Congruence, Similarity and Circle Theorems: 12 Applications, Worked Step by Step

Two Triangular Sails Cut to the Same Pattern

methodTwo Sides and the Angle Between Them Match, So the Sails Are Congruent by SAS and the Unmeasured Edge Is Read from the Other Sail

A sailmaker has cut two triangular sails. On sail ABC, the edge AB is 8 m long, the edge BC is 5 m long and the angle between them, angle ABC, is 60^°. The third edge AC has been measured as 7 m. On sail PQR, the edge PQ is 8 m long, the edge QR is 5 m long and angle PQR is 60^°, but the edge PR has not been measured. (a) How long is PR? (b) A tape is to be sewn along all three edges of sail PQR. How many meters of tape are needed?

The two sails share two sides and the angle between them: 8 m, 5 m and 60^°.

Compare the two sails. AB and PQ are both 8 m, BC and QR are both 5 m, and the angles between these pairs of sides, angle ABC and angle PQR, are both 60^°.

step 1 of 4

Two triangles are congruent when they are the same shape and the same size. The SAS test says that two sides and the angle between them are enough to fix a triangle, so if those three match in the two sails, every other length matches as well.

Compare the two sails. AB and PQ are both 8 m, BC and QR are both 5 m, and the angles between these pairs of sides, angle ABC and angle PQR, are both 60^°.
Two sides and the angle between them are equal in the two triangles, so triangle ABC is congruent to triangle PQR by the SAS test. The corners match in the order A to P, B to Q and C to R.
In congruent triangles every matching side is equal. AC is opposite the 60^° angle at B, and PR is opposite the 60^° angle at Q, so PR matches AC. (a) PR = AC = 7 m.
The tape runs along PQ, QR and PR. (b) The tape needed is 8 + 5 + 7 = 20 m. Check: the cosine rule on sail ABC gives AC² = 8² + 5² − 2 × 8 × 5 × cos 60^° = 64 + 25 − 40 = 49, so AC = 7 m, as measured.

answer(a) 7 m; (b) 20 m

techniqueCongruence Tests

examsO-Level · SAT · GCSE Higher

Common pitfalls

Saying the sails cannot be compared because only two sides of sail PQR are known. Two sides and the angle between them fix a triangle completely, so the third side is already decided.
Matching PR with AB because both are edges that meet at the 60^° angle. PR is the edge opposite the 60^° angle, so it matches AC, the edge opposite the 60^° angle in the first sail.

The Crossed Legs of an Ironing Board

methodParallel Board and Floor Give Alternate Angles, and the Legs Cross in Vertically Opposite Angles, So the Two Triangles Are Similar by AA

The two legs of an ironing board cross at a pivot X. The board AB rests on the tops of the legs and is parallel to the floor, and the feet C and D stand on the floor. The leg from A runs down through X to the foot D, and the leg from B runs down through X to the foot C. XA = XB = 39 cm, XC = XD = 65 cm, and the feet C and D are 50 cm apart. (a) How far apart are the tops of the legs, A and B? (b) The pivot X is 60 cm above the floor. How high is the board above the floor?

The board is parallel to the floor, so the alternate angles at A and D are equal, and the angles at X are vertically opposite: triangle XAB is similar to triangle XDC.

AB is parallel to DC, and the leg AD crosses both, so angle XAB is equal to angle XDC because they are alternate angles. Angle AXB is equal to angle DXC because they are vertically opposite angles. Two angles of triangle XAB are equal to two angles of triangle XDC, so the triangles are similar by the AA test.

step 1 of 5

Two triangles are similar when two angles of one are equal to two angles of the other. The board and the floor are parallel, and the legs cross, so the small triangle above the pivot and the large triangle below it have the same angles and their sides are in one ratio.

AB is parallel to DC, and the leg AD crosses both, so angle XAB is equal to angle XDC because they are alternate angles. Angle AXB is equal to angle DXC because they are vertically opposite angles. Two angles of triangle XAB are equal to two angles of triangle XDC, so the triangles are similar by the AA test.
XA corresponds to XD, so the scale factor from the large triangle to the small one is XAXD = 3965 = 35.
AB corresponds to DC. (a) The tops of the legs are AB = 35 × 50 = 30 cm apart.
The height of each triangle is measured from X. The height of the large triangle is the 60 cm from X down to the floor, so the height of the small triangle, from X up to the board, is 35 × 60 = 36 cm.
(b) The board is 60 + 36 = 96 cm above the floor. Check: 3050 = 3660 = 35, the same ratio as 3965.

answer(a) 30 cm; (b) 96 cm

techniqueSimilarity Conditions

examsO-Level · SAT · GCSE Higher

Common pitfalls

Giving 36 cm as the height of the board. That is the height of the small triangle, measured up from the pivot. The board is 36 cm above X, and X is already 60 cm above the floor.
Using 3965 − 39 = 3926 as the scale factor. The sides that correspond are XA and the whole of XD, from the pivot to the foot, not the difference between them.

The Braces of a Garden Swing Frame

methodA Brace Joining the Midpoints of Two Legs Is Parallel to the Third Side and Half Its Length, by the Midpoint Theorem

The end frame of a garden swing is a triangle ABC. The legs AB and AC are each 3 m long and the feet B and C stand 3.6 m apart on level ground. A brace MN joins M, the midpoint of AB, to N, the midpoint of AC. (a) How long is the brace MN, and why is it level? (b) A second brace joins N to P, the midpoint of BC. How long is NP?

M and N are the midpoints of AB and AC, so by the midpoint theorem MN is parallel to BC and half of it.

M is the midpoint of AB and N is the midpoint of AC. By the midpoint theorem, MN is parallel to the third side BC and MN = 12 BC.

step 1 of 4

The midpoint theorem says that the line joining the midpoints of two sides of a triangle is parallel to the third side and half as long as it. Each brace joins two midpoints, so its length is half of the side it does not touch.

M is the midpoint of AB and N is the midpoint of AC. By the midpoint theorem, MN is parallel to the third side BC and MN = 12 BC.
(a) MN = 12 × 3.6 = 1.8 m. The brace is level because it is parallel to BC, and BC lies along the level ground.
For the second brace, N is the midpoint of CA and P is the midpoint of CB, so the third side of the triangle is AB.
(b) NP = 12 AB = 12 × 3 = 1.5 m, and NP is parallel to the leg AB. Check: triangle AMN has sides 1.5 m, 1.5 m and 1.8 m, each exactly half of 3 m, 3 m and 3.6 m.

answer(a) 1.8 m; (b) 1.5 m

techniqueThe Midpoint Theorem

examsO-Level

Common pitfalls

Halving the wrong side: taking MN to be half of AB because M lies on AB. The brace MN is parallel to and half of the side it does not touch, which is BC.
Measuring NP as half of BC because P is the midpoint of BC. The third side for the brace NP is AB, the side that neither N nor P lies on.

Two Similar Juice Bottles: the Label and the Juice Inside

methodLengths Are in the Ratio k, Areas in the Ratio k Squared and Volumes in the Ratio k Cubed

A drinks company sells its juice in two bottles that are similar in shape. The small bottle is 12 cm tall and the large bottle is 18 cm tall. The small bottle holds 240 ml of juice, and its label has an area of 36 cm². (a) What is the area of the label on the large bottle? (b) How much juice does the large bottle hold?

The heights correspond, so the scale factor is k = 1812 = 32.

The heights correspond, so the scale factor from the small bottle to the large one is k = 1812 = 32.

step 1 of 5

When two solids are similar with scale factor k, every length is multiplied by k, every area by k² and every volume by k³. The heights give k, and the label is an area while the juice is a volume.

The heights correspond, so the scale factor from the small bottle to the large one is k = 1812 = 32.
The label is an area, so it is multiplied by k² = (32)² = 94.
(a) The label on the large bottle has an area of 36 × 94 = 81 cm².
The juice fills a volume, so it is multiplied by k³ = (32)³ = 278.
(b) The large bottle holds 240 × 278 = 30 × 27 = 810 ml. Check: 810240 = 3.375 = 278 and 8136 = 2.25 = 94.

answer(a) 81 cm²; (b) 810 ml

techniqueRatio of Volumes · Ratio of Areas

examsO-Level · SAT · GCSE Higher

Common pitfalls

Multiplying the volume by 32 to get 360 ml. A bottle one and a half times as tall is also one and a half times as wide and one and a half times as deep, so its volume is multiplied three times by 32.
Multiplying the label area by 278 as well. A label is a flat area, so it grows by k², not k³.

The Width of the Water in a Round Drain Pipe

methodThe Water Surface Is a Chord, and the Perpendicular from the Center Bisects It, So Half the Width Comes from Pythagoras

A drain pipe has a circular cross-section of radius 25 cm and center O. Water lies in the pipe to a depth of 10 cm at the deepest point. (a) How wide is the surface of the water? (b) After rain the water is 18 cm deep. How wide is the surface now?

The water surface AB is a chord. The perpendicular from the center O meets it at M and bisects it.

Call the ends of the water surface A and B. The surface AB is a chord. Draw the perpendicular from O to AB, meeting it at M. The perpendicular from the center bisects the chord, so AM = MB.

step 1 of 5

The surface of the water is a straight line across the circle, which is a chord. The line from the center perpendicular to a chord cuts the chord in half, and it makes a right-angled triangle with a radius as its hypotenuse.

Call the ends of the water surface A and B. The surface AB is a chord. Draw the perpendicular from O to AB, meeting it at M. The perpendicular from the center bisects the chord, so AM = MB.
The lowest point of the pipe is 25 cm below O, and the water surface is 10 cm above the lowest point, so OM = 25 − 10 = 15 cm.
Triangle OMA is right-angled at M with hypotenuse OA = 25 cm. By Pythagoras' theorem, AM² = 25² − 15² = 625 − 225 = 400, so AM = 20 cm.
(a) The surface is AB = 2 × 20 = 40 cm wide.
When the depth is 18 cm, OM = 25 − 18 = 7 cm, so AM² = 625 − 49 = 576 and AM = 24 cm. (b) The surface is now 2 × 24 = 48 cm wide. Check: 7² + 24² = 49 + 576 = 625 = 25².

answer(a) 40 cm; (b) 48 cm

techniqueChords and the Center

examsO-Level · SAT · GCSE Higher

Common pitfalls

Using the depth 10 cm as the distance OM from the center to the surface. The depth is measured from the bottom of the pipe, and the center is a radius above the bottom, so OM = 25 − 10 = 15 cm.
Giving 20 cm as the width. Pythagoras' theorem gives half the chord, from M to one end, so the full width is twice that.

The Concrete Below the Road in a Round Tunnel

methodThe Filled Part Is a Segment: the Sector Minus the Triangle on the Same Two Radii

A road tunnel has a circular cross-section of radius 4 m and center O. The flat road surface runs from A to B across the circle, and angle AOB is 90^°. The part of the circle below the road is filled with concrete. Take π = 3.14. (a) Find the area of the concrete in the cross-section. (b) Find the area of the cross-section that is left open above the road.

The concrete fills the segment below the chord AB: the sector AOB with the triangle AOB taken away.

The concrete is the minor segment cut off by the chord AB. Its area is the area of sector AOB minus the area of triangle AOB.

step 1 of 5

The road AB is a chord, and the concrete fills the segment between the chord and the arc below it. A segment is what is left of a sector when the triangle made by the two radii and the chord is taken away.

The concrete is the minor segment cut off by the chord AB. Its area is the area of sector AOB minus the area of triangle AOB.
Angle AOB is 90^°, a quarter turn, so the sector is a quarter of the circle: 90360 × π × 4² = 14 × 3.14 × 16 = 12.56 m².
OA and OB are at right angles, so triangle AOB has base 4 m and height 4 m: its area is 12 × 4 × 4 = 8 m².
(a) The concrete has an area of 12.56 − 8 = 4.56 m².
The whole cross-section is π × 4² = 3.14 × 16 = 50.24 m². (b) The open part is 50.24 − 4.56 = 45.68 m². Check: the open part is three quarters of the circle plus the triangle, 37.68 + 8 = 45.68 m².

answer(a) 4.56 m²; (b) 45.68 m²

techniqueThe Area of a Segment · Chords, Arcs, Sectors and Segments

examsO-Level

Common pitfalls

Taking the concrete to be the whole sector AOB, 12.56 m². The sector reaches up to the center O, which is above the road; the concrete stops at the chord, so the triangle AOB must be taken away.
Using 12 × 4 × 4 for the triangle without checking the angle. The base and height of the triangle are the two radii only because the angle between them is 90^°.

A Rear Wiper Sweeping Across a Car Window

methodThe Tip Travels Along an Arc and the Blade Wipes the Sector Between Two Radii: Take the Fraction of the Circle That the Angle Gives

The rear wiper of a car turns about a pivot O through an angle of 120^° in one sweep. The arm is 45 cm long from O to its tip, and the rubber blade covers the outer 30 cm of the arm. (a) How far does the tip of the arm travel in one sweep? (b) What area of glass does the blade wipe in one sweep? Leave π in both answers.

The tip is 45 cm from the pivot, and 120^° is 13 of a full turn.

The tip is 45 cm from O, so it moves on a circle of radius 45 cm. The sweep is 120360 = 13 of a full turn.

step 1 of 5

A point on a turning arm moves along an arc of a circle, and the arm sweeps out a sector. An angle of 120^° is 120360 = 13 of a full turn, so the arc is a third of a circumference and the sector is a third of a circle.

The tip is 45 cm from O, so it moves on a circle of radius 45 cm. The sweep is 120360 = 13 of a full turn.
(a) The tip travels along an arc of length 13 × 2 π × 45 = 30π cm.
The blade covers the outer 30 cm, so its inner end is 45 − 30 = 15 cm from O. The wiped region is the sector of radius 45 cm with the sector of radius 15 cm removed from it.
The large sector has area 13 × π × 45² = 13 × 2025π = 675π cm², and the small sector has area 13 × π × 15² = 75π cm².
(b) The blade wipes 675π − 75π = 600π cm². Check: 13 × π × (45² − 15²) = 13 × 1800π = 600π.

answer(a) 30π cm; (b) 600π cm²

techniqueArcs and Sectors · Chords, Arcs, Sectors and Segments

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

Using the blade's length, 30 cm, as a radius. The blade lies from 15 cm to 45 cm along the arm, so the region it wipes is the difference of two sectors, and 30 cm is not the radius of either.
Taking the tip's path to be a third of the area rather than a third of the circumference. Distance traveled is a length, so it comes from 2π r, not from π r².

Security Cameras on the Wall of a Round Hall

methodA Camera on the Wall Sees the Doorway Under Half the Angle It Makes at the Center, and a Second Camera on the Same Arc Sees the Same Angle

A round hall has center O. Its doorway runs from A to B along the wall, and angle AOB is 84^°. A security camera C is fixed to the wall on the far side of the hall from the doorway. A second camera D is fixed to the wall at the point directly opposite A, so that AD is a diameter of the hall. (a) What angle does the doorway fill in the view of camera C, that is, what is angle ACB? (b) Find angle DAB, the angle between the diameter AD and the doorway AB.

(a) The angle at the center is twice the angle at the wall: ACB = 12 × 84^° = 42^°.

Angle AOB at the center and angle ACB at the wall both stand on the arc AB, the doorway. The angle at the center is twice the angle at the circumference, so ACB = 12 × 84^° = 42^°. (a) The doorway fills 42^° of camera C's view.

step 1 of 4

The angle that an arc makes at the center of a circle is twice the angle it makes at any point on the rest of the circle. Every point on the same side of the chord sees the arc under the same angle, and a diameter is seen from the circle under a right angle.

Angle AOB at the center and angle ACB at the wall both stand on the arc AB, the doorway. The angle at the center is twice the angle at the circumference, so ACB = 12 × 84^° = 42^°. (a) The doorway fills 42^° of camera C's view.
Camera D is on the same side of the doorway as camera C, so angles ACB and ADB are angles in the same segment: ADB = 42^° as well.
AD is a diameter, so angle ABD, the angle in a semicircle, is 90^°.
(b) In triangle ABD the angles add to 180^°: DAB = 180^° − 90^° − 42^° = 48^°. Check: triangle OAB is isosceles with OA = OB, so OAB = 12(180^° − 84^°) = 48^°, and OA lies along the diameter AD.

answer(a) 42^°; (b) 48^°

techniqueAngle at the Center · Angles in the Same Segment · Angle in a Semicircle

examsO-Level · SAT · GCSE Higher

Common pitfalls

Giving 84^° as the angle at the camera. The doorway makes 84^° at the center of the hall; from the wall, further away, it fills only half of that.
Doubling instead of halving, to get 168^°. The angle at the center is the larger one, so the angle at the wall is found by halving 84^°.

Four Fence Posts Round a Circular Pond

methodFour Posts on One Circle Make a Cyclic Quadrilateral, So Opposite Angles Add to 180 Degrees and an Exterior Angle Equals the Opposite Interior Angle

Four fence posts A, B, C and D stand in that order on the edge of a circular pond, and straight rails join A to B, B to C, C to D and D to A. Angle DAB is 78^° and angle ABC is 105^°. (a) Find angle BCD and angle CDA. (b) The rail AB is continued in a straight line past B to a lamp at E. Find angle CBE.

All four posts stand on one circle, so ABCD is a cyclic quadrilateral and its opposite angles add to 180^°.

All four posts lie on the edge of the pond, a circle, so ABCD is a cyclic quadrilateral and its opposite angles add to 180^°.

step 1 of 5

The four posts all lie on one circle, so ABCD is a cyclic quadrilateral. In a cyclic quadrilateral each pair of opposite angles adds to 180^°, and it follows that an exterior angle at any corner is equal to the interior angle at the opposite corner.

All four posts lie on the edge of the pond, a circle, so ABCD is a cyclic quadrilateral and its opposite angles add to 180^°.
Angle BCD is opposite angle DAB: BCD = 180^° − 78^° = 102^°.
Angle CDA is opposite angle ABC: CDA = 180^° − 105^° = 75^°. (a) The angles are 102^° at C and 75^° at D.
ABE is a straight line, so angles ABC and CBE add to 180^°: CBE = 180^° − 105^° = 75^°.
(b) Angle CBE is 75^°, the same as the interior angle at the opposite corner D. Check: the four angles of the quadrilateral add to 78^° + 105^° + 102^° + 75^° = 360^°.

answer(a) 102^° and 75^°; (b) 75^°

techniqueCyclic Quadrilaterals

examsO-Level · SAT · GCSE Higher

Common pitfalls

Pairing the angles that are next to each other, 78^° and 105^°, and expecting them to add to 180^°. It is the opposite angles of a cyclic quadrilateral that add to 180^°; neighboring angles need not.
Taking angle CBE to be equal to angle ABC. The two angles lie on a straight line, so they add to 180^°; the exterior angle equals the opposite interior angle at D, not the interior angle at B.

Two Paths from a Lamp Post That Touch a Round Pond

methodA Path That Just Touches the Pond Is a Tangent, at Right Angles to the Radius, and the Two Tangents from the Post Are Equal

A circular pond has center O and radius 8 m. A lamp post stands at a point P that is 17 m from O. Two straight paths run from P and each just touches the edge of the pond, one at A and the other at B. (a) How long is the path PA? (b) A low fence runs from A to P and on from P to B. How long is the fence?

PA just touches the pond, so it is a tangent and the radius OA is at right angles to it.

The path PA just touches the pond at A, so PA is a tangent and the radius OA is perpendicular to it: angle OAP is 90^°.

step 1 of 5

A line that touches a circle at one point is a tangent, and the radius drawn to that point is at right angles to it. That right angle puts the radius, the tangent and the line to the center in a right-angled triangle, and the two tangents drawn from one outside point are always equal in length.

The path PA just touches the pond at A, so PA is a tangent and the radius OA is perpendicular to it: angle OAP is 90^°.
Triangle OAP is right-angled at A, with hypotenuse OP = 17 m and OA = 8 m. By Pythagoras' theorem, PA² = 17² − 8² = 289 − 64 = 225.
(a) PA = √225 = 15 m.
PA and PB are the two tangents from the same point P, so they are equal: PB = PA = 15 m. In the same way, triangle OBP is right-angled at B with OP = 17 m and OB = 8 m, which gives PB = 15 m directly.
(b) The fence is PA + PB = 15 + 15 = 30 m long. Check: 8² + 15² = 64 + 225 = 289 = 17².

answer(a) 15 m; (b) 30 m

techniqueTangent Properties · Congruence Tests

examsO-Level · SAT · GCSE Higher

Common pitfalls

Taking OP as a shorter side and adding the squares: 17² + 8² = 353. The right angle is at A, where the tangent meets the radius, so OP is the hypotenuse and the squares must be subtracted.
Measuring the fence as 15 + 8 m by running it through the center. The fence runs along the two tangents, from A to P and from P to B, and both are 15 m.

A Model Car That Leaves a Circular Track Along the Tangent

methodThe Angle Between the Tangent and a Chord Equals the Angle in the Alternate Segment

A model car runs round a circular track and leaves it at the point T, driving on in a straight line TX along the tangent to the track at T. Two marker cones stand on the track at A and at B, on the other side of the chord TA from X. The angle between the car's path TX and the chord TA is 52^°, and angle ATB is 63^°. (a) A camera at B is pointed along BT and must turn to face A. Through what angle does it turn, that is, what is angle TBA? (b) Find angle TAB.

TX is the tangent at T and TA is a chord from T; the angle between them is 52^°.

TX is the tangent at T and TA is a chord from the point of contact. The angle between them is 52^°.

step 1 of 4

The alternate segment theorem says that the angle between a tangent and a chord drawn from the point of contact is equal to the angle that the chord makes at the circumference in the segment on the other side of the chord.

TX is the tangent at T and TA is a chord from the point of contact. The angle between them is 52^°.
B lies in the alternate segment, on the other side of the chord TA from X. By the alternate segment theorem, angle TBA is equal to the angle between the tangent and the chord. (a) The camera turns through TBA = 52^°.
In triangle TAB, the angles add to 180^°: TAB = 180^° − 63^° − 52^°.
(b) TAB = 65^°. Check: continue the tangent the other way from T to a point Y. The angle between TY and the chord TB is 180^° − 52^° − 63^° = 65^°, and A lies in the alternate segment for that angle, so the theorem gives TAB = 65^° by a second route.

answer(a) 52^°; (b) 65^°

techniqueThe Alternate Segment Theorem

examsO-Level · SAT · GCSE Higher

Common pitfalls

Matching the 52^° with the angle at A instead of the angle at B. The angle between the tangent and the chord TA equals the angle that TA makes at a point on the far side of TA, and that point is B; A is an end of the chord.
Using 63^° as the angle at B. Angle ATB is the angle at T between the two chords; the alternate segment theorem gives the angle at B from the tangent-chord angle of 52^°, not from 63^°.

The Lead Strips of a Round Window, Marked in x on the Plan

methodThe Angle at the Center Is Twice the Angle at the Rim, So the Two Expressions in x Make One Equation

A round stained-glass window has center O. Lead strips run from O to two points A and B on the rim, and from a third point C on the rim, on the far side of the window from A and B, to A and to B. On the maker's plan, angle AOB is marked as (5x − 6)^° and angle ACB is marked as (2x + 9)^°. (a) Find x. (b) Find angle ACB and angle AOB.

Both angles stand on the arc AB, and the angle at the center is twice the angle at the rim: 5x − 6 = 2(2x + 9).

Both angles stand on the arc AB. The angle at the center is twice the angle at the circumference, so 5x − 6 = 2(2x + 9).

step 1 of 5

A circle theorem gives a relation between two angles, and when the angles are written in terms of x that relation becomes an equation. Angle AOB at the center and angle ACB at the rim both stand on the arc AB, so one is twice the other.

Both angles stand on the arc AB. The angle at the center is twice the angle at the circumference, so 5x − 6 = 2(2x + 9).
Expand the bracket: 5x − 6 = 4x + 18.
Subtract 4x from both sides and add 6 to both sides: x = 24. (a) x = 24.
Angle ACB is 2 × 24 + 9 = 57^° and angle AOB is 5 × 24 − 6 = 114^°. (b) The angles are 57^° at C and 114^° at O.
Check: 114 = 2 × 57, so the angle at the center is twice the angle at the rim, and both angles are between 0^° and 180^°, as angles in a triangle must be.

answer(a) x = 24; (b) 57^° and 114^°

techniqueCircle Theorems with Algebra · Angle at the Center

examsO-Level · SAT · GCSE Higher

Common pitfalls

Setting the two expressions equal, 5x − 6 = 2x + 9, which gives x = 5. The two angles are not equal: the angle at the center is twice the angle at the rim, so the factor 2 must go on the angle at C.
Putting the factor 2 on the wrong side, 2(5x − 6) = 2x + 9, which gives x = 2.625. The angle at the center is the larger one, so it is the angle at the rim that is doubled.