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Triangles
EXERCISE 6.3
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
All the congruent figures are similar but the converse is not true.
Step-by-Step Solution
Key Concept: Congruent figures have exactly the same size and shape (all corresponding sides and angles are equal). Similar figures have the same shape (all corresponding angles are equal) but may differ in size; therefore every congruent pair is automatically similar, whereas two similar figures need not be congruent.
1. Recall the definitions
- Two figures are *congruent* if their corresponding sides are equal in length and the corresponding angles are equal in measure. Symbolically, \(\triangle ABC \cong \triangle DEF\) means \(AB = DE, BC = EF, CA = FD\) and \(\angle A = \angle D, \angle B = \angle E, \angle C = \angle F\).
- Two figures are *similar* if their corresponding angles are equal and the corresponding sides are in proportion. Symbolically, \(\triangle ABC \sim \triangle DEF\) means \(\angle A = \angle D, \angle B = \angle E, \angle C = \angle F\) and \(\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=k\) for some constant \(k>0\).
2. Show that congruence \(\Rightarrow\) similarity
- If \(\triangle ABC \cong \triangle DEF\), then by definition all corresponding sides are equal, i.e., \(AB = DE, BC = EF, CA = FD\). Hence the ratio of any pair of corresponding sides is \(\frac{AB}{DE}=1\), \(\frac{BC}{EF}=1\), \(\frac{CA}{FD}=1\).
- Since the ratios are equal (all equal to 1) and the corresponding angles are equal, the condition for similarity is satisfied. Therefore, congruent triangles are always similar.
3. Explain why the converse is not true
- The converse would state: "If two figures are similar, then they are congruent." This is false because similarity allows a common scale factor \(k
eq 1\). When \(k
eq 1\), the corresponding sides are not equal, so the figures are not congruent.
- Counter‑example: Consider two right‑angled triangles \(\triangle ABC\) and \(\triangle DEF\) where \(\angle A = \angle D = 90^{\circ}\), \(\angle B = \angle E = 30^{\circ}\), \(\angle C = \angle F = 60^{\circ}\). Let the sides of \(\triangle ABC\) be \(AB = 6\,\text{cm}, BC = 3\,\text{cm}, AC = 3\sqrt{3}\,\text{cm}\). Let the sides of \(\triangle DEF\) be \(DE = 12\,\text{cm}, EF = 6\,\text{cm}, DF = 6\sqrt{3}\,\text{cm}\).
- Here, \(\frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC}=2\). All corresponding angles are equal, so the triangles are similar with scale factor \(k=2\). However, the side lengths are not equal; hence the triangles are not congruent.
4. Conclusion
- Every pair of congruent figures is automatically similar (scale factor \(k=1\)).
- The converse fails because similarity permits a non‑unit scale factor, leading to figures that have the same shape but different sizes.
Hence, the statement is proved and a counter‑example is provided.
- Two figures are *congruent* if their corresponding sides are equal in length and the corresponding angles are equal in measure. Symbolically, \(\triangle ABC \cong \triangle DEF\) means \(AB = DE, BC = EF, CA = FD\) and \(\angle A = \angle D, \angle B = \angle E, \angle C = \angle F\).
- Two figures are *similar* if their corresponding angles are equal and the corresponding sides are in proportion. Symbolically, \(\triangle ABC \sim \triangle DEF\) means \(\angle A = \angle D, \angle B = \angle E, \angle C = \angle F\) and \(\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=k\) for some constant \(k>0\).
2. Show that congruence \(\Rightarrow\) similarity
- If \(\triangle ABC \cong \triangle DEF\), then by definition all corresponding sides are equal, i.e., \(AB = DE, BC = EF, CA = FD\). Hence the ratio of any pair of corresponding sides is \(\frac{AB}{DE}=1\), \(\frac{BC}{EF}=1\), \(\frac{CA}{FD}=1\).
- Since the ratios are equal (all equal to 1) and the corresponding angles are equal, the condition for similarity is satisfied. Therefore, congruent triangles are always similar.
3. Explain why the converse is not true
- The converse would state: "If two figures are similar, then they are congruent." This is false because similarity allows a common scale factor \(k
eq 1\). When \(k
eq 1\), the corresponding sides are not equal, so the figures are not congruent.
- Counter‑example: Consider two right‑angled triangles \(\triangle ABC\) and \(\triangle DEF\) where \(\angle A = \angle D = 90^{\circ}\), \(\angle B = \angle E = 30^{\circ}\), \(\angle C = \angle F = 60^{\circ}\). Let the sides of \(\triangle ABC\) be \(AB = 6\,\text{cm}, BC = 3\,\text{cm}, AC = 3\sqrt{3}\,\text{cm}\). Let the sides of \(\triangle DEF\) be \(DE = 12\,\text{cm}, EF = 6\,\text{cm}, DF = 6\sqrt{3}\,\text{cm}\).
- Here, \(\frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC}=2\). All corresponding angles are equal, so the triangles are similar with scale factor \(k=2\). However, the side lengths are not equal; hence the triangles are not congruent.
4. Conclusion
- Every pair of congruent figures is automatically similar (scale factor \(k=1\)).
- The converse fails because similarity permits a non‑unit scale factor, leading to figures that have the same shape but different sizes.
Hence, the statement is proved and a counter‑example is provided.
Correct Answer: The statement is true. Congruent figures are always similar because they have equal corresponding sides (scale factor = 1) and equal angles. The converse is false; similar figures need not be congruent. A counter‑example is two right‑angled triangles with angles \(90^{\circ},30^{\circ},60^{\circ}\) where one triangle’s sides are twice those of the other. They are similar (same angles) but not congruent (different side lengths).
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