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NCERT Exemplar · Class 9 Mathematics Quadrilaterals

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EXERCISE 8.3 1–10 (part 5 of 7)

  1. Exercise 1

    One angle of a quadrilateral is of 108\displaystyle 108° and the remaining three angles are equal. Find each of the three equal angles.

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    NCERT’s answer
    $\displaystyle 84$°
    \[\angle A + \angle B + \angle C + \angle D = 360^\circ \quad \text{(angle sum of a quadrilateral)} \] \[108^\circ + 3x = 360^\circ \] \[3x = 252^\circ \] \[x = 84^\circ \]Answer: each of the three equal angles is \(\displaystyle 84^\circ\).

  2. Exercise 2

    ABCD is a trapezium in which AB∥DC\displaystyle \mathrm{AB} \| \mathrm{DC} and ∠A=∠B=45∘\displaystyle \angle \mathrm{A}=\angle \mathrm{B}=45^{\circ}. Find angles C and D of the trapezium.

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    NCERT’s answer
    $\displaystyle 135$° each
    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-3_Q2\[\angle A + \angle D = 180^\circ \quad \text{(co-interior angles, AB} \parallel \text{DC)} \] \[\angle D = 180^\circ - 45^\circ = 135^\circ \] \[\angle B + \angle C = 180^\circ \quad \text{(co-interior angles, AB} \parallel \text{DC)} \] \[\angle C = 180^\circ - 45^\circ = 135^\circ \]Answer: \(\displaystyle \angle C = \angle D = 135^\circ\).

  3. Exercise 3

    The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60\displaystyle 60°. Find the angles of the parallelogram.

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    NCERT’s answer
    $\displaystyle 120$°, $\displaystyle 60$°, $\displaystyle 120$°, $\displaystyle 60$° $\displaystyle 4.120$°, $\displaystyle 60$°, $\displaystyle 120$°, $\displaystyle 60$°
    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-3_Q3 \[\angle ALC = \angle AMC = 90^\circ \quad \text{(AL} \perp \text{DC, AM} \perp \text{BC)} \] \[\angle LAM + \angle ALC + \angle LCM + \angle AMC = 360^\circ \quad \text{(angle sum of quadrilateral ALCM)} \] \[60^\circ + 90^\circ + \angle LCM + 90^\circ = 360^\circ \] \[\angle LCM = 120^\circ \] \[\angle A = \angle C = \angle LCM = 120^\circ \quad \text{(opposite angles of a parallelogram)} \] \[\angle B = \angle D = 180^\circ - 120^\circ = 60^\circ \quad \text{(co-interior angles, AB} \parallel \text{DC)} \]Answer: \(\displaystyle \angle A = \angle C = 120^\circ\), \(\displaystyle \angle B = \angle D = 60^\circ\).

  4. Exercise 4

    ABCD is a rhombus in which altitude from D to side AB bisects AB . Find the angles of the rhombus.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-3_Q4 \[DM \perp AB, \ M \text{ is the midpoint of } AB \Rightarrow DA = DB \quad \text{(perpendicular bisector of AB)} \] \[DA = AB \quad \text{(sides of a rhombus)} \] \[\Rightarrow DA = AB = DB \Rightarrow \triangle ABD \text{ is equilateral} \] \[\angle A = 60^\circ \quad \text{(equilateral } \triangle ABD\text{)} \] \[\angle B = 180^\circ - \angle A = 120^\circ \quad \text{(adjacent angles of a rhombus)} \] \[\angle C = \angle A = 60^\circ, \quad \angle D = \angle B = 120^\circ \quad \text{(opposite angles of a rhombus)} \]Answer: \(\displaystyle \angle A = \angle C = 60^\circ\), \(\displaystyle \angle B = \angle D = 120^\circ\).

  5. Exercise 5

    E and F are points on diagonal AC of a parallelogram ABCD such that AE=CF\displaystyle \mathrm{AE}=\mathrm{CF}. Show that BFDE is a parallelogram.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-3_Q5 \[AC, BD \text{ bisect each other at } O \quad \text{(diagonals of parallelogram ABCD)} \] \[OA = OC, \quad OB = OD \] \[AE = CF \quad \text{(given)} \] \[OA - AE = OC - CF \Rightarrow OE = OF \] \[OB = OD, \ OE = OF \Rightarrow BD, EF \text{ bisect each other at } O \] \[\Rightarrow BFDE \text{ is a parallelogram} \quad \text{(diagonals bisect each other)} \]Answer: BFDE is a parallelogram, since its diagonals BD and EF bisect each other at O.

  6. Exercise 6

    E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC. [Hint: Join AC]

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-3_Q6 \[EF \parallel AB, \quad AB \parallel DC \Rightarrow EG \parallel DC \] \[E \text{ is the midpoint of } AD, \ EG \parallel DC \text{ in } \triangle ACD \Rightarrow AG = GC \quad \text{(converse of midpoint theorem)} \] \[G \text{ is the midpoint of } AC, \ GF \parallel AB \text{ in } \triangle ABC \Rightarrow BF = FC \quad \text{(converse of midpoint theorem)} \]Answer: F is the midpoint of BC.

  7. Exercise 7

    Through A, B and C, lines RQ, PR and QP have been drawn, respectively parallel to sides BC,CA\displaystyle \mathrm{BC}, \mathrm{CA} and AB of a △ABC\displaystyle \triangle \mathrm{ABC} as shown in Fig.8.5. Show that BC=12QR\displaystyle \mathrm{BC}=\frac{1}{2} \mathrm{QR}. NCERT_Question_Class9_Maths_Exemplar_Ch8_Ex8-3_Q7

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    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-3_Q7 \[AR \parallel BC, \quad RB \parallel CA \Rightarrow ARBC \text{ is a parallelogram} \] \[AR = BC \quad \text{(opposite sides of parallelogram ARBC)} \] \[AQ \parallel BC, \quad CQ \parallel AB \Rightarrow ABCQ \text{ is a parallelogram} \] \[AQ = BC \quad \text{(opposite sides of parallelogram ABCQ)} \] \[QR = QA + AR = BC + BC = 2\,BC \]Answer: \(\displaystyle BC = \dfrac{1}{2}QR\).

  8. Exercise 8

    D, E and F are the mid-points of the sides BC , CA and AB , respectively of an equilateral triangle ABC. Show that Δ\displaystyle \Delta DEF is also an equilateral triangle.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-3_Q8 \[D, E \text{ midpoints of } BC, CA \Rightarrow DE = \tfrac{1}{2}AB \quad \text{(midpoint theorem)} \] \[E, F \text{ midpoints of } CA, AB \Rightarrow EF = \tfrac{1}{2}BC \quad \text{(midpoint theorem)} \] \[F, D \text{ midpoints of } AB, BC \Rightarrow FD = \tfrac{1}{2}CA \quad \text{(midpoint theorem)} \] \[AB = BC = CA \quad \text{(} \triangle ABC \text{ equilateral)} \] \[\Rightarrow DE = EF = FD \]Answer: \(\displaystyle DE = EF = FD\), so \(\displaystyle \triangle DEF\) is equilateral.

  9. Exercise 9

    Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP=CQ\displaystyle \mathrm{AP}=\mathrm{CQ} (Fig. 8.6\displaystyle 8.6). Show that AC and PQ bisect each other. NCERT_Question_Class9_Maths_Exemplar_Ch8_Ex8-3_Q9

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-3_Q9 Let \(\displaystyle O\) be the point where \(\displaystyle PQ\) meets diagonal \(\displaystyle AC\). \[AB \parallel DC,\ AC \text{ transversal} \implies \angle PAO = \angle QCO \quad \text{(alternate angles)} \] \[AP = CQ \quad \text{(given)} \] \[\angle AOP = \angle COQ \quad \text{(vertically opposite angles)} \] \[\triangle AOP \cong \triangle COQ \quad \text{(AAS)} \] \[OA = OC, \quad OP = OQ \quad \text{(CPCT)} \] Answer: \(\displaystyle O\) bisects both \(\displaystyle AC\) and \(\displaystyle PQ\).

  10. Exercise 10

    In Fig. 8.7\displaystyle 8.7, P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP=∠DAP\displaystyle \angle \mathrm{BAP}=\angle \mathrm{DAP}. Prove that AD=2CD\displaystyle \mathrm{AD}=2 \mathrm{CD}. NCERT_Question_Class9_Maths_Exemplar_Ch8_Ex8-3_Q10

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-3_Q10 \[AD \parallel BC \implies \angle DAP = \angle APB \quad \text{(alternate angles, } AP \text{ transversal)} \] \[\angle BAP = \angle DAP \quad \text{(given)} \] \[\angle BAP = \angle APB \implies AB = BP \quad \text{(sides opp. equal angles, } \triangle ABP\text{)} \] \[P \text{ midpoint of } BC \implies BP = \tfrac12 BC = \tfrac12 AD \] \[AB = CD \quad \text{(opposite sides of parallelogram)} \] \[CD = AB = BP = \tfrac12 AD \] Answer: \(\displaystyle AD = 2\,CD\).