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NCERT Exemplar · Class 9 Mathematics Triangles

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EXERCISE 7.1 1–11 (part 1 of 5)

  1. In each of the following, write the correct answer:

    Exercise 1

    Which of the following is not a criterion for congruence of triangles? (A) SAS (B) ASA (C) SSA (D) SSS

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    NCERT’s answer
    (C)
    (C) \(\displaystyle \mathrm{SSA}\)Two sides and a non-included angle can be satisfied by two different triangles (the ambiguous case), so SSA does not fix a unique triangle.

  2. Exercise 2

    If AB=QR,BC=PR\displaystyle \mathrm{AB}=\mathrm{QR}, \mathrm{BC}=\mathrm{PR} and CA=PQ\displaystyle \mathrm{CA}=\mathrm{PQ}, then (A) ΔABC≅ΔPQR\displaystyle \Delta \mathrm{ABC} \cong \Delta \mathrm{PQR} (B) ΔCBA≅ΔPRQ\displaystyle \Delta \mathrm{CBA} \cong \Delta \mathrm{PRQ} (C) ΔBAC≅ΔRPQ\displaystyle \Delta \mathrm{BAC} \cong \Delta \mathrm{RPQ} (D) ΔPQR≅ΔBCA\displaystyle \Delta \mathrm{PQR} \cong \Delta \mathrm{BCA}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle \Delta CBA \cong \Delta PRQ\)NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-1_Q2\[AB=QR,\quad BC=PR,\quad CA=PQ \]The vertex common to \(\displaystyle AB,BC\) is \(\displaystyle B\), matching the vertex common to \(\displaystyle QR,PR\), which is \(\displaystyle R\); continuing round the triangle gives \(\displaystyle C\leftrightarrow P\) and \(\displaystyle A\leftrightarrow Q\).

  3. Exercise 3

    In △ABC,AB=AC\displaystyle \triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC} and ∠B=50∘\displaystyle \angle \mathrm{B}=50^{\circ}. Then ∠C\displaystyle \angle \mathrm{C} is equal to (A) 40\displaystyle 40° (B) 50\displaystyle 50° (C) 80\displaystyle 80° (D) 130\displaystyle 130°

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 50^{\circ}\)NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-1_Q3\[AB=AC \quad\Rightarrow\quad \angle C=\angle B \quad\text{(angles opposite equal sides)} \] \[\angle C=\angle B=50^{\circ} \]

  4. Exercise 4

    In △ABC,BC=AB\displaystyle \triangle \mathrm{ABC}, \mathrm{BC}=\mathrm{AB} and ∠B=80∘\displaystyle \angle \mathrm{B}=80^{\circ}. Then ∠A\displaystyle \angle \mathrm{A} is equal to (A) 80\displaystyle 80° (B) 40\displaystyle 40° (C) 50\displaystyle 50° (D) 100\displaystyle 100°

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 50^{\circ}\)NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-1_Q4\[AB=BC \quad\Rightarrow\quad \angle C=\angle A \quad\text{(angles opposite equal sides)} \] \[\angle A+\angle B+\angle C=180^{\circ} \] \[2\angle A+80^{\circ}=180^{\circ}\quad\Rightarrow\quad \angle A=50^{\circ} \]

  5. Exercise 5

    In ΔPQR,∠R=∠P\displaystyle \Delta \mathrm{PQR}, \angle \mathrm{R}=\angle \mathrm{P} and QR=4 cm\displaystyle \mathrm{QR}=4 \mathrm{~cm} and PR=5 cm\displaystyle \mathrm{PR}=5 \mathrm{~cm}. Then the length of PQ is (A) 4\displaystyle 4 cm (B) 5\displaystyle 5 cm (C) 2\displaystyle 2 cm (D) 2.5\displaystyle 2.5 cm

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 4\ \text{cm}\)NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-1_Q5\[\angle R=\angle P \quad\Rightarrow\quad PQ=QR \quad\text{(sides opposite equal angles)} \] \[PQ=QR=4\ \text{cm} \]

  6. Exercise 6

    D is a point on the side BC of a ΔABC\displaystyle \Delta \mathrm{ABC} such that AD bisects ∠BAC\displaystyle \angle \mathrm{BAC}. Then (A) BD=CD\displaystyle \mathrm{BD}=\mathrm{CD} (B) BA>BD\displaystyle \mathrm{BA}>\mathrm{BD} (C) BD>BA\displaystyle \mathrm{BD}>\mathrm{BA} (D) CD>CA\displaystyle \mathrm{CD}>\mathrm{CA}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle BA>BD\)NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-1_Q6\[\angle ADB=\angle DAC+\angle ACD \quad\text{(exterior angle theorem)} \] \[\angle ADB>\angle DAC=\angle BAD \quad\text{(}AD\text{ bisects }\angle A\text{)} \]In \(\displaystyle \triangle ABD\), the greater angle \(\displaystyle \angle ADB\) faces side \(\displaystyle BA\), so \(\displaystyle BA>BD\).

  7. Exercise 7

    It is given that ΔABC≅ΔFDE\displaystyle \Delta \mathrm{ABC} \cong \Delta \mathrm{FDE} and AB=5 cm,∠ B=40∘\displaystyle \mathrm{AB}=5 \mathrm{~cm}, \angle \mathrm{~B}=40^{\circ} and ∠A=80∘\displaystyle \angle \mathrm{A}=80^{\circ}. Then which of the following is true? (A) DF=5 cm,∠ F=60∘\displaystyle \mathrm{DF}=5 \mathrm{~cm}, \angle \mathrm{~F}=60^{\circ} (B) DF=5 cm,∠E=60∘\displaystyle \mathrm{DF}=5 \mathrm{~cm}, \angle \mathrm{E}=60^{\circ} (C) DE=5 cm,∠E=60∘\displaystyle \mathrm{DE}=5 \mathrm{~cm}, \angle \mathrm{E}=60^{\circ} (D) DE=5 cm,∠D=40∘\displaystyle \mathrm{DE}=5 \mathrm{~cm}, \angle \mathrm{D}=40^{\circ}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle DF=5\ \text{cm},\ \angle E=60^{\circ}\)NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-1_Q7\[\Delta ABC\cong\Delta FDE \quad\Rightarrow\quad A\leftrightarrow F,\ B\leftrightarrow D,\ C\leftrightarrow E \] \[\angle C=180^{\circ}-\angle A-\angle B=180^{\circ}-80^{\circ}-40^{\circ}=60^{\circ} \] \[DF=AB=5\ \text{cm},\quad \angle E=\angle C=60^{\circ} \]

  8. Exercise 8

    Two sides of a triangle are of lengths 5\displaystyle 5 cm and 1.5\displaystyle 1.5 cm. The length of the third side of the triangle cannot be (A) 3.6\displaystyle 3.6 cm (B) 4.1\displaystyle 4.1 cm (C) 3.8\displaystyle 3.8 cm (D) 3.4\displaystyle 3.4 cm

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 3.4\ \text{cm}\)\[|5-1.5|<\text{third side}<5+1.5 \] \[3.5\ \text{cm}<\text{third side}<6.5\ \text{cm} \]$\displaystyle 3.4$ cm lies below $\displaystyle 3.5$ cm, so it cannot be the third side.

  9. Exercise 9

    In ΔPQR\displaystyle \Delta \mathrm{PQR}, if ∠R>∠Q\displaystyle \angle \mathrm{R}>\angle \mathrm{Q}, then (A) QR>PR\displaystyle \mathrm{QR}>\mathrm{PR} (B) PQ>PR\displaystyle \mathrm{PQ}>\mathrm{PR} (C) PQ < PR (D) QR<PR\displaystyle \mathrm{QR}<\mathrm{PR}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle PQ>PR\)NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-1_Q9\[\angle R>\angle Q \quad\Rightarrow\quad PQ>PR \quad\text{(side opposite the greater angle is longer)} \]

  10. Exercise 10

    In triangles ABC and PQR,AB=AC,∠C=∠P\displaystyle \mathrm{PQR}, \mathrm{AB}=\mathrm{AC}, \angle \mathrm{C}=\angle \mathrm{P} and ∠B=∠Q\displaystyle \angle \mathrm{B}=\angle \mathrm{Q}. The two triangles are (A) isosceles but not congruent (B) isosceles and congruent (C) congruent but not isosceles (D) neither congruent nor isosceles

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    (A)
    (A) isosceles but not congruentNCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-1_Q10\[AB=AC \quad\Rightarrow\quad \angle B=\angle C \quad\text{(angles opposite equal sides)} \] \[\angle C=\angle P,\ \angle B=\angle Q \quad\Rightarrow\quad \angle P=\angle Q \] \[\angle P=\angle Q \quad\Rightarrow\quad RQ=RP \quad\text{(sides opposite equal angles)} \]Equal angles fix shape (AAA), not size; with no side given equal between the two triangles, both are isosceles but need not be congruent.

  11. Exercise 11

    In triangles ABC and DEF,AB=FD\displaystyle \mathrm{DEF}, \mathrm{AB}=\mathrm{FD} and ∠A=∠D\displaystyle \angle \mathrm{A}=\angle \mathrm{D}. The two triangles will be congruent by SAS axiom if (A) BC=EF\displaystyle \mathrm{BC}=\mathrm{EF} (B) AC=DE\displaystyle \mathrm{AC}=\mathrm{DE} (C) AC=EF\displaystyle \mathrm{AC}=\mathrm{EF} (D) BC=DE\displaystyle \mathrm{BC}=\mathrm{DE}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle AC=DE\)NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-1_Q11\[AB=FD,\quad \angle A=\angle D \quad\text{(given)} \] \[AC=DE \quad\Rightarrow\quad \Delta ABC\cong\Delta DFE \quad\text{(SAS)} \]