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NCERT Exemplar · Class 9 Mathematics Quadrilaterals

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EXERCISE 8.4 1–10 (part 6 of 7)

  1. Exercise 1

    A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

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

    Let square \(\displaystyle APQR\) share vertex \(\displaystyle A\) with the right angle of \(\displaystyle \triangle ABC\) (\(\displaystyle AB=AC\)), with \(\displaystyle P\) on \(\displaystyle AB\), \(\displaystyle R\) on \(\displaystyle AC\), and \(\displaystyle Q\) on \(\displaystyle BC\). \[AP = AR \quad \text{(sides of a square)} \] \[AB = AC \quad \text{(given, isosceles right triangle)} \] \[\Rightarrow BP = AB-AP = AC-AR = CR \] In \(\displaystyle \triangle BPQ\) and \(\displaystyle \triangle CRQ\): \[BP = CR, \quad PQ = QR \quad \text{(sides of the square)}, \quad \angle BPQ = \angle CRQ = 90^\circ \] \[\Rightarrow \triangle BPQ \cong \triangle CRQ \quad \text{(SAS)} \] \[\Rightarrow BQ = CQ \] NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q1 \(\displaystyle Q\) lies on \(\displaystyle BC\), so it bisects the hypotenuse. Answer: \(\displaystyle Q\), the vertex opposite \(\displaystyle A\), is the midpoint of \(\displaystyle BC\).

  2. Exercise 2

    In a parallelogram ABCD,AB=10 cm\displaystyle \mathrm{ABCD}, \mathrm{AB}=10 \mathrm{~cm} and AD=6 cm\displaystyle \mathrm{AD}=6 \mathrm{~cm}. The bisector of ∠A\displaystyle \angle \mathrm{A} meets DC in E . AE and BC produced meet at F . Find the length of CF .

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    NCERT’s answer
    $\displaystyle 4$ cm.
    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q2\[AD\parallel BC \quad \text{(opposite sides of a parallelogram)} \]\[\angle DAF=\angle AFB \quad \text{(alternate angles, transversal AF)} \]\[\angle DAF=\angle DAE=\angle EAB=\angle FAB \quad \text{(AE bisects }\angle A\text{; }A,E,F\text{ collinear)} \]\[\angle AFB=\angle FAB \implies BF=AB=10 \text{ cm} \quad \text{(sides opposite equal angles in }\triangle ABF\text{)} \]\[CF=BF-BC=10-6 \]Answer: \(\displaystyle CF=4\) cm.

  3. Exercise 3

    P,Q,R\displaystyle \mathrm{P}, \mathrm{Q}, \mathrm{R} and S are respectively the mid-points of the sides AB,BC,CD\displaystyle \mathrm{AB}, \mathrm{BC}, \mathrm{CD} and DA of a quadrilateral ABCD in which AC=BD\displaystyle \mathrm{AC}=\mathrm{BD}. Prove that PQRS is a rhombus.

    Not cross-checked

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

    \[PQ \parallel AC, \; PQ = \frac{1}{2}AC \quad \text{(midpoint theorem, } \triangle ABC\text{)} \] \[SR \parallel AC, \; SR = \frac{1}{2}AC \quad \text{(midpoint theorem, } \triangle ADC\text{)} \] \[QR \parallel BD, \; QR = \frac{1}{2}BD \quad \text{(midpoint theorem, } \triangle BCD\text{)} \] \[PS \parallel BD, \; PS = \frac{1}{2}BD \quad \text{(midpoint theorem, } \triangle ABD\text{)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q3 \(\displaystyle PQRS\) is a parallelogram, since \(\displaystyle PQ\parallel RS\) and \(\displaystyle QR\parallel PS\). \[PQ = SR = \frac{1}{2}AC, \quad QR = PS = \frac{1}{2}BD \] \[AC = BD \quad \text{(given)} \Rightarrow PQ = QR = RS = SP \] Answer: \(\displaystyle PQRS\) is a rhombus, as all four sides are equal.

  4. Exercise 4

    P,Q,R\displaystyle \mathrm{P}, \mathrm{Q}, \mathrm{R} and S are respectively the mid-points of the sides AB,BC,CD\displaystyle \mathrm{AB}, \mathrm{BC}, \mathrm{CD} and DA of a quadrilateral ABCD such that AC⊥BD\displaystyle \mathrm{AC} \perp \mathrm{BD}. Prove that PQRS is a rectangle.

    Not cross-checked

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

    \[PQ \parallel AC, \; QR \parallel BD, \; RS \parallel AC, \; SP \parallel BD \quad \text{(midpoint theorem)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q4 \(\displaystyle PQRS\) is a parallelogram, with \(\displaystyle PQ\parallel RS\) and \(\displaystyle QR \parallel SP\). \[AC \perp BD \quad \text{(given)} \] \[PQ \parallel AC, \; QR \parallel BD \Rightarrow \angle PQR = 90^\circ \] Answer: \(\displaystyle PQRS\) is a rectangle, since it is a parallelogram with a right angle.

  5. Exercise 5

    P,Q,R\displaystyle \mathrm{P}, \mathrm{Q}, \mathrm{R} and S are respectively the mid-points of sides AB,BC,CD\displaystyle \mathrm{AB}, \mathrm{BC}, \mathrm{CD} and DA of quadrilateral ABCD in which AC=BD\displaystyle \mathrm{AC}=\mathrm{BD} and AC⊥BD\displaystyle \mathrm{AC} \perp \mathrm{BD}. Prove that PQRS is a square.

    Not cross-checked

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

    From the midpoint theorem in \(\displaystyle \triangle ABC, \triangle BCD, \triangle ACD, \triangle ABD\): \[PQ = RS = \frac{1}{2}AC, \quad QR = SP = \frac{1}{2}BD, \quad PQ\parallel RS \parallel AC, \quad QR \parallel SP \parallel BD \] NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q5 \(\displaystyle PQRS\) is a parallelogram. \[AC = BD \quad \text{(given)} \Rightarrow PQ=QR=RS=SP \quad \text{(rhombus)} \] \[AC \perp BD \quad \text{(given)}, \; PQ\parallel AC, \; QR\parallel BD \Rightarrow \angle PQR = 90^\circ \quad \text{(rectangle)} \] Answer: \(\displaystyle PQRS\) is a square, being both a rhombus and a rectangle.

  6. Exercise 6

    A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

    Not cross-checked

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

    \[AB \parallel DC \quad \text{(opposite sides of the parallelogram)} \] \[\angle BAC = \angle DCA \quad \text{(alternate angles, transversal AC)} \] \[\angle DAC = \angle BAC \quad \text{(AC bisects } \angle A\text{)} \] \[\Rightarrow \angle DAC = \angle DCA \] NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q6 In \(\displaystyle \triangle ADC\), the side opposite \(\displaystyle \angle DAC\) equals the side opposite \(\displaystyle \angle DCA\): \[DC = AD \] \[AB = DC, \; AD = BC \quad \text{(opposite sides of the parallelogram)} \] \[\Rightarrow AB = BC = CD = DA \] Answer: \(\displaystyle ABCD\) is a rhombus, as all its sides are equal.

  7. Exercise 7

    P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.

    Not cross-checked

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

    \[AP=PB,\quad DQ=QC \quad \text{(P, Q are mid-points)} \]NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q7\[AB\parallel DC,\ AB=DC \quad \text{(given)} \implies AP\parallel QC,\ AP=QC \implies APCQ \text{ is a parallelogram} \implies AQ\parallel PC \] \[DQ\parallel PB,\ DQ=PB \implies DQBP \text{ is a parallelogram} \implies DP\parallel QB \] S lies on AQ and DP; R lies on CP and BQ, so \[SQ\parallel PR \quad (\subset AQ, PC), \qquad PS\parallel RQ \quad (\subset DP, QB) \]Answer: \(\displaystyle PRQS\) is a parallelogram.

  8. Exercise 8

    ABCD is a quadrilateral in which AB∥DC\displaystyle \mathrm{AB} \| \mathrm{DC} and AD=BC\displaystyle \mathrm{AD}=\mathrm{BC}. Prove that ∠A=∠B\displaystyle \angle \mathrm{A}=\angle \mathrm{B} and ∠C=∠D\displaystyle \angle \mathrm{C}=\angle \mathrm{D}.

    Not cross-checked

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

    Draw \(\displaystyle CE\parallel AD\), with \(\displaystyle E\) on \(\displaystyle AB\).NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q8\[AE\parallel DC,\ CE\parallel AD \implies AECD \text{ is a parallelogram} \implies CE=AD \] \[AD=BC \quad \text{(given)} \implies CE=BC \implies \angle CEB=\angle CBE \quad (\triangle BCE \text{ isosceles}) \] \[AD\parallel CE \implies \angle A+\angle CEA=180^\circ \quad \text{(co-interior angles)} \] \[\angle CEA+\angle CEB=180^\circ \quad \text{(linear pair)} \implies \angle A=\angle CEB \] E lies on AB, so \(\displaystyle \angle CBE=\angle B\). \[\implies \angle A=\angle CEB=\angle CBE=\angle B \] \[AB\parallel DC \implies \angle A+\angle D=180^\circ,\ \angle B+\angle C=180^\circ \implies \angle D=\angle C \]Answer: \(\displaystyle \angle A=\angle B\) and \(\displaystyle \angle C=\angle D\).

  9. Exercise 9

    In Fig. 8.11\displaystyle 8.11, AB∥DE,AB=DE,AC∥DF\displaystyle \mathrm{AB}\|\mathrm{DE}, \mathrm{AB}=\mathrm{DE}, \mathrm{AC}\| \mathrm{DF} and AC=DF\displaystyle \mathrm{AC}=\mathrm{DF}. Prove that BC∥EF\displaystyle \mathrm{BC} \| \mathrm{EF} and BC=EF\displaystyle \mathrm{BC}=\mathrm{EF}. NCERT_Question_Class9_Maths_Exemplar_Ch8_Ex8-4_Q9

    Not cross-checked

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

    \[AB\parallel DE,\ AB=DE \implies ABED \text{ is a parallelogram} \implies AD\parallel BE,\ AD=BE \]NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q9\[AC\parallel DF,\ AC=DF \implies ACFD \text{ is a parallelogram} \implies AD\parallel CF,\ AD=CF \] \[\implies BE\parallel CF,\ BE=CF \quad (\text{both}=AD) \implies BEFC \text{ is a parallelogram} \]Answer: \(\displaystyle BC\parallel EF\) and \(\displaystyle BC=EF\).

  10. Exercise 10

    E is the mid-point of a median AD of △ABC\displaystyle \triangle \mathrm{ABC} and BE is produced to meet AC at F. Show that AF=13AC\displaystyle \mathrm{AF}=\frac{1}{3} \mathrm{AC}.

    Not cross-checked

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

    Draw \(\displaystyle DG\parallel BF\), meeting \(\displaystyle AC\) at \(\displaystyle G\).NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q10\[\text{In } \triangle BCF:\ D \text{ mid-point of } BC,\ DG\parallel BF \implies G \text{ mid-point of } CF \quad \text{(converse, midpoint theorem)} \] \[CG=GF \] \[\text{In } \triangle ADG:\ E \text{ mid-point of } AD,\ EF\parallel DG \implies F \text{ mid-point of } AG \quad \text{(converse, midpoint theorem)} \] \[AF=FG \] \[AF=FG=GC \implies AC=AF+FG+GC=3\,AF \]Answer: \(\displaystyle AF=\dfrac{1}{3}AC\).