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NCERT Exemplar · Class 9 Mathematics Quadrilaterals

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EXERCISE 8.2 1–10 (part 3 of 7)

  1. Exercise 1

    Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA=3 cm\displaystyle \mathrm{OA}=3 \mathrm{~cm} and OD=2 cm\displaystyle \mathrm{OD}=2 \mathrm{~cm}, determine the lengths of AC and BD .

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    $\displaystyle 6$ cm, $\displaystyle 4$ cm; Diagonals of a parallelogram bisect each other.
    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-2_Q1 \[OA = OC = 3\text{ cm}, \quad OB = OD = 2\text{ cm} \quad \text{(diagonals of a parallelogram bisect each other)} \] \[AC = OA + OC = 3 + 3 = 6\text{ cm} \] \[BD = OB + OD = 2 + 2 = 4\text{ cm} \] Answer: \(\displaystyle AC = 6\text{ cm}, \ BD = 4\text{ cm} \).

  2. Exercise 2

    Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reason for your answer.

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    No; Diagonals of a parallelogram bisect each other.
    False.Diagonals of a parallelogram bisect each other, but they are perpendicular only in a special case, a rhombus — not for a general parallelogram.NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-2_Q2Answer: False.

  3. Exercise 3

    Can the angles 110\displaystyle 110°, 80\displaystyle 80°, 70\displaystyle 70° and 95\displaystyle 95° be the angles of a quadrilateral? Why or why not?

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    No; Angle sum must be $\displaystyle 360$°.
    \[110^\circ + 80^\circ + 70^\circ + 95^\circ = 355^\circ \] The angle sum property of a quadrilateral requires the four angles to total \(\displaystyle 360^\circ \), and \(\displaystyle 355^\circ \neq 360^\circ \). Answer: No, these angles cannot be the angles of a quadrilateral.

  4. Exercise 4

    In quadrilateral ABCD,∠A+∠D=180∘\displaystyle \mathrm{ABCD}, \angle \mathrm{A}+\angle \mathrm{D}=180^{\circ}. What special name can be given to this quadrilateral?

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    NCERT’s answer
    Trapezium.
    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-2_Q4 \[\angle A + \angle D = 180^\circ \] \[\Rightarrow AB \parallel DC \quad \text{(co-interior angles on transversal AD)} \] Answer: ABCD is a trapezium.

  5. Exercise 5

    All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?

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    NCERT’s answer
    Rectangle.
    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-2_Q5 \[4\angle A = 360^\circ \quad \text{(angle sum property of a quadrilateral)} \] \[\angle A = 90^\circ \] Each angle is a right angle, but the sides need not be equal. Answer: Rectangle.

  6. Exercise 6

    Diagonals of a rectangle are equal and perpendicular. Is this statement true? Give reason for your answer.

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    NCERT’s answer
    No; Diagonals of a rectangle need not be perpendicular.
    False. NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-2_Q6 A rectangle's diagonals are equal, but they are perpendicular only when the rectangle is a square.Answer: False.

  7. Exercise 7

    Can all the four angles of a quadrilateral be obtuse angles? Give reason for your answer.

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    NCERT’s answer
    No; sum of the angles of a quadrilateral is $\displaystyle 360$°.
    \[\angle A + \angle B + \angle C + \angle D = 360^\circ \quad \text{(angle sum property of a quadrilateral)} \] If all four angles were obtuse, each exceeds \(\displaystyle 90^\circ \) and their sum would exceed \(\displaystyle 360^\circ \), which is impossible. Answer: No, all four angles of a quadrilateral cannot be obtuse.

  8. Exercise 8

    In △ABC,AB=5 cm,BC=8 cm\displaystyle \triangle \mathrm{ABC}, \mathrm{AB}=5 \mathrm{~cm}, \mathrm{BC}=8 \mathrm{~cm} and CA=7 cm\displaystyle \mathrm{CA}=7 \mathrm{~cm}. If D and E are respectively the mid-points of AB and BC, determine the length of DE.

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    NCERT’s answer
    3.$\displaystyle 5$ cm, as $\displaystyle \mathrm{DE}=\frac{1}{2} \mathrm{AC}$.
    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-2_Q8 \[AD = DB, \quad BE = EC \quad \text{(D, E are mid-points)} \] \[DE = \tfrac{1}{2}AC \quad \text{(mid-point theorem)} \] \[DE = \tfrac{1}{2} \times 7 = 3.5\text{ cm} \] Answer: \(\displaystyle DE = 3.5\text{ cm} \).

  9. Exercise 9

    In Fig.8.1, it is given that BDEF and FDCE are parallelograms. Can you say that BD=CD\displaystyle \mathrm{BD}=\mathrm{CD} ? Why or why not? NCERT_Question_Class9_Maths_Exemplar_Ch8_Ex8-2_Q9

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    NCERT’s answer
    Yes; because $\displaystyle \mathrm{BD}=\mathrm{EF}$ and $\displaystyle \mathrm{CD}=\mathrm{EF}$.
    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-2_Q9\[BD = FE \quad \text{(opposite sides of parallelogram } BDEF\text{)} \] \[DC = FE \quad \text{(opposite sides of parallelogram } FDCE\text{)} \]Both equal \(\displaystyle FE\), so \(\displaystyle BD = DC\).Answer: Yes, \(\displaystyle BD = DC\).

  10. Exercise 10

    In Fig.8.2, ABCD and AEFG are two parallelograms. If ∠C=55∘\displaystyle \angle \mathrm{C}=55^{\circ}, determine ∠F\displaystyle \angle \mathrm{F}. NCERT_Question_Class9_Maths_Exemplar_Ch8_Ex8-2_Q10

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    NCERT’s answer
    $\displaystyle 55^{\circ}, \angle \mathrm{F}=\angle \mathrm{A}$ and $\displaystyle \angle \mathrm{A}=\angle \mathrm{C}$.
    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-2_Q10\[\angle A = \angle C = 55^\circ \quad \text{(opposite angles of parallelogram } ABCD\text{)} \]\(\displaystyle AE\) lies along \(\displaystyle AB\) and \(\displaystyle AG\) along \(\displaystyle AD\), so \(\displaystyle \angle GAE = \angle DAB\).\[\angle F = \angle A \quad \text{(opposite angles of parallelogram } AEFG\text{)} \] \[\angle F = 55^\circ \]Answer: \(\displaystyle \angle F = 55^\circ\).