Triangles | ICSE Class 9 Maths Notes

This note covers triangle notation, congruence and corresponding parts, the SSS, SAS, AAS and RHS tests, equal sides and angles, triangle inequalities, perpendicular distance, the mid-point theorem and its converse, equal intercepts, and Pythagoras’ theorem with its area proof and converse.

What does congruence of triangles mean?

How do we read the notation?

A triangle is a closed figure formed by joining three points that do not lie on one straight line. These points are its vertices, or corners, and the joining line segments are its sides. A line segment is the part of a straight line between two endpoints.

In △ABC, the symbol △ means triangle, and A, B and C name the vertices. AB denotes the segment joining A to B, or its length when used in a calculation. The symbol = means “is equal to”.

An angle measures the turn between two rays meeting at a vertex. A ray starts at a point and continues in one direction. In ∠ABC, the symbol ∠ means angle and the middle letter B identifies its vertex.

We may write ∠B when its meaning is clear. Angles here are measured in degrees, denoted by °. The three interior angles, meaning angles inside a triangle, add to 180°. A right angle measures 90°; a right-angled triangle contains a right angle.

How do corresponding parts match?

Definition: Congruent triangles have the same shape and size. They cover one another exactly when placed one over the other. This placing of one figure over another is called superposition.

The symbol ≅ means “is congruent to”. In △ABC ≅ △PQR, the vertices P, Q and R match A, B and C respectively. These matching vertices, sides and angles are called corresponding parts. Their order matters.

Trace and cut out the two triangles, then move, turn or turn over one cutout to match the other. The correspondence follows the vertices that coincide, meaning occupy the same position. A different orientation on the page does not prevent congruence.

After proving congruence, use CPCT, meaning corresponding parts of congruent triangles, to justify equal matching sides or angles. Congruence tests provide sufficient conditions: conditions that establish congruence without checking all six pairs of parts separately.

How do the SSS and SAS congruence tests work?

Result: The SSS test

SSS means side-side-side. If the three sides of one triangle equal the three corresponding sides of another, the triangles are congruent. Match the sides first, then identify the common vertex of each pair of matched sides to determine the vertex order.

Worked example 1. Triangles ABC and PQR have AB = 3.5 cm, BC = 7.1 cm, AC = 5 cm, PQ = 7.1 cm, QR = 5 cm and PR = 3.5 cm. Here cm means centimetres. Establish congruence and write the correct correspondence.

Answer: AB = RP = 3.5 cm, BC = PQ = 7.1 cm and AC = RQ = 5 cm. Therefore △ABC ≅ △RPQ by SSS. Vertex A matches R, B matches P and C matches Q.

The order △RPQ follows from the equalities; it is not chosen alphabetically. In particular, AB and AC meet at A, while their matching sides RP and RQ meet at R. This fixes the first pair of corresponding vertices.

Result: The SAS test

SAS means side-angle-side. Two corresponding sides and their included angles must be equal. The included angle is the angle between the two named sides. For sides AB and BC, it is ∠ABC because those sides meet at B.

Worked example 2. In △ABC, AB = 7 cm, BC = 5 cm and ∠B = 50°. In △DEF, whose vertices are D, E and F, DE = 5 cm, EF = 7 cm and ∠E = 50°. Determine the congruence relation.

Answer: AB = FE = 7 cm, BC = ED = 5 cm and the included angles ∠B and ∠E both equal 50°. Therefore △ABC ≅ △FED by SAS. The equal angle lies between the two matched sides in each triangle.

A common side belongs to both triangles being compared. Its length equals itself and can supply one pair of equal sides. It should be stated explicitly in a proof, even when the two triangles visibly share it.

Note: Two sides and an angle not included between them do not, in general, establish congruence. Before naming SAS, identify the vertex where the two given sides meet and check that the given angle is at that vertex.

How do AAS and RHS establish congruence?

Result: The AAS test

AAS means angle-angle-side. Two corresponding angles and a corresponding non-included side establish congruence. A non-included side is a side other than the one joining the vertices of the two given angles.

ASA, meaning angle-side-angle, uses the included side between two equal corresponding angles. The angle sum of 180° connects the two forms: equality of two angle pairs gives equality of the remaining pair, allowing the appropriate included side to be used.

Worked example 3. Segments AB and CD intersect at O, their common point. A and B lie on opposite sides of O, as do C and D. Given ∠ACO = ∠BDO = 70°, ∠AOC = 30° and AC = BD, prove △AOC ≅ △BOD.

Answer: ∠BOD = ∠AOC = 30° because vertically opposite angles, the opposite angles formed by intersecting lines, are equal. Together with ∠ACO = ∠BDO = 70° and AC = BD, this gives AAS. Equivalently, ∠CAO = ∠DBO = 180° − 70° − 30° = 80°, so ASA also proves congruence. The symbol − means subtraction.

The side must correspond under the same angle matching. Equal lengths taken from unrelated positions do not complete the test. A useful check is to write the angle correspondence first and then identify the matching endpoints of the given side.

Result: The RHS test

RHS means right angle-hypotenuse-side. The hypotenuse is the side opposite the right angle. The other two sides are the legs. Two right-angled triangles are congruent when their hypotenuses and one pair of corresponding legs are equal.

Worked example 4. In △ABC, ∠B = 90°, AC = 8 cm and AB = 3 cm. In △PQR, ∠P = 90°, QR = 8 cm and PR = 3 cm. Prove the triangles congruent.

Answer: ∠B = ∠P = 90°. The hypotenuses AC and RQ both measure 8 cm, and the legs AB and RP both measure 3 cm. Therefore △ABC ≅ △RPQ by RHS. The right-angle vertices correspond to one another.

RHS requires both triangles to be right-angled; equal hypotenuses cannot simply be assumed from a drawing. If both corresponding legs and their right angles are given, SAS also applies. Identify which data are actually available before choosing a test.

Three equal corresponding angles alone do not establish congruence. One triangle can have the same angles as an enlarged copy while its sides have different lengths. A congruence argument must contain suitable information fixing size as well as shape.

What links equal sides and equal angles?

Theorem: Angles opposite equal sides

An isosceles triangle has two equal sides. In △ABC with AB = AC, the angles opposite those sides are ∠C and ∠B respectively. They are equal. The remaining side BC is called the base in this description, and ∠B and ∠C are its base angles.

A side is opposite an angle when it does not contain the angle’s vertex. Thus BC is opposite ∠A. Locating the opposite side correctly is essential: the theorem relates an angle to the side across from it, rather than to either side forming it.

What does the converse say?

A converse interchanges the condition and conclusion of a statement. Here it says that sides opposite equal angles in a triangle are equal. Thus ∠B = ∠C gives AC = AB. This converse is true, but a statement’s converse needs its own justification.

An angle bisector divides an angle into two equal angles. If AD bisects ∠BAC and meets BC at D, then ∠BAD = ∠CAD. For AB = AC, triangles ADB and ADC share AD and are congruent by SAS.

Corresponding parts then give ∠ABD = ∠ACD. Since D lies on BC, these are the base angles ∠ABC and ∠ACB. This also illustrates how a larger triangle can contain two smaller triangles useful for a congruence argument.

Read the condition before deciding which direction to use. Starting from equal sides calls for the theorem; starting from equal angles calls for its converse. Do not treat a side or angle as equal merely because it looks equal in a sketch.

How do triangle inequalities compare sides, angles and distances?

Theorem: The greater side faces the greater angle

An inequality compares quantities that are not equal. The symbol > means “is greater than”, and < means “is less than”. If two sides of a triangle are unequal, the greater angle is opposite the greater side.

In △ABC, AB > AC gives ∠C > ∠B. The converse also holds: ∠C > ∠B gives AB > AC. Both comparisons belong to the same triangle. The result does not compare sides of unrelated triangles solely from their angles.

Theorem: The sum of two sides exceeds the third

The triangle inequality says that the sum of any two sides of a triangle is greater than the third side. For △ABC, all three comparisons hold: AB + BC > AC, BC + CA > AB and CA + AB > BC. The symbol + denotes addition.

Equality does not form a triangle with three non-collinear vertices, meaning vertices not on the same straight line. When checking three proposed positive side lengths, the sum of the two shorter lengths must exceed the longest. The other two inequalities then follow.

For the side lengths 3, 4 and 5 units, the comparisons are 3 + 4 > 5, 4 + 5 > 3 and 5 + 3 > 4. These lengths satisfy the condition. A “unit” here means a common chosen unit of length.

Property: The perpendicular is shortest

Lines are perpendicular when they meet at a right angle; the symbol ⊥ denotes this relationship. From a point outside a given line, the perpendicular segment to the line is shorter than every other segment joining that point to the line.

Its length is the distance from the point to the line. The meeting point is called the foot of the perpendicular. An oblique segment, meaning one meeting the line at an angle other than a right angle, does not measure this shortest distance.

To see the comparison, join the outside point to another point on the line. The perpendicular and the part of the line between the two meeting points form the legs of a right-angled triangle. The oblique segment is its hypotenuse and is longer than either leg.

What does the mid-point theorem state and how is it proved?

Theorem: Joining two midpoints

A midpoint divides a segment into two equal parts. Lines in the same plane are parallel if they do not meet however far extended; the symbol ∥ means parallel to. The mid-point theorem links equal divisions of triangle sides to parallel lines.

Definition: The segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length. In △ABC, if P and Q are midpoints of AB and AC, then PQ ∥ BC and PQ = BC/2. The slash / denotes division.

Both conclusions matter. Knowing the direction of PQ does not state its length, and stating its length does not explicitly establish its direction. A complete application identifies the two midpoints before using either conclusion.

Triangle ABC has P on AB and Q on AC. Line PQ continues through Q to R. A line through C and R is parallel to AB. Equal marks show the two halves of AB and AC.

See Fig. 12.14 in your NCERT textbook

How does the construction prove both conclusions?

A parallelogram is a quadrilateral, or four-sided closed figure, with both pairs of opposite sides parallel. We use the fact that a quadrilateral with one pair of opposite sides equal and parallel is a parallelogram; its opposite sides are equal.

1. Through C draw a line parallel to BA, meeting the extended line PQ at R. Thus CR ∥ AP.
2. In △APQ and △CRQ, AQ = CQ because Q is a midpoint. Also ∠AQP = ∠CQR, since these angles are vertically opposite.
3. ∠APQ = ∠CRQ because AP and CR are parallel. These are alternate angles: angles on opposite sides of a crossing line and between the parallel lines.
4. Therefore △APQ ≅ △CRQ by AAS. Corresponding parts give PQ = QR and AP = CR.
5. As AP = PB, we have PB = CR as well as PB ∥ CR. Thus BCRP is a parallelogram.
6. Consequently PR ∥ BC and PR = BC. Since PQ = QR, PQ = PR/2 = BC/2. Also PQ ∥ BC because PQ lies on PR.

The proof uses construction, meaning an added line or point chosen to support an argument. The new point R creates triangles with enough known equal parts to apply congruence, then a parallelogram supplies the required parallelism and length.

How do we use the converse of the mid-point theorem?

Theorem: A parallel through a midpoint bisects another side

To bisect a segment is to divide it into two equal parts. The converse says that a line through the midpoint of one side of a triangle, parallel to another side, bisects the third side. The midpoint and parallelism are both necessary hypotheses, or starting conditions.

In △ABC, let P be the midpoint of AB. Draw PQ parallel to BC, meeting AC at Q. The conclusion is AQ = QC. Once Q is known to be a midpoint, the direct theorem also gives PQ = BC/2.

How can the direct theorem prove the converse?

1. Let M be the midpoint of AC. Join P to M.
2. Because P and M are midpoints of AB and AC, the direct mid-point theorem gives PM ∥ BC.
3. Through P there is a unique line parallel to BC. Hence PM and PQ are the same line.
4. That line meets AC at the same point, so M and Q coincide. Therefore AQ = QC.

The extra fact about a unique parallel justifies the converse. We have not assumed that reversing a true statement automatically produces another true statement. The point M is introduced independently as a midpoint and then shown to be Q.

Worked example 5. In △ABC, P, Q and R are the midpoints of AB, AC and BC respectively. Show that △PQR ≅ △QPA.

Answer: PQ = QP is common. Also PR = AC/2 = AQ and QR = AB/2 = AP, by the mid-point theorem and the midpoint definitions. Hence △PQR ≅ △QPA by SSS. Each equality matches the stated vertex order.

Worked example 6. In △ABC, M and N are midpoints of AB and AC. Let D lie on BC and let MN meet AD at E. Show that E bisects AD.

Answer: MN ∥ BC by the mid-point theorem. For D distinct from B, apply the converse in △ABD: M is the midpoint of AB and ME ∥ BD, so AE = ED = AD/2. If D is B, E is M and the same conclusion follows directly.

In this last application, naming the smaller triangle is important. The known midpoint lies on AB, and the side being bisected is AD. The relevant parallel side is BD, which lies along BC.

What is the equal intercept theorem?

Theorem: Equal divisions transfer between transversals

A transversal is a line crossing two or more lines at distinct points. An intercept is the segment cut off on a transversal between two of those lines. The equal intercept theorem concerns a family of parallel lines crossed by two transversals.

If three or more parallel lines cut off equal intercepts on one transversal, they cut off equal intercepts on another transversal. Equal lengths are compared along each transversal. The theorem does not say that an intercept on one transversal equals an intercept on the other.

How can we prove the result for three parallels?

Let three parallel lines meet one transversal at A, B and C, in that order, and another at D, E and F, in that order. Suppose AB = BC. We shall prove DE = EF using the converse mid-point theorem.

1. Join A to F, and let this segment meet the middle parallel at G. Thus G is a newly constructed point on the line through B and E.
2. In △ACF, B is the midpoint of AC because AB = BC. Also BG ∥ CF because these segments lie on the middle and last parallels.
3. The converse mid-point theorem gives AG = GF, so G is the midpoint of AF.
4. In △FAD, GE ∥ AD because these segments lie on the middle and first parallels. A parallel through midpoint G therefore bisects FD.
5. Hence FE = ED. If the transversals are parallel, the quadrilaterals ABED and BCEF are parallelograms instead; then DE = AB = BC = EF.

For more parallel lines, apply this three-line argument to successive groups of three. Each application passes equality from one pair of consecutive intercepts to the corresponding pair on the second transversal.

The construction above applies when the transversals have distinct intersection points on all three parallels. If the transversals meet on an outer parallel, the result follows directly by the converse mid-point theorem in the resulting triangle. If they meet on the middle parallel, B and E coincide, so BG and GE have zero length and the construction above cannot be used. Instead, compare △ABD and △CBF: AB = BC, ∠ABD = ∠CBF as vertically opposite angles, and ∠BAD = ∠BCF as alternate angles between the outer parallels. Thus △ABD ≅ △CBF by ASA, giving BD = BF. Since B and E coincide, DE = EF. Together with the parallel-transversal case, these arguments cover all placements of two distinct transversals.

Note: Keep the roles of the theorems separate. The mid-point theorem joins two known midpoints. Its converse begins with one midpoint and a parallel. The equal intercept theorem transfers equal divisions between transversals across a set of parallel lines.

What is Pythagoras’ theorem and how does an area proof work?

Theorem: Squares of sides in a right-angled triangle

For a right-angled triangle, let a and b be the lengths of its perpendicular legs and let c be its hypotenuse length. Pythagoras’ theorem states a² + b² = c². The superscript ² means the square of a number, its product with itself.

The right-angle condition identifies which side is c. The formula cannot be applied to an arbitrary triangle without establishing that condition. For positive side lengths, the hypotenuse is longer than either leg because its square exceeds the square of each leg.

How does rearranging area prove the formula?

Area measures the region enclosed by a figure in square units. A square has four equal sides and four right angles, and its area is its side length squared. A right-angled triangle with legs a and b has area ab/2, where ab means a multiplied by b.

Construct a square of side a + b. Place four congruent copies of the right-angled triangle inside it, with their right angles at the four corners. Along each outer side, one leg of length a and one leg of length b fill the side.

The four hypotenuses bound a central quadrilateral. Each side of this quadrilateral has length c. Each of its angles is a straight angle minus the two acute angles of the original triangle. An acute angle is less than 90°; those two angles sum to 90°.

Thus every central angle is 90°, so the central quadrilateral is a square of side c. This establishes its shape before using c² for its area. A straight angle measures 180°.

1. The outer square has area (a + b)². Parentheses group a + b before squaring.
2. The four triangles have total area 4 × ab/2 = 2ab. The symbol × denotes multiplication.
3. The central square has area c², so (a + b)² = 2ab + c².
4. Expanding the left side gives a² + 2ab + b² = 2ab + c².
5. Subtracting 2ab from both sides gives a² + b² = c².

This is an area proof: the same outer region is measured as one square and as the sum of its non-overlapping pieces. The cancellation removes the four triangle areas and leaves the relationship between the three squared side lengths.

When finding a leg, subtract the square of the known leg from the hypotenuse square. When finding the hypotenuse, add the leg squares. Finally take the positive square root, meaning the positive number whose square gives the calculated value, because lengths are positive.

How do the converse and worked calculations test a right angle?

Theorem: Converse of Pythagoras’ theorem

If the side lengths a, b and c of a triangle satisfy a² + b² = c², the angle opposite c is a right angle. Check the longest side as c. The direct theorem begins with a right angle; the converse begins with the squared-length equality.

For a proof, take △ABC with BC = a, CA = b and AB = c. Construct a right-angled triangle XYZ, whose vertices are X, Y and Z, with YZ = a, XZ = b and a right angle at Z.

1. By Pythagoras’ theorem in △XYZ, XY² = a² + b².
2. The given relation says a² + b² = c², so XY² = c² and XY = c because both lengths are positive.
3. Now BC = YZ, CA = ZX and AB = YX. Therefore △ABC ≅ △XYZ by SSS.
4. Corresponding angles give ∠ACB = ∠XZY = 90°. Thus the angle opposite AB is a right angle.

Worked example 7. A triangle has sides 3, 4 and 5 units. Determine whether it is right-angled and find its area.

Answer: The longest side is 5 units. Since 3² + 4² = 9 + 16 = 25 = 5², the triangle is right-angled by the converse. Its perpendicular legs are 3 and 4 units, so its area is 3 × 4/2 = 6 square units.

The triangle is labelled A at the top, B at the lower left and C at the lower right. AB is marked 4, BC is marked 3 and AC is marked 5.

See Fig. 6.25 in your NCERT textbook

Worked example 8. A right-angled triangle has area 54 cm² and one leg 12 cm. Find its perimeter. Here cm² means square centimetres, and perimeter means the total length of the boundary.

Answer: Let b be the other leg length in centimetres. The area equation is 12b/2 = 54, giving b = 9. The hypotenuse square is 12² + 9² = 144 + 81 = 225, so its length is 15 cm. The perimeter is 12 + 9 + 15 = 36 cm.

Worked example 9. A triangle has sides 7 cm, 24 cm and 25 cm. Use a right-angle test to find its area.

Answer: 7² + 24² = 49 + 576 = 625 = 25². Hence the triangle is right-angled, with legs 7 cm and 24 cm. Its area is 7 × 24/2 = 84 cm².

These calculations distinguish length from area. The sides and perimeter use units of length; the area uses square units. Squared lengths appear temporarily in Pythagoras’ equation, but the required side length comes after taking the positive square root.

Glossary

• Congruent triangles — Triangles with the same shape and size that cover each other exactly when superposed.
• Corresponding parts — Matching sides, angles or vertices identified by a stated correspondence between two triangles.
• Included angle — The angle formed at the common endpoint of the two sides being considered.
• Hypotenuse — The side opposite the right angle in a right-angled triangle.
• Isosceles triangle — A triangle with two equal sides and equal angles opposite those sides.
• Converse — A statement obtained by interchanging the condition and conclusion of another statement.
• Triangle inequality — The sum of any two side lengths exceeds the third side length.
• Perpendicular distance — The shortest distance from a point to a line, measured along a perpendicular segment.
• Midpoint — A point on a segment that divides it into two equal lengths.
• Angle bisector — A ray that divides a given angle into two equal angles.
• Transversal — A line that intersects two or more other lines at distinct points.
• Intercept — The segment cut off on a transversal between two of the intersected lines.
• Parallelogram — A quadrilateral in which both pairs of opposite sides are parallel.
• Perimeter — The total length of the boundary of a closed plane figure.

Common errors and misconceptions

• Misconception: Vertex order does not matter in a congruence statement. Correct: The order identifies corresponding vertices, sides and angles; check it against the given equalities.
• Misconception: Any two sides and an angle establish SAS. Correct: SAS requires the equal angle to be included between the two matched sides.
• Misconception: Equal corresponding angles guarantee congruence. Correct: Angles alone do not fix size; an enlarged triangle can have the same angles.
• Misconception: RHS applies whenever two side pairs are equal. Correct: Both triangles must be right-angled, with equal hypotenuses and a pair of equal corresponding legs.
• Misconception: The sum of two sides may equal the third. Correct: A triangle with non-collinear vertices requires the strict triangle inequality.
• Misconception: Any parallel to one side bisects the other sides. Correct: The converse mid-point theorem requires the parallel to pass through a known midpoint.
• Misconception: Equal intercepts on two transversals have the same length across both transversals. Correct: The theorem establishes equality of consecutive intercepts on each transversal separately.
• Misconception: Pythagoras’ formula applies to every triangle. Correct: Use it for a known right-angled triangle, or test the squared-length equality to establish a right angle by its converse.

Exam-style questions with model answers

Q1. State the relationship between the angles opposite equal sides of a triangle, and state its converse. [2 marks]

1. Angles opposite equal sides of a triangle are equal.
2. Conversely, sides opposite equal angles of a triangle are equal.

Q2. In △ABC, AB = 3.5 cm, BC = 7.1 cm and AC = 5 cm. In △PQR, PQ = 7.1 cm, QR = 5 cm and PR = 3.5 cm. Prove congruence and give the correct vertex order. [3 marks]

1. Match AB with RP: both measure 3.5 cm. Match BC with PQ: both measure 7.1 cm.
2. The third pair also matches: AC = RQ = 5 cm. Thus all three corresponding side pairs are equal, satisfying the SSS congruence test.
3. The matching vertices are A with R, B with P and C with Q. Therefore △ABC ≅ △RPQ.

Q3. Segments AB and CD intersect at O, with A and B on opposite sides of O and C and D on opposite sides of O. Given ∠ACO = ∠BDO = 70°, ∠AOC = 30° and AC = BD, prove △AOC ≅ △BOD. [4 marks]

1. The intersecting lines give vertically opposite angles ∠AOC = ∠BOD. As ∠AOC = 30°, ∠BOD is also 30°.
2. The other specified angles satisfy ∠ACO = ∠BDO = 70°, as given in the question.
3. The sides AC and BD are equal by the given condition. They correspond under the matching A with B, O with O and C with D.
4. Two corresponding angles and a corresponding non-included side are equal. Hence △AOC ≅ △BOD by AAS.

Q4. In △ABC, ∠B = 90°, AC = 8 cm and AB = 3 cm. In △PQR, ∠P = 90°, QR = 8 cm and PR = 3 cm. Prove congruence using RHS. [3 marks]

1. The angles ∠B and ∠P are both 90°, so both triangles are right-angled. Their opposite sides AC and RQ are the hypotenuses.
2. The hypotenuses satisfy AC = RQ = 8 cm. The corresponding legs satisfy AB = RP = 3 cm, giving the required two length equalities.
3. By RHS, △ABC ≅ △RPQ, with the right-angle vertex B corresponding to P and the other vertices A to R and C to Q.

Q5. In △ABC, P and Q are the midpoints of AB and AC respectively. Prove PQ ∥ BC and PQ = BC/2. You may use the test that a quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. [5 marks]

1. Draw through C a line parallel to BA, meeting PQ extended beyond Q at R. Then CR ∥ AP, while AQ = CQ because Q is the midpoint of AC.
2. In △APQ and △CRQ, ∠AQP = ∠CQR as vertically opposite angles, and ∠APQ = ∠CRQ as alternate angles between parallel lines. Together with AQ = CQ, AAS gives congruence.
3. Corresponding parts yield PQ = QR and AP = CR. Since P is the midpoint of AB, AP = PB, so PB = CR.
4. Now PB and CR are equal and parallel. Therefore BCRP is a parallelogram, giving PR ∥ BC and PR = BC.
5. As PQ = QR, PQ = PR/2 = BC/2. Also PQ lies on PR, so PQ ∥ BC. Both required conclusions follow.

Q6. Prove Pythagoras’ theorem for a right-angled triangle whose perpendicular leg lengths are a and b and whose hypotenuse length is c. Use four congruent copies placed at the corners of a square of side a + b. [5 marks]

1. Place the four right angles at the four outer corners, with legs a and b filling each side of the outer square. Its area is (a + b)².
2. The central quadrilateral has four sides of length c. Its angles are 180° minus the two acute triangle angles, whose sum is 90°. Thus it is a square.
3. The central square has area c². Each of the four triangles has area ab/2, so together their area is 2ab.
4. Adding the non-overlapping areas gives (a + b)² = 2ab + c². Expanding the square produces a² + 2ab + b² = 2ab + c².
5. Subtract 2ab from both sides. The result is a² + b² = c², which proves the required relation.

Q7. A triangle has sides 7 cm, 24 cm and 25 cm. Show that it is right-angled and calculate its area. [3 marks]

1. The longest side is 25 cm. The squares of the other two sides add to 7² + 24² = 49 + 576 = 625.
2. As 25² = 625, the squared-length equality holds. By the converse of Pythagoras’ theorem, the angle opposite the 25 cm side is a right angle.
3. The perpendicular legs are therefore 7 cm and 24 cm. Their product divided by two gives the area: 7 × 24/2 = 84 cm².

Q8. A right-angled triangle has area 54 cm² and one perpendicular leg of length 12 cm. Find the other leg, hypotenuse and perimeter. [4 marks]

1. Let b be the other perpendicular leg length in centimetres. The area formula gives 12b/2 = 54.
2. Hence 6b = 54 and b = 9. The other perpendicular leg therefore measures 9 cm.
3. Pythagoras’ theorem gives the hypotenuse square as 12² + 9² = 225. The hypotenuse is the positive square root of 225, which is 15 cm.
4. The perimeter is the sum of the three side lengths: 12 + 9 + 15 = 36 cm.

Key takeaways

• Congruence means the same shape and size; the order of vertices identifies every pair of corresponding parts.
• SSS uses three side pairs, while SAS requires two side pairs and the equal included angles.
• AAS uses two angle pairs and a corresponding side; RHS additionally requires two right-angled triangles.
• Equal sides face equal angles, while a greater side faces a greater angle in the same triangle.
• Any two triangle sides together exceed the third, and a perpendicular gives the shortest point-to-line distance.
• The mid-point theorem gives both parallelism and half-length; its converse uses a midpoint and a parallel.
• Parallel lines cutting equal intercepts on one transversal cut equal intercepts on another transversal as well.
• Pythagoras’ theorem starts with a right angle; its converse uses the squared-length equality to establish that angle.

Test yourself

What does △ABC ≅ △RPQ say about corresponding vertices?

It matches A with R, B with P and C with Q.

Which angle is included between sides AB and BC?

It is ∠ABC, the angle at their common vertex B.

Why can two equal corresponding angles help establish the third angle pair?

The angles in each triangle total 180°, so subtracting equal angle pairs leaves equal third angles.

What must be checked before using RHS?

Both triangles must be right-angled, with equal hypotenuses and one equal pair of corresponding legs.

In △ABC, AB > AC. Which of ∠B and ∠C is greater?

∠C is greater because it lies opposite the greater side AB.

If P and Q are midpoints of AB and AC in △ABC, what follows?

The mid-point theorem gives PQ ∥ BC and PQ = BC/2.

Does a true statement automatically have a true converse?

No. The converse is a separate statement whose truth needs its own justification.

Which side should be tested as c when checking a² + b² = c²?

Use the longest side as c, opposite the proposed right angle.