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NCERT Exemplar · Class 9 Mathematics Quadrilaterals

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EXERCISE 8.4 11–18 (part 7 of 7)

  1. Exercise 11

    Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.

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

    Let \(\displaystyle P,Q,R,S\) be the mid-points of \(\displaystyle AB,BC,CD,DA\) of square \(\displaystyle ABCD\).NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q11\[\text{In } \triangle ABC:\ PQ\parallel AC,\ PQ=\tfrac12AC \quad \text{(midpoint theorem)} \] \[\text{In } \triangle ADC:\ SR\parallel AC,\ SR=\tfrac12AC \quad \text{(midpoint theorem)} \implies PQ\parallel SR,\ PQ=SR \implies PQRS \text{ is a parallelogram} \] \[\text{In } \triangle ABD,\ \triangle CBD:\ PS\parallel BD,\ PS=\tfrac12BD; \quad QR\parallel BD,\ QR=\tfrac12BD \quad \text{(midpoint theorem)} \] \[AC=BD \quad \text{(diagonals of a square)} \implies PQ=QR=RS=SP \] \[AC\perp BD \quad \text{(diagonals of a square)},\ PQ\parallel AC,\ QR\parallel BD \implies \angle PQR=90^\circ \]Answer: \(\displaystyle PQRS\) is a square (a rhombus with a right angle).

  2. Exercise 12

    E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD . Prove that EF∥AB\displaystyle \mathrm{EF} \| \mathrm{AB} and EF=12(AB+CD)\displaystyle \mathrm{EF}=\frac{1}{2}(\mathrm{AB}+\mathrm{CD}). [Hint: Join BE and produce it to meet CD produced at G.]

    Not cross-checked

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

    Join \(\displaystyle BE\) and produce it to meet \(\displaystyle CD\) produced at \(\displaystyle G\).NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q12\[\text{In } \triangle AEB,\ \triangle DEG:\ AE=ED\ (E \text{ mid-point}),\ \angle AEB=\angle DEG\ \text{(vert. opp.)},\ \angle EAB=\angle EDG\ \text{(alt., } AB\parallel CD) \] \[\triangle AEB\cong\triangle DEG \quad \text{(ASA)} \implies EB=EG,\ AB=DG \] \[\text{In } \triangle BGC:\ E \text{ mid-point of } BG,\ F \text{ mid-point of } BC \implies EF\parallel GC,\ EF=\tfrac12GC \quad \text{(midpoint theorem)} \] \[GC=GD+DC=AB+CD \] \[EF=\tfrac12(AB+CD),\quad EF\parallel GC\parallel AB \quad (CD\parallel AB) \]Answer: \(\displaystyle EF\parallel AB\) and \(\displaystyle EF=\dfrac12(AB+CD)\).

  3. Exercise 13

    Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram 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.

    Let the bisectors of \(\displaystyle \angle A,\angle B,\angle C,\angle D\) meet at \(\displaystyle P,Q,R,S\) as shown.NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q13\[\angle A+\angle B=180^\circ,\quad \angle B+\angle C=180^\circ,\quad \angle C+\angle D=180^\circ,\quad \angle D+\angle A=180^\circ \quad \text{(co-interior angles)} \]\[\text{In } \triangle APB:\ \angle APB=180^\circ-\tfrac12(\angle A+\angle B)=180^\circ-90^\circ=90^\circ \]\[\angle SPQ=\angle APB=90^\circ \quad \text{(same angle: } S \text{ lies on } PA,\ Q \text{ lies on } PB\text{)} \]\[\angle PQR=\angle QRS=\angle RSP=90^\circ \quad \text{(same argument at each vertex)} \]Answer: all four angles of \(\displaystyle PQRS\) are \(\displaystyle 90^\circ\), so \(\displaystyle PQRS\) is a rectangle.

  4. Exercise 14

    P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD . Show that PQ is bisected at O.

    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_Ex8-4_Q14\[AO=CO \quad \text{(diagonals of a parallelogram bisect each other)} \]\[\angle PAO=\angle QCO \quad \text{(alternate angles, } AD\parallel BC\text{, transversal } AC\text{)} \]\[\angle AOP=\angle COQ \quad \text{(vertically opposite angles)} \]\[\triangle AOP\cong\triangle COQ \quad \text{(ASA)} \]Answer: \(\displaystyle OP=OQ\), so \(\displaystyle PQ\) is bisected at \(\displaystyle O\).

  5. Exercise 15

    ABCD is a rectangle in which diagonal BD bisects ∠B. Show that ABCD 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.

    NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q15\[\angle B=90^\circ \quad \text{(angle of a rectangle)} \]\[\angle ABD=\tfrac12\angle B=45^\circ \quad \text{(}BD\text{ bisects }\angle B\text{)} \]\[\angle A=90^\circ \quad \text{(angle of a rectangle)} \]\[\angle ADB=180^\circ-\angle A-\angle ABD=180^\circ-90^\circ-45^\circ=45^\circ \quad \text{(angle sum, } \triangle ABD\text{)} \]\[\angle ABD=\angle ADB \implies AB=AD \quad \text{(sides opposite equal angles)} \]Answer: \(\displaystyle AB=AD\); a rectangle with a pair of equal adjacent sides is a square, so \(\displaystyle ABCD\) is a square.

  6. Exercise 16

    D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles.

    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_Ex8-4_Q16\[FD=\tfrac12BC=BE=EC,\quad DE=\tfrac12AC=AF=FC,\quad EF=\tfrac12AB=AD=DB \quad \text{(midpoint theorem, applied to each side)} \]\[\triangle ADF\cong\triangle EFD \quad (AD=EF,\ DF=FD,\ FA=DE;\ \text{SSS}) \]\[\triangle BDE\cong\triangle FED \quad (BD=EF,\ BE=DF,\ DE=DE;\ \text{SSS}) \]\[\triangle CEF\cong\triangle DFE \quad (CE=DF,\ CF=DE,\ EF=EF;\ \text{SSS}) \]Answer: \(\displaystyle \triangle ADF\), \(\displaystyle \triangle BDE\), \(\displaystyle \triangle CEF\) and \(\displaystyle \triangle DEF\) are all congruent (SSS).

  7. Exercise 17

    Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.

    Not cross-checked

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

    Let \(\displaystyle ABCD\) be a trapezium with \(\displaystyle AB\parallel DC\), and let \(\displaystyle E,F\) be the mid-points of diagonals \(\displaystyle AC,BD\). Let \(\displaystyle G\) be the mid-point of \(\displaystyle BC\).NCERT_Solution_Class9_Maths_Exemplar_Ch8_Ex8-4_Q17\[\text{In } \triangle ABC:\ EG\parallel AB \quad \text{(midpoint theorem; } E,G \text{ mid-points of } AC,BC\text{)} \]\[\text{In } \triangle DBC:\ FG\parallel DC \quad \text{(midpoint theorem; } F,G \text{ mid-points of } BD,BC\text{)} \]\[AB\parallel DC \quad \text{(given)} \implies EG\parallel FG \implies E,G,F \text{ are collinear} \]Answer: \(\displaystyle EF\parallel AB\parallel DC\).

  8. Exercise 18

    P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA=AR\displaystyle \mathrm{DA}=\mathrm{AR} and CQ=QR\displaystyle \mathrm{CQ}=\mathrm{QR}.

    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_Ex8-4_Q18\[\text{In } \triangle DRC:\ PA\parallel CR,\ P \text{ mid-point of } DC \implies A \text{ mid-point of } DR \quad \text{(converse, midpoint theorem)} \]\[DA=AR \]\[AB\parallel DC \implies AQ\parallel DC \quad (Q\in AB) \]\[\text{In } \triangle DRC:\ A \text{ mid-point of } DR,\ AQ\parallel DC \implies Q \text{ mid-point of } RC \quad \text{(converse, midpoint theorem)} \]\[CQ=QR \]Answer: \(\displaystyle DA=AR\) and \(\displaystyle CQ=QR\).