This MCQ module is based on: Isosceles, Equilateral Triangles and Exercises
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Isosceles, Equilateral Triangles and Exercises
🎓 Class 7
Mathematics
CBSE
Theory
Ch 1 — Triangles
⏱ ~40 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Isosceles, Equilateral Triangles and Exercises
Targeting Class 7 level in General Mathematics, with Basic difficulty.
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1.3 Angles of Isosceles and Equilateral Triangles
Congruence is a very powerful tool for studying properties of geometric figures. Let us use it to discover an important property of isosceles triangles?.
\(\triangle ABC\) is isosceles with AB = AC, and \(\angle A = 80°\). What can we say about \(\angle B\) and \(\angle C\)?
Isosceles \(\triangle ABC\) with AB = AC (tick marks) and \(\angle A = 80°\). Altitude from A to BC drawn.
Construct the altitude from A to BC. We have:
- AB = AC (given)
- \(\angle ADB = \angle ADC = 90°\) (from construction)
- AD is common to both \(\triangle ADB\) and \(\triangle ADC\)
Thus, the triangles satisfy the RHS condition. Hence, \(\triangle ADB \cong \triangle ADC\).
This shows that \(\angle B = \angle C\), as they are corresponding parts of congruent triangles.
Can you use this fact to find \(\angle B\) and \(\angle C\)?
Since \(\angle A = 80°\) and \(\angle B = \angle C\), we have \(80° + 2\angle B = 180°\), giving \(\angle B = \angle C = 50°\).
Property of Isosceles Triangles
In an isosceles triangle, the angles opposite to the equal sides are equal. If AB = AC, then \(\angle B = \angle C\).
Angles in an Equilateral Triangle
Equilateral triangles? are those in which all three sides have equal lengths.
What can we say about the angles of an equilateral triangle?
Equilateral \(\triangle ABC\) — all sides equal, all angles = 60°
We can use the recently discovered fact that angles opposite to equal sides are equal:
- The sides AB and AC are equal. So \(\angle B = \angle C\).
- Similarly, the sides AB and BC are equal. So \(\angle A = \angle C\).
So all three angles of an equilateral triangle are equal, just like their sides. As the three angles should add up to 180°, we have:
\(3 \times \text{angle} = 180° \implies \text{each angle} = 60°\)
Property of Equilateral Triangles
All three angles of an equilateral triangle are equal, and each angle is 60°.
Congruent Triangles in Real Life
Congruent triangles can be seen in various constructions and designs from ancient to modern times. Here are a few examples:
Louvre Museum, Paris
Glass pyramid with congruent triangular panels
Glass pyramid with congruent triangular panels
Egyptian Pyramids, Giza
Four congruent triangular faces
Four congruent triangular faces
Geodesic Dome
Network of congruent triangles
Network of congruent triangles
Rangoli Design
Symmetric congruent triangle patterns
Symmetric congruent triangle patterns
Rabindra Setu (Howrah Bridge)
Congruent triangular trusses
Congruent triangular trusses
Describe the congruent triangles you see in each picture.
Figure it Out (Section 1.3 — Exercises)
Q1. \(\triangle AIR \cong \triangle FLY\). Identify the corresponding vertices, sides, and angles.
Answer:
Corresponding Vertices: A ↔ F, I ↔ L, R ↔ Y
Corresponding Sides: AI = FL, IR = LY, AR = FY
Corresponding Angles: \(\angle A = \angle F\), \(\angle I = \angle L\), \(\angle R = \angle Y\)
Corresponding Vertices: A ↔ F, I ↔ L, R ↔ Y
Corresponding Sides: AI = FL, IR = LY, AR = FY
Corresponding Angles: \(\angle A = \angle F\), \(\angle I = \angle L\), \(\angle R = \angle Y\)
Q2. Each of the following cases contains certain measurements taken from two triangles. Identify the pairs in which the triangles are congruent to each other, with reason. Express the congruence whenever they are congruent.
(a) AB = DE, BC = EF, CA = DF
(b) AB = EF, \(\angle A = \angle E\), AC = ED
(c) AB = DF, \(\angle B = \angle D = 90°\), AC = FE
(d) \(\angle A = \angle D\), \(\angle B = \angle E\), AC = DF
(e) AB = DF, \(\angle B = \angle F\), AC = DE
(b) AB = EF, \(\angle A = \angle E\), AC = ED
(c) AB = DF, \(\angle B = \angle D = 90°\), AC = FE
(d) \(\angle A = \angle D\), \(\angle B = \angle E\), AC = DF
(e) AB = DF, \(\angle B = \angle F\), AC = DE
Answers:
(a) AB = DE, BC = EF, CA = DF → All three sides match. SSS condition: \(\triangle ABC \cong \triangle DEF\) ✅
(b) AB = EF, \(\angle A = \angle E\), AC = ED → Two sides and included angle (the angle is at A/E, between AB-AC and EF-ED). SAS condition: \(\triangle ABC \cong \triangle EFD\) ✅
(c) AB = DF, \(\angle B = \angle D = 90°\), AC = FE → Right angle, hypotenuse AC = FE, side AB = DF. RHS condition: \(\triangle ABC \cong \triangle DFE\) ✅
(d) \(\angle A = \angle D\), \(\angle B = \angle E\), AC = DF → Two angles + non-included side. AAS condition: \(\triangle ABC \cong \triangle DEF\) ✅
(e) AB = DF, \(\angle B = \angle F\), AC = DE → Two sides and a non-included angle (SSA). SSA does NOT guarantee congruence. ❌ Not necessarily congruent.
(a) AB = DE, BC = EF, CA = DF → All three sides match. SSS condition: \(\triangle ABC \cong \triangle DEF\) ✅
(b) AB = EF, \(\angle A = \angle E\), AC = ED → Two sides and included angle (the angle is at A/E, between AB-AC and EF-ED). SAS condition: \(\triangle ABC \cong \triangle EFD\) ✅
(c) AB = DF, \(\angle B = \angle D = 90°\), AC = FE → Right angle, hypotenuse AC = FE, side AB = DF. RHS condition: \(\triangle ABC \cong \triangle DFE\) ✅
(d) \(\angle A = \angle D\), \(\angle B = \angle E\), AC = DF → Two angles + non-included side. AAS condition: \(\triangle ABC \cong \triangle DEF\) ✅
(e) AB = DF, \(\angle B = \angle F\), AC = DE → Two sides and a non-included angle (SSA). SSA does NOT guarantee congruence. ❌ Not necessarily congruent.
Q3. It is given that OB = OC, and OA = OD. Show that AB is parallel to CD. (Hint: AD is a transversal for these two lines. Are there any equal alternate angles?)
Answer: In \(\triangle AOB\) and \(\triangle DOC\):
OA = OD (given) ✓
OB = OC (given) ✓
\(\angle AOB = \angle DOC\) (vertically opposite angles) ✓
By SAS condition: \(\triangle AOB \cong \triangle DOC\)
By CPCT: \(\angle OAB = \angle ODC\)
These are alternate interior angles with AD as a transversal cutting lines AB and CD.
Since alternate interior angles are equal, AB ∥ CD. ∎
OA = OD (given) ✓
OB = OC (given) ✓
\(\angle AOB = \angle DOC\) (vertically opposite angles) ✓
By SAS condition: \(\triangle AOB \cong \triangle DOC\)
By CPCT: \(\angle OAB = \angle ODC\)
These are alternate interior angles with AD as a transversal cutting lines AB and CD.
Since alternate interior angles are equal, AB ∥ CD. ∎
Q4. ABCD is a square. Show that \(\triangle ABC \cong \triangle ADC\). Is \(\triangle ABC\) also congruent to \(\triangle CDA\)?
Answer: In \(\triangle ABC\) and \(\triangle ADC\):
AB = AD (sides of a square) ✓
BC = DC (sides of a square) ✓
AC = AC (common diagonal) ✓
By SSS: \(\triangle ABC \cong \triangle ADC\) ✅
Is \(\triangle ABC \cong \triangle CDA\)? Note that \(\triangle CDA\) is the same triangle as \(\triangle ADC\) (just written starting from C). The correspondence here is A↔C, B↔D, C↔A. Checking: AB = CD ✓, BC = DA ✓, AC = CA ✓. Yes, \(\triangle ABC \cong \triangle CDA\) as well. In fact, in a square, the diagonal creates two congruent triangles that can be expressed in multiple equivalent ways.
AB = AD (sides of a square) ✓
BC = DC (sides of a square) ✓
AC = AC (common diagonal) ✓
By SSS: \(\triangle ABC \cong \triangle ADC\) ✅
Is \(\triangle ABC \cong \triangle CDA\)? Note that \(\triangle CDA\) is the same triangle as \(\triangle ADC\) (just written starting from C). The correspondence here is A↔C, B↔D, C↔A. Checking: AB = CD ✓, BC = DA ✓, AC = CA ✓. Yes, \(\triangle ABC \cong \triangle CDA\) as well. In fact, in a square, the diagonal creates two congruent triangles that can be expressed in multiple equivalent ways.
Q5. Find \(\angle B\) and \(\angle C\), if A is the centre of the circle and the angle at A in the triangle is 130°.
Answer: Since A is the centre of the circle, AB and AC are radii, so AB = AC. This makes \(\triangle ABC\) isosceles.
By the isosceles triangle property: \(\angle B = \angle C\).
Using angle sum: \(\angle A + \angle B + \angle C = 180°\)
\(130° + 2\angle B = 180°\)
\(2\angle B = 50°\)
\(\angle B = \angle C = \mathbf{25°}\)
By the isosceles triangle property: \(\angle B = \angle C\).
Using angle sum: \(\angle A + \angle B + \angle C = 180°\)
\(130° + 2\angle B = 180°\)
\(2\angle B = 50°\)
\(\angle B = \angle C = \mathbf{25°}\)
Q6. Find the missing angles in the complex figure below. All line segments marked with a single '|' are equal to each other, and those marked with '||' are equal to each other, etc.
Answer: This complex figure requires systematically identifying isosceles triangles (sides with equal tick marks) and applying the property that angles opposite equal sides are equal. Then use the angle sum property (180° per triangle) and vertically opposite angles to find all missing values.
Strategy:
1. Identify all triangles formed by the diagonals and internal lines.
2. For each triangle, check which sides have equal marks (single | or double ||).
3. Use isosceles triangle property: angles opposite equal sides are equal.
4. Use angle sum property: angles in each triangle sum to 180°.
5. Use vertically opposite angles at intersection points.
6. Work outward from triangles with the most known angles, filling in the missing values step by step.
The key angles: 34°, 44°, 40°, 90°, 36°, 56°, 98°, 68°, 30° are distributed throughout, and each missing angle can be found using these properties in sequence.
Strategy:
1. Identify all triangles formed by the diagonals and internal lines.
2. For each triangle, check which sides have equal marks (single | or double ||).
3. Use isosceles triangle property: angles opposite equal sides are equal.
4. Use angle sum property: angles in each triangle sum to 180°.
5. Use vertically opposite angles at intersection points.
6. Work outward from triangles with the most known angles, filling in the missing values step by step.
The key angles: 34°, 44°, 40°, 90°, 36°, 56°, 98°, 68°, 30° are distributed throughout, and each missing angle can be found using these properties in sequence.
Puzzle: Expression Engineer
Puzzle: Draw lines and split the white region (7×7 grid minus the green centre square) into 6 smaller congruent regions. Each region must have the same shape and size!
Need a hint?
Hint: The total area of the white region = 7 × 7 − 1 × 1 = 49 − 1 = 48 unit squares. Dividing into 6 congruent regions means each region = 8 unit squares. Try using L-shaped or T-shaped pieces of 8 squares each. The symmetry of the figure (with the hole in the centre) is your friend — think about rotational symmetry!
Chapter Summary
- Figures that have the same shape and size are said to be congruent. These figures can be superimposed so that one fits exactly over the other via rotation/flipping.
- When two triangles have the same sidelengths, we say the SSS (Side Side Side) condition is satisfied. The SSS condition guarantees congruence.
- When two sides and the included angle of one triangle are equal to the two sides and the included angle of another triangle, we say the SAS (Side Angle Side) condition is satisfied. The SAS condition also guarantees congruence.
- When two angles and the included side of one triangle are equal to the two angles and the included side of another triangle, we say the ASA (Angle Side Angle) condition is satisfied. Congruence holds even if the side is not included between the two angles — the AAS (Angle Angle Side) condition. Congruence holds because knowing two angles determines the third.
- In a right-angled triangle, the side opposite to the right angle is called the hypotenuse. When a side and a hypotenuse of a right-angled triangle are equal to a side and the hypotenuse of another right-angled triangle, we say the RHS (Right Hypotenuse Side) condition is satisfied. This also guarantees congruence.
- Two triangles need not be congruent if two sides and a non-included angle are equal. The SSA condition does not guarantee congruence.
- In a triangle, angles opposite to equal sides are equal.
- The angles in an equilateral triangle are all 60°.
Activity: Finding Congruent Triangles Around You
L3 Apply
Predict: Can you find at least 5 examples of congruent triangles in your surroundings? Where do you expect to find them — in nature, architecture, or everyday objects?
- Look around your classroom, home, or neighbourhood for congruent triangles.
- For each example, describe where you found the triangles and which congruence condition (SSS, SAS, ASA, AAS, RHS) would verify their congruence.
- Take photographs or make sketches of at least 3 examples.
- For one example, measure the sides/angles and verify the congruence condition numerically.
Examples you might find:
- Butterfly wings: The triangular patterns on opposite wings are congruent (SSS)
- Roof trusses: The triangular supports on both sides of a roof are congruent (SAS — same two beams and included angle)
- Floor tiles: Triangular tiles are manufactured to be identical (SSS)
- Bridge trusses: Like the Howrah Bridge — repeated congruent triangles for structural strength
- Folded paper: Folding a rectangular sheet along its diagonal creates two congruent right triangles (RHS)
Competency-Based Questions
Scenario: An architect is designing a symmetric entrance gate for a park. The gate has two identical triangular panels on either side of a vertical central pillar. Each panel is an isosceles triangle with the base along the pillar. The pillar height is 3 metres, and each panel has a base of 2 metres and two equal slanting edges of 2.5 metres each.
Q1. Which congruence condition proves the two panels are identical?
L3 Apply
Answer: (a) SSS. Both panels have base = 2 m, and two equal slanting edges = 2.5 m each. All three sides match, so SSS directly applies.
Q2. The architect claims that each panel is an isosceles triangle, so the two base angles must be equal. Analyse this claim using the congruence property learned in Section 1.3.
L4 Analyse
Answer: The claim is correct. Since each panel is isosceles (two sides = 2.5 m), we can drop an altitude from the apex to the base. This creates two right triangles that are congruent by RHS (same hypotenuse = 2.5 m, same height, right angle). By CPCT, the base angles are equal. This is exactly the property proved in Section 1.3: in an isosceles triangle, angles opposite equal sides are equal.
Q3. If the architect changes one panel to have a base angle of 55° while keeping the sides the same, would the gate still be symmetric? Evaluate whether the panels would still be congruent.
L5 Evaluate
Answer: If all three sides are kept the same (2 m, 2.5 m, 2.5 m), then by SSS the triangles must be congruent regardless of the architect's intention. The base angles are completely determined by the side lengths — they cannot be independently changed. A triangle with sides 2, 2.5, 2.5 will always have the same angles. So either the architect is mistaken about changing the angle (SSS fixes all angles), or the sides would have to change too (breaking congruence). The gate's symmetry depends on congruence, which depends on SSS.
Q4. Design a verification procedure that a site engineer could use to confirm both panels are perfectly congruent without removing them from the gate. List the minimum measurements needed.
L6 Create
Answer — Verification Procedure:
Option A (SSS — 6 measurements):
Measure all 3 sides of each panel (3 + 3 = 6 measurements). If corresponding sides match: congruent. ✅ Most reliable.
Option B (SAS — 4 measurements):
Measure 2 sides + the included angle on each panel (2+1 per panel × 2 = 6 measurements). More efficient if angles are easier to measure.
Option C (Minimum — 3 measurements):
Since both panels share the same vertical pillar (common base), measure: (1) the base of panel 1, (2) the base of panel 2, and (3) one slanting edge of each. If bases are equal and slanting edges are equal, and both are isosceles (verify the other slanting edge), then SSS confirms congruence. Minimum: 3 distinct measurements (base + both slanting edges of one panel, then verify the other panel matches).
Option A (SSS — 6 measurements):
Measure all 3 sides of each panel (3 + 3 = 6 measurements). If corresponding sides match: congruent. ✅ Most reliable.
Option B (SAS — 4 measurements):
Measure 2 sides + the included angle on each panel (2+1 per panel × 2 = 6 measurements). More efficient if angles are easier to measure.
Option C (Minimum — 3 measurements):
Since both panels share the same vertical pillar (common base), measure: (1) the base of panel 1, (2) the base of panel 2, and (3) one slanting edge of each. If bases are equal and slanting edges are equal, and both are isosceles (verify the other slanting edge), then SSS confirms congruence. Minimum: 3 distinct measurements (base + both slanting edges of one panel, then verify the other panel matches).
Assertion–Reason Questions
Assertion (A): In an isosceles triangle with AB = AC, the angles \(\angle B\) and \(\angle C\) are equal.
Reason (R): The altitude from A to BC divides the triangle into two congruent triangles by the RHS condition.
Reason (R): The altitude from A to BC divides the triangle into two congruent triangles by the RHS condition.
Answer: (a) — Both are true. The altitude creates two right triangles with the same hypotenuse (AB = AC) and a common side (the altitude). By RHS, they are congruent, and by CPCT, \(\angle B = \angle C\). The reason correctly explains the assertion.
Assertion (A): An equilateral triangle has all angles equal to 60°.
Reason (R): In an equilateral triangle, all sides are equal, so by the isosceles triangle property, all pairs of opposite angles are equal, and since they sum to 180°, each must be 60°.
Reason (R): In an equilateral triangle, all sides are equal, so by the isosceles triangle property, all pairs of opposite angles are equal, and since they sum to 180°, each must be 60°.
Answer: (a) — Both are true. Since AB = AC, we get \(\angle B = \angle C\). Since AB = BC, we get \(\angle A = \angle C\). So \(\angle A = \angle B = \angle C\). With \(3x = 180°\), each angle = 60°. The reason is the correct explanation.
Frequently Asked Questions
What is the base angles theorem for isosceles triangles?
The base angles theorem states that in an isosceles triangle, the angles opposite the equal sides are themselves equal. If a triangle has AB equal to AC, then angle B equals angle C. This is proved by drawing a perpendicular from A to BC and using SAS congruence. NCERT Class 7 Part II Chapter 1 proves this theorem.
Why are all angles of an equilateral triangle 60 degrees?
In an equilateral triangle, all three sides are equal. By the base angles theorem, the angles opposite equal sides are equal. Since all three sides are equal, all three angles must be equal. The sum of angles in any triangle is 180 degrees, so each angle equals 60 degrees.
How do you prove two triangles congruent in exercises?
To prove congruence: identify given information, choose the appropriate criterion (SSS, SAS, ASA, AAS or RHS), list all three required equalities with reasons, and write the congruence statement with correct vertex correspondence. NCERT Class 7 Chapter 1 exercises require this structured approach.
What is the relationship between sides and angles in triangles?
In any triangle, the larger side is opposite the larger angle and vice versa. In an isosceles triangle, equal sides are opposite equal angles. These relationships help determine unknown sides and angles. NCERT Class 7 Ganita Prakash Part II explores these connections through congruence.
What types of questions come in Class 7 congruence exercises?
Class 7 congruence exercises include identifying congruent triangle pairs, stating the criterion used, finding unknown sides and angles using congruence, proving properties of isosceles and equilateral triangles, and applying congruence to solve geometric problems from NCERT Ganita Prakash Part II.
Frequently Asked Questions — Chapter 1
What is Isosceles, Equilateral Triangles and Exercises in NCERT Class 7 Mathematics?
Isosceles, Equilateral Triangles and Exercises is a key concept covered in NCERT Class 7 Mathematics, Chapter 1: Chapter 1. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Isosceles, Equilateral Triangles and Exercises step by step?
To solve problems on Isosceles, Equilateral Triangles and Exercises, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 1: Chapter 1?
The essential formulas of Chapter 1 (Chapter 1) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Isosceles, Equilateral Triangles and Exercises important for the Class 7 board exam?
Isosceles, Equilateral Triangles and Exercises is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Isosceles, Equilateral Triangles and Exercises?
Common mistakes in Isosceles, Equilateral Triangles and Exercises include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Isosceles, Equilateral Triangles and Exercises?
End-of-chapter NCERT exercises for Isosceles, Equilateral Triangles and Exercises cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.
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