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NCERT Exemplar · Class 9 Mathematics Quadrilaterals

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EXERCISE 8.1 11–14 (part 2 of 7)

  1. Write the correct answer in each of the following:

    Exercise 11

    The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, (A) ABCD is a rhombus (B) diagonals of ABCD are equal (C) diagonals of ABCD are equal and perpendicular (D) diagonals of ABCD are perpendicular.

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    NCERT’s answer
    (C)
    (C) diagonals of ABCD are equal and perpendicularNCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-1_Q11\[WX=XY=YZ=ZW=\tfrac12 AC=\tfrac12 BD \quad \text{(midpoint theorem, }AC=BD\text{)} \] \[WX\parallel AC,\ XY\parallel BD,\ AC\perp BD\ \Rightarrow\ WX\perp XY \] Equal sides make WXYZ a rhombus; the right angle makes it a square, needing both diagonal conditions together.

  2. Exercise 12

    The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O . If ∠DAC=32∘\displaystyle \angle \mathrm{DAC}=32^{\circ} and ∠AOB=70∘\displaystyle \angle \mathrm{AOB}=70^{\circ}, then ∠DBC\displaystyle \angle \mathrm{DBC} is equal to (A) 24\displaystyle 24º (B) 86\displaystyle 86° (C) 38\displaystyle 38° (D) 32\displaystyle 32°

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 38^{\circ}\)NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-1_Q12\[\angle AOD=180^{\circ}-\angle AOB=180^{\circ}-70^{\circ}=110^{\circ} \quad \text{(linear pair on diagonal }BD\text{)} \] \[\angle ADB=180^{\circ}-\angle DAC-\angle AOD=180^{\circ}-32^{\circ}-110^{\circ}=38^{\circ} \quad \text{(angle sum of }\triangle AOD\text{)} \] \[\angle DBC=\angle ADB=38^{\circ} \quad \text{(alternate angles, }AD\parallel BC\text{)} \]

  3. Exercise 13

    Which of the following is not true for a parallelogram? (A) opposite sides are equal (B) opposite angles are equal (C) opposite angles are bisected by the diagonals (D) diagonals bisect each other.

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    NCERT’s answer
    (C)
    (C) opposite angles are bisected by the diagonalsNCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-1_Q13\[AD \parallel BC \implies \angle DAC = \angle BCA \] \[AB \parallel DC \implies \angle BAC = \angle DCA \]Diagonal \(\displaystyle AC\) bisects \(\displaystyle \angle A\) only if \(\displaystyle \angle BCA=\angle DCA\), true only when \(\displaystyle AB=AD\) — not for every parallelogram.

  4. Exercise 14

    D and E are the mid-points of the sides AB and AC respectively of ΔABC\displaystyle \Delta \mathrm{ABC}. DE is produced to F . To prove that CF is equal and parallel to DA , we need an additional information which is (A) ∠DAE=∠EFC\displaystyle \angle \mathrm{DAE}=\angle \mathrm{EFC} (B) AE=EF\displaystyle \mathrm{AE}=\mathrm{EF} (C) DE=EF\displaystyle \mathrm{DE}=\mathrm{EF} (D) ∠ADE=∠ECF\displaystyle \angle \mathrm{ADE}=\angle \mathrm{ECF}.

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    NCERT’s answer
    (C)
    (C) \(\displaystyle DE = EF\)NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-1_Q14\[AE = EC \quad \text{(E is midpoint of AC)} \] \[\angle AED = \angle CEF \quad \text{(vertically opposite angles)} \] \[DE = EF \implies \triangle ADE \cong \triangle CFE \quad \text{(SAS)} \] \[\implies AD = CF, \ \angle ADE = \angle CFE \]These are alternate angles for transversal \(\displaystyle DF\), so \(\displaystyle AB \parallel CF\), giving \(\displaystyle DA\) equal and parallel to \(\displaystyle CF\).