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NCERT Exemplar · Class 9 Mathematics Triangles

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EXERCISE 7.2 1–12 (part 2 of 5)

  1. Exercise 1

    In triangles ABC and PQR,∠A=∠Q\displaystyle \mathrm{PQR}, \angle \mathrm{A}=\angle \mathrm{Q} and ∠B=∠R\displaystyle \angle \mathrm{B}=\angle \mathrm{R}. Which side of ΔPQR\displaystyle \Delta \mathrm{PQR} should be equal to side AB of ΔABC\displaystyle \Delta \mathrm{ABC} so that the two triangles are congruent? Give reason for your answer.

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    NCERT’s answer
    QR; They will be congruent by ASA.
    The equal angles fix the correspondence \(\displaystyle A\leftrightarrow Q,\ B\leftrightarrow R,\ C\leftrightarrow P\). Side \(\displaystyle AB\) lies between the two given angles, so its partner is the side between \(\displaystyle Q\) and \(\displaystyle R\). NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-2_Q1 \[AB=QR \implies \triangle ABC \cong \triangle QRP \quad \text{(ASA)} \] Answer: \(\displaystyle QR\).

  2. Exercise 2

    In triangles ABC and PQR,∠A=∠Q\displaystyle \mathrm{PQR}, \angle \mathrm{A}=\angle \mathrm{Q} and ∠B=∠R\displaystyle \angle \mathrm{B}=\angle \mathrm{R}. Which side of △PQR\displaystyle \triangle \mathrm{PQR} should be equal to side BC of ΔABC\displaystyle \Delta \mathrm{ABC} so that the two triangles are congruent? Give reason for your answer.

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    NCERT’s answer
    RP; They will be congruent by AAS.
    The correspondence \(\displaystyle A\leftrightarrow Q,\ B\leftrightarrow R,\ C\leftrightarrow P\) sends \(\displaystyle B\) to \(\displaystyle R\) and \(\displaystyle C\) to \(\displaystyle P\), so \(\displaystyle BC\), a side not between the two given angles, pairs with \(\displaystyle RP\). NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-2_Q2 \[BC=RP \implies \triangle ABC \cong \triangle QRP \quad \text{(AAS)} \] Answer: \(\displaystyle RP\).

  3. Exercise 3

    "If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent." Is the statement true? Why?

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    NCERT’s answer
    No; Angles must be included angles.
    False. The equal angle must be included between the two sides (SAS); an angle opposite one is ambiguous — SSA fits two different triangles. Answer: False — SSA is not a valid congruence test.

  4. Exercise 4

    "If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent." Is the statement true? Why?

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    NCERT’s answer
    No; Sides must be corresponding sides.
    False. The equal side must be a corresponding side. Counterexample: \(\displaystyle 30^\circ,60^\circ,90^\circ\) triangles with sides \[(1,\ \sqrt3,\ 2) \quad\text{and}\quad \Big(\tfrac{1}{\sqrt3},\ 1,\ \tfrac{2}{\sqrt3}\Big) \] share equal angles and a side of length \(\displaystyle 1\), yet \(\displaystyle 2\neq\dfrac{2}{\sqrt3}\): not congruent. Answer: False — the equal side must correspond to the same pair of angles in both triangles.

  5. Exercise 5

    Is it possible to construct a triangle with lengths of its sides as 4\displaystyle 4 cm, 3\displaystyle 3 cm and 7\displaystyle 7 cm? Give reason for your answer.

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    NCERT’s answer
    No; Sum of the two sides = the third side.
    The two shorter sides must together exceed the longest one. \[4+3=7 \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-2_Q5 \[4+3 \not> 7 \] Answer: No — equality makes the third vertex fall exactly on the opposite side, collapsing the triangle into a straight line.

  6. Exercise 6

    It is given that ΔABC≅ΔRPQ\displaystyle \Delta \mathrm{ABC} \cong \Delta \mathrm{RPQ}. Is it true to say that BC=QR\displaystyle \mathrm{BC}=\mathrm{QR} ? Why?

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    NCERT’s answer
    No; BC = PQ.
    False. \(\displaystyle \triangle ABC\cong\triangle RPQ\) pairs \(\displaystyle A\leftrightarrow R,\ B\leftrightarrow P,\ C\leftrightarrow Q\), so by CPCT \[AB=RP,\quad BC=PQ,\quad CA=QR \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-2_Q6 \(\displaystyle BC\) matches \(\displaystyle PQ\), not \(\displaystyle QR\). Answer: False — BC corresponds to PQ, not QR.

  7. Exercise 7

    If ΔPQR≅ΔEDF\displaystyle \Delta \mathrm{PQR} \cong \Delta \mathrm{EDF}, then is it true to say that PR=EF\displaystyle \mathrm{PR}=\mathrm{EF} ? Give reason for your answer.

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    NCERT’s answer
    Yes; They are corresponding sides.
    True. \(\displaystyle \triangle PQR\cong\triangle EDF\) pairs \(\displaystyle P\leftrightarrow E,\ Q\leftrightarrow D,\ R\leftrightarrow F\), so by CPCT \[PQ=ED,\quad QR=DF,\quad RP=EF \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-2_Q7 \(\displaystyle RP\), i.e. \(\displaystyle PR\), matches \(\displaystyle EF\) exactly. Answer: True — PR corresponds to EF.

  8. Exercise 8

    In ΔPQR,∠P=70∘\displaystyle \Delta \mathrm{PQR}, \angle \mathrm{P}=70^{\circ} and ∠R=30∘\displaystyle \angle \mathrm{R}=30^{\circ}. Which side of this triangle is the longest? Give reason for your answer.

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    NCERT’s answer
    PR; Side opposite the greater angle is longer.
    \[\angle Q = 180^\circ-\angle P-\angle R = 180^\circ-70^\circ-30^\circ = 80^\circ \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-2_Q8 \[\angle Q>\angle P>\angle R \implies PR>QR>PQ \] Answer: \(\displaystyle PR\) — the side opposite the greatest angle, \(\displaystyle \angle Q=80^\circ\), is the longest.

  9. Exercise 9

    AD is a median of the triangle ABC. Is it true that AB + BC + CA > 2\displaystyle 2 AD? Give reason for your answer.

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    NCERT’s answer
    Yes; $\displaystyle \mathrm{AB}+\mathrm{BD}>\mathrm{AD}$ and $\displaystyle \mathrm{AC}+\mathrm{CD}>\mathrm{AD}$.
    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-2_Q9 \[AB + BD > AD \quad \text{(triangle inequality, } \triangle ABD\text{)} \] \[AC + CD > AD \quad \text{(triangle inequality, } \triangle ACD\text{)} \] \[AB + AC + (BD+CD) > 2AD \] \[BD + DC = BC \quad \text{(D is the midpoint of BC)} \] \[\Rightarrow AB + BC + CA > 2AD \] Answer: Yes; \(\displaystyle AB+BC+CA>2AD\).

  10. Exercise 10

    M is a point on side BC of a triangle ABC such that AM is the bisector of ∠BAC\displaystyle \angle \mathrm{BAC}. Is it true to say that perimeter of the triangle is greater than 2\displaystyle 2 AM? Give reason for your answer.

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    NCERT’s answer
    Yes; $\displaystyle \mathrm{AB}+\mathrm{BM}>\mathrm{AM}$ and $\displaystyle \mathrm{AC}+\mathrm{CM}>\mathrm{AM}$.
    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-2_Q10 \[AB + BM > AM \quad \text{(triangle inequality, } \triangle ABM\text{)} \] \[AC + CM > AM \quad \text{(triangle inequality, } \triangle ACM\text{)} \] \[AB + AC + (BM+CM) > 2AM \] \[BM + MC = BC \quad \text{(M lies on BC)} \] \[\Rightarrow AB + BC + CA > 2AM \] Answer: Yes; \(\displaystyle AB+BC+CA>2AM\).

  11. Exercise 11

    Is it possible to construct a triangle with lengths of its sides as 9\displaystyle 9 cm, 7\displaystyle 7 cm and 17\displaystyle 17 cm? Give reason for your answer.

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    NCERT’s answer
    No; Sum of two sides is less than the third side.
    \[9+7=16 \] \[16<17 \quad \text{(triangle inequality: sum of two sides must exceed the third)} \] Answer: No; \(\displaystyle 9+7=16<17\), so no triangle has these side lengths.

  12. Exercise 12

    Is it possible to construct a triangle with lengths of its sides as 8\displaystyle 8 cm, 7\displaystyle 7 cm and 4\displaystyle 4 cm? Give reason for your answer.

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    NCERT’s answer
    Yes, because in each case the sum of two sides is greater than the third side.
    \[8+7=15>4 \] \[7+4=11>8 \] \[8+4=12>7 \quad \text{(triangle inequality holds for every pair)} \] Answer: Yes; sides $\displaystyle 8$ cm, $\displaystyle 7$ cm, $\displaystyle 4$ cm satisfy the triangle inequality, so the triangle can be constructed.