How much further from the wall is its foot now?

2.5 m0.7 mhthe ladder is the hypotenuse: 2.5 m
The wall and the ground meet at a right angle, so the ladder is the hypotenuse.
The wall is vertical and the ground is level, so the wall, the ground and the ladder make a right-angled triangle. The ladder is opposite the right angle, so it is the hypotenuse, 2.5 m. Let the height that the ladder reaches be h m.
step 1 of 5

A vertical wall and level ground meet at a right angle, so the wall, the ground and the ladder make a right-angled triangle. The ladder does not change length when it slips, so it is the hypotenuse of both triangles. A shorter side is found by subtracting the known square from the square of the hypotenuse.

  1. The wall is vertical and the ground is level, so the wall, the ground and the ladder make a right-angled triangle. The ladder is opposite the right angle, so it is the hypotenuse, 2.5 m. Let the height that the ladder reaches be h m.
  2. By Pythagoras' theorem, h2 + 0.72 = 2.52. To find a shorter side, subtract the known square from the square of the hypotenuse: h2 = 2.52 − 0.72 = 6.25 − 0.49 = 5.76.
  3. Take the square root: h = √5.76 = 2.4. (a) The ladder reaches 2.4 m up the wall. Check: 2.4 m is less than the length of the ladder, as a shorter side must be.
  4. After the slip the ladder is still 2.5 m long, and its top is 2.0 m high. Let the foot be d m from the wall. Then d2 = 2.52 − 2.02 = 6.25 − 4 = 2.25, so d = √2.25 = 1.5.
  5. (b) The foot has moved from 0.7 m to 1.5 m from the wall, so it is 1.5 − 0.7 = 0.8 m further away. Check: 1.52 + 2.02 = 2.25 + 4 = 6.25 = 2.52.

answer(a) 2.4 m; (b) 0.8 m further

techniqueFinding a Shorter Side with Pythagoras’ Theorem · Pythagoras’ Theorem

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the squares, 2.52 + 0.72, to find the height. The squares are added only to find the hypotenuse. The ladder is the hypotenuse here, so the height is a shorter side and the known square is subtracted.
  • Saying that the foot moves out by 0.4 m because the top moves down by 0.4 m. The two distances are linked through their squares, not directly, so each position has to be worked out with Pythagoras' theorem.
02

A Footpath Along the Diagonal of a Rectangular Field

methodA Corner of a Rectangle Is a Right Angle, So the Diagonal Is a Hypotenuse: Add the Squares and Take the Square Root

A rectangular field is 120 m long and 50 m wide. A footpath runs in a straight line from one corner to the opposite corner. (a) How long is the footpath? (b) Mei walks from one corner to the opposite corner along the footpath. Ravi walks between the same two corners along two sides of the field. How much shorter is Mei's walk?

120 m50 mcthe footpath c is the hypotenuse
A corner of a rectangle is a right angle, so the footpath is a hypotenuse.
Every corner of a rectangle is a right angle, so the length, the width and the footpath make a right-angled triangle. The footpath is opposite the right angle, so it is the hypotenuse. Let its length be c m.
step 1 of 5

A diagonal cuts a rectangle into two right-angled triangles. The length and the width of the field are the two shorter sides of one of these triangles, and the footpath is its hypotenuse, so Pythagoras' theorem gives the length of the footpath.

  1. Every corner of a rectangle is a right angle, so the length, the width and the footpath make a right-angled triangle. The footpath is opposite the right angle, so it is the hypotenuse. Let its length be c m.
  2. By Pythagoras' theorem, c2 = 1202 + 502 = 14400 + 2500 = 16900.
  3. Take the square root: c = √16900 = 130. (a) The footpath is 130 m long. Check: 130 m is longer than either side of the field, as a hypotenuse must be, and shorter than the two sides added together.
  4. Ravi walks one length and one width of the field: 120 + 50 = 170 m.
  5. (b) Mei's walk is 170 − 130 = 40 m shorter.

answer(a) 130 m; (b) 40 m shorter

techniquePythagoras’ Theorem

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the sides and then squaring, (120 + 50)2, or taking √120 + 50. Pythagoras' theorem adds the squares of the sides, 1202 + 502, and the square root is taken only after the two squares are added.
  • Stopping at c2 = 16900 and giving 16900 m as the length. That number is the square of the length, so its square root still has to be taken.
03

Two Square Plots Replaced by One Square Plot of the Same Total Area

methodThe Squares on the Two Shorter Sides Together Equal the Square on the Hypotenuse, So the New Side Is a Hypotenuse

A gardener has two square vegetable plots, one with sides of 6 m and one with sides of 8 m. She wants to replace them with a single square plot that has the same total area. (a) Find the length of a side of the new plot. (b) Each plot has a fence all the way round it. How much less fencing does the single plot need than the two plots together?

36 m264 m26 m8 mnew area: 36 + 64 = 100 m2
The new plot must have an area of 36 + 64 = 100 m2.
The areas of the two plots are 62 = 36 m2 and 82 = 64 m2, so the new plot must have an area of 36 + 64 = 100 m2.
step 1 of 5

Pythagoras' theorem is a statement about areas: the squares drawn on the two shorter sides of a right-angled triangle have, together, the same area as the square drawn on the hypotenuse. So the two plots can be placed on the shorter sides of a right-angled triangle, and the new plot is the square on its hypotenuse.

  1. The areas of the two plots are 62 = 36 m2 and 82 = 64 m2, so the new plot must have an area of 36 + 64 = 100 m2.
  2. Draw a right-angled triangle whose two shorter sides are 6 m and 8 m. The two plots are the squares on these sides. By Pythagoras' theorem the two squares together have the same area as the square on the hypotenuse, so the new plot is the square on the hypotenuse.
  3. Let the hypotenuse be c m. Then c2 = 62 + 82 = 100, so c = √100 = 10. (a) Each side of the new plot is 10 m long. Check: 10 × 10 = 100 m2.
  4. A square has four equal sides. The two old plots need 4 × 6 + 4 × 8 = 24 + 32 = 56 m of fencing, and the new plot needs 4 × 10 = 40 m.
  5. (b) The single plot needs 56 − 40 = 16 m less fencing.

answer(a) 10 m; (b) 16 m less

techniqueProving Pythagoras by Dissection · Pythagoras’ Theorem

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the sides to get a new side of 6 + 8 = 14 m. A square with sides of 14 m has an area of 196 m2, which is almost twice the 100 m2 that is needed. The areas are added, not the sides.
  • Expecting the same total area to need the same length of fencing. The area stays at 100 m2, but the perimeter falls from 56 m to 40 m, because one large square has less edge for its area than two smaller squares.
04

Checking Whether the Corner of a Floor Is a Right Angle

methodThe Converse of Pythagoras' Theorem: Compare the Square of the Longest Side with the Sum of the Squares of the Other Two

A builder checks whether the corners of a new floor are right angles. From a corner she marks a point 2.4 m along one wall and a point 3.2 m along the other wall, and then she measures the straight distance between the two marks. (a) At the first corner the distance is 4.0 m. Is this corner a right angle? (b) At the second corner the marks are made in the same way, but the distance is 4.1 m. Is this corner a right angle? If it is not, is the angle larger or smaller than 90°?

3.2 m2.4 m4.0 mfirst corner3.2 m2.4 m4.1 msecond cornera right angle needs c2= a2+ b2
The corner is a right angle if the square of the longest side is equal to the sum of the squares of the other two sides.
The corner and the two marks make a triangle with sides 2.4 m, 3.2 m and the measured distance, and the corner is opposite the measured distance. By the converse of Pythagoras' theorem, the corner is a right angle if the square of the longest side is equal to the sum of the squares of the other two sides.
step 1 of 5

Pythagoras' theorem starts from a right angle and gives a fact about the sides. Its converse goes the other way: if the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite the longest side is a right angle. The builder knows three lengths, so the converse is the test to use.

  1. The corner and the two marks make a triangle with sides 2.4 m, 3.2 m and the measured distance, and the corner is opposite the measured distance. By the converse of Pythagoras' theorem, the corner is a right angle if the square of the longest side is equal to the sum of the squares of the other two sides.
  2. Add the squares of the two shorter sides: 2.42 + 3.22 = 5.76 + 10.24 = 16.
  3. At the first corner the longest side is 4.0 m, and 4.02 = 16. The two results are equal. (a) Yes, the first corner is a right angle, because 2.42 + 3.22 = 4.02.
  4. At the second corner the longest side is 4.1 m, and 4.12 = 16.81. This is not equal to 16, so the second corner is not a right angle.
  5. (b) The distance between the marks is longer than the 4.0 m that a right angle gives. The marks are further apart because the two walls open wider, so the angle at the second corner is larger than 90°.

answer(a) Yes: 2.42 + 3.22 = 16 = 4.02; (b) No: 4.12 = 16.81, which is more than 16, so the angle is larger than 90°

techniqueThe Converse of Pythagoras’ Theorem

examsO-Level · SAT

Common pitfalls

  • Comparing the lengths instead of their squares: 2.4 + 3.2 = 5.6 is not 4.0, so the corner is said not to be a right angle. The test uses the squares of the sides, and 5.76 + 10.24 = 16 = 4.02.
  • Deciding that 4.1 m is close enough to 4.0 m. The converse needs the two results to be equal. A difference of 0.1 m over 4 m means that the angle is about 93°, and tiles laid from that corner would drift away from the wall.
05

The Shortest Path for an Ant Over the Surface of a Box

methodUnfold Two Faces into One Flat Rectangle, So That the Path Is a Straight Diagonal, and Use Pythagoras' Theorem

A closed box is 8 cm long, 3 cm wide and 3 cm high. An ant at a bottom corner A walks over the surface of the box to the opposite top corner B. (a) The ant walks in a straight line up the front face and then across the top face. How long is this path? (b) A second ant walks from A across the front face and then across the end face to B. How long is that path, to 1 decimal place, and which of the two paths is shorter?

fronttopAB8 cm3 + 3 = 6 cmpfront and top, unfolded: 8 cm by 6 cm
Unfold the top face to lie flat with the front face: one rectangle, 8 cm by 6 cm, with A and B at opposite corners.
Unfold the box: lift the top face until it lies flat with the front face. The two faces make one rectangle that is 8 cm long and 3 + 3 = 6 cm high, with A at one corner and B at the opposite corner.
step 1 of 5

On a flat surface the shortest path between two points is a straight line. A path over two faces of a box bends at the edge, but when the two faces are unfolded to lie flat, the shortest path becomes one straight line across a rectangle. That line is the hypotenuse of a right-angled triangle.

  1. Unfold the box: lift the top face until it lies flat with the front face. The two faces make one rectangle that is 8 cm long and 3 + 3 = 6 cm high, with A at one corner and B at the opposite corner.
  2. The path is the diagonal of this rectangle, which is the hypotenuse of a right-angled triangle with shorter sides of 8 cm and 6 cm. Let the path be p cm long. By Pythagoras' theorem, p2 = 82 + 62 = 64 + 36 = 100.
  3. (a) The path over the top is √100 = 10 cm long.
  4. For the second ant, unfold the end face until it lies flat beside the front face. The two faces make a rectangle that is 8 + 3 = 11 cm long and 3 cm high. Let this path be q cm long: q2 = 112 + 32 = 121 + 9 = 130.
  5. (b) The second path is √130 = 11.4 cm long, to 1 decimal place, so the path over the top, 10 cm, is the shorter one. Check: both paths are shorter than walking along three edges, which is 8 + 3 + 3 = 14 cm.

answer(a) 10 cm; (b) 11.4 cm, so the path over the top is shorter

techniquePythagoras’ Theorem

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Using Pythagoras' theorem on each face separately and adding the two diagonals: √82 + 32 + 3 ≈ 11.5 cm, for example. A path that bends at a corner of a face is longer than it needs to be. The faces are unfolded first, so that the whole path is one straight line.
  • Assuming that every way of unfolding gives the same length. The two rectangles here are 8 cm by 6 cm and 11 cm by 3 cm, and their diagonals are 10 cm and about 11.4 cm, so each unfolding has to be worked out.
06

Two Ships That Leave a Port on Perpendicular Courses

methodNorth and East Meet at a Right Angle: Find Each Distance from Its Speed, Then the Distance Apart Is a Hypotenuse

Two ships leave a port at the same time. One sails due north at 24 km/h and the other sails due east at 18 km/h. (a) How far apart are the two ships after 2 hours? (b) The ships' radios work over a distance of up to 90 km. For how many hours after they leave the port can the two ships talk to each other by radio?

northeastportthe right angle is at the port
North and east meet at a right angle at the port, so the distance between the ships is a hypotenuse.
North and east are at right angles, so the port and the two ships are the corners of a right-angled triangle. The right angle is at the port, and the distance between the ships is the hypotenuse.
step 1 of 5

North and east are at right angles, so at every moment the port and the two ships are the corners of a right-angled triangle. Each shorter side is a speed multiplied by the time, and the distance between the ships is the hypotenuse.

  1. North and east are at right angles, so the port and the two ships are the corners of a right-angled triangle. The right angle is at the port, and the distance between the ships is the hypotenuse.
  2. Distance is speed multiplied by time. After 2 hours the first ship is 24 × 2 = 48 km north of the port and the second ship is 18 × 2 = 36 km east of it.
  3. Let the distance between the ships be d km. By Pythagoras' theorem, d2 = 482 + 362 = 2304 + 1296 = 3600, so d = √3600 = 60. (a) The ships are 60 km apart.
  4. After n hours the ships are 24n km and 18n km from the port, so d2 = (24n)2 + (18n)2 = 576n2 + 324n2 = 900n2, and d = 30n. The distance between the ships grows by 30 km every hour.
  5. (b) Set 30n = 90 and divide both sides by 30: n = 3. The ships can talk by radio for 3 hours. Check: after 3 hours the ships are 72 km and 54 km from the port, and 722 + 542 = 5184 + 2916 = 8100 = 902.

answer(a) 60 km; (b) 3 hours

techniquePythagoras’ Theorem

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the two distances, 48 + 36 = 84 km. That is the distance from one ship to the other by way of the port. The straight distance between the ships is the hypotenuse of the triangle, which is 60 km.
  • Using the speeds as if the ships moved apart at 24 + 18 = 42 km/h. The ships sail at right angles to each other, not in opposite directions, so the distance between them grows by √242 + 182 = 30 km every hour.
07

The Height of a Tree from the Length of Its Shadow

methodParallel Rays of the Sun Make Similar Right-Angled Triangles: Find the Scale Factor from the Two Shadows

At the same moment on a sunny day, a vertical pole 1.5 m tall casts a shadow 2 m long on level ground, and a tree casts a shadow 16 m long. (a) How tall is the tree? (b) A bird flies in a straight line from the top of the tree to the tip of the tree's shadow. How far does the bird fly?

16 m2 m1.5 m?equal angles, so the two triangles are similar
The rays of the sun are parallel and both objects are vertical, so the two triangles have equal angles and are similar.
The rays of the sun are parallel, so they meet the ground at the same angle at the pole and at the tree. The pole and the tree are vertical, so each makes a right angle with the ground. The two triangles have equal angles, so they are similar.
step 1 of 5

The rays of the sun are parallel, so at one moment they meet level ground at the same angle everywhere. A vertical object, its shadow and the ray past its top make a right-angled triangle, and the triangles for the pole and for the tree have equal angles, so they are similar. Corresponding sides of similar triangles are in the same ratio.

  1. The rays of the sun are parallel, so they meet the ground at the same angle at the pole and at the tree. The pole and the tree are vertical, so each makes a right angle with the ground. The two triangles have equal angles, so they are similar.
  2. In similar triangles, corresponding sides are in the same ratio. The two shadows correspond, so the scale factor from the pole's triangle to the tree's triangle is 16 ÷ 2 = 8.
  3. The two heights correspond as well. (a) The tree is 1.5 × 8 = 12 m tall.
  4. The bird flies along the hypotenuse of the tree's triangle. Let this distance be c m. By Pythagoras' theorem, c2 = 122 + 162 = 144 + 256 = 400, so c = √400 = 20.
  5. (b) The bird flies 20 m. Check with the pole's triangle: its hypotenuse is √1.52 + 22 = √6.25 = 2.5 m, and 2.5 × 8 = 20 m.

answer(a) 12 m; (b) 20 m

techniqueSimilar Shapes · Pythagoras’ Theorem

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the same amount instead of multiplying: the tree's shadow is 14 m longer than the pole's shadow, so the tree is said to be 1.5 + 14 = 15.5 m tall. Similar shapes keep the ratio of their sides, not the difference between them.
  • Matching sides that do not correspond, such as 1.516 = 2h. A height must be matched with a height and a shadow with a shadow: h1.5 = 162.
08

A Logo Enlarged for a Banner: the New Height and the Ink It Uses

methodLengths Are Multiplied by the Scale Factor k, and Areas Are Multiplied by k Squared

A rectangular school logo is 12 cm wide and 8 cm high. It is enlarged to make a banner that is 36 cm wide, and the banner is similar to the logo. (a) How high is the banner? (b) Printing the logo uses 20 ml of ink, and the amount of ink is proportional to the area that is printed. How much ink does the banner use?

logo12 cm8 cm36 cm?k = 36 divided by 12 = 3
The widths correspond, so the scale factor is k = 36 ÷ 12 = 3.
The banner is similar to the logo, so every length is multiplied by the same scale factor. The two widths correspond, so the scale factor is k = 36 ÷ 12 = 3.
step 1 of 4

When a shape is enlarged to a similar shape, every length is multiplied by the same scale factor k. An area is a length multiplied by a length, so it is multiplied by k twice, which is k2.

  1. The banner is similar to the logo, so every length is multiplied by the same scale factor. The two widths correspond, so the scale factor is k = 36 ÷ 12 = 3.
  2. The height is multiplied by 3 as well. (a) The banner is 8 × 3 = 24 cm high.
  3. An area is a length multiplied by a length. Both lengths are multiplied by 3, so the area is multiplied by 3 × 3 = 32 = 9. The banner holds 9 copies of the logo, in 3 rows of 3.
  4. (b) The ink is proportional to the area, so the banner uses 20 × 9 = 180 ml of ink. Check with the areas: the logo is 12 × 8 = 96 cm2 and the banner is 36 × 24 = 864 cm2, and 864 ÷ 96 = 9.

answer(a) 24 cm; (b) 180 ml

techniqueSimilar Shapes

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Multiplying the ink by the scale factor for lengths, 20 × 3 = 60 ml. The ink covers an area, and the area is multiplied by 32 = 9, not by 3.
  • Adding 24 cm to the height because the width grew by 24 cm, which gives a banner 36 cm by 32 cm. That rectangle is not similar to the logo, because 36 ÷ 12 = 3 but 32 ÷ 8 = 4. An enlargement multiplies every length by the same number.
09

A Horizontal Beam Across the Triangular End Wall of a Roof

methodA Line Parallel to One Side Cuts Off a Similar Triangle: Compare the Small Triangle with the Whole Triangle

The end wall of a roof is a triangle ABC. The base BC is 6 m long, the top A is 4 m above BC, and the sloping edges AB and AC are each 5 m long. A horizontal beam DE joins the two sloping edges, with D on AB and E on AC, so that DE is parallel to BC. The point D is 3 m from A. (a) How long is the beam DE? (b) How high is the beam above the base BC?

ABCDE6 m3 mAC = AB = 5 m4 m?equal angles: triangle ADE is similar to triangle ABC
DE is parallel to BC, so the corresponding angles are equal and triangle ADE is similar to triangle ABC.
DE is parallel to BC, so angle ADE is equal to angle ABC, and angle AED is equal to angle ACB, because they are corresponding angles. The angle at A belongs to both triangles. Triangle ADE and triangle ABC have equal angles, so they are similar.
step 1 of 5

A line drawn parallel to one side of a triangle cuts off a smaller triangle at the opposite corner. The small triangle and the whole triangle have equal angles, so they are similar, and every length of the small triangle is the same fraction of the matching length of the whole triangle.

  1. DE is parallel to BC, so angle ADE is equal to angle ABC, and angle AED is equal to angle ACB, because they are corresponding angles. The angle at A belongs to both triangles. Triangle ADE and triangle ABC have equal angles, so they are similar.
  2. AD corresponds to AB, so the scale factor from triangle ABC to triangle ADE is ADAB = 35.
  3. DE corresponds to BC. (a) The beam is DE = 35 × 6 = 3.6 m long.
  4. A height is a length, so the heights are in the same ratio. The height of triangle ADE, measured down from A to the beam, is 35 × 4 = 2.4 m.
  5. (b) The beam is 4 − 2.4 = 1.6 m above BC. Check: DEBC = 3.66 = 35 and 2.44 = 35.

answer(a) 3.6 m; (b) 1.6 m

techniqueFinding Similar Triangles in a Figure · Similar Shapes

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Using the ratio of the two parts of AB, ADDB = 32, as the scale factor. The similar triangles are ADE and ABC, so the sides that correspond are AD and the whole of AB, and the ratio is 35.
  • Giving 2.4 m as the answer to part (b). That is the height of the small triangle, measured down from A. The question asks for the height of the beam above BC, which is the rest of the 4 m.
10

The Width of a River Found by Sighting a Tree on the Far Bank

methodTwo Right Angles and a Pair of Vertically Opposite Angles Give Similar Triangles: Scale the Small Triangle on Land up to the Triangle Across the River

Aini wants to find the width of a river without crossing it. A tree T stands at the edge of the far bank. She stands at A on the near bank, directly opposite T, walks 20 m along the bank to a stake at C, and then walks 5 m further along the bank to D. At D she turns through a right angle and walks away from the river until, at E, the stake C is exactly in line with the tree T. She measures DE as 12 m. (a) How wide is the river? (b) How far is E from the tree T in a straight line?

riverACDET20 m5 m12 m?two equal angles: triangle TAC is similar to triangle EDC
There are right angles at A and at D, and the angles at C are vertically opposite, so triangle TAC is similar to triangle EDC.
TA crosses the river at right angles to the bank, and DE leaves the bank at right angles, so angle TAC and angle EDC are both 90°. The lines TE and AD cross at C, so angle ACT and angle DCE are equal, because they are vertically opposite angles. Triangle TAC and triangle EDC have equal angles, so they are similar.
step 1 of 5

The width of the river is a side of a triangle that cannot be measured, but Aini has made a small triangle on land with the same angles. Once the two triangles are shown to be similar, the scale factor comes from the two sides along the bank, and it turns every measured length of the small triangle into a length of the large triangle.

  1. TA crosses the river at right angles to the bank, and DE leaves the bank at right angles, so angle TAC and angle EDC are both 90°. The lines TE and AD cross at C, so angle ACT and angle DCE are equal, because they are vertically opposite angles. Triangle TAC and triangle EDC have equal angles, so they are similar.
  2. AC corresponds to DC, so the scale factor from triangle EDC to triangle TAC is 20 ÷ 5 = 4.
  3. TA corresponds to ED. (a) The river is 12 × 4 = 48 m wide.
  4. In the small triangle, CE is the hypotenuse: CE2 = 52 + 122 = 25 + 144 = 169, so CE = √169 = 13 m. CT corresponds to CE, so CT = 13 × 4 = 52 m.
  5. (b) E, C and T are in one straight line, so ET = 13 + 52 = 65 m. Check: 202 + 482 = 400 + 2304 = 2704 = 522.

answer(a) 48 m; (b) 65 m

techniqueFinding Similar Triangles in a Figure · Similar Shapes · Pythagoras’ Theorem

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Matching TA with DC because both are drawn the same way on the page. Corresponding sides are opposite equal angles: TA and ED are each opposite the angle at C, and AC and DC each join the right angle to C.
  • Using the whole walk, AD = 25 m, as a side of the large triangle. The large triangle is TAC, and its side along the bank is AC = 20 m. The last 5 m belongs to the small triangle.
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