Quadrilaterals CBSE Class 9 Mathematics Worksheets Set 07

Practice Set

Question 1. Prove that the area of a rhombus is half the product of its diagonals.
Answer:
The diagonals of a rhombus bisect each other at right angles and divide it into four congruent right triangles with legs \( \frac{d_1}2 \) and \( \frac{d_2}2 \).
Area \( =4\times\frac12\cdot\frac{d_1}2\cdot\frac{d_2}2=\frac12d_1d_2 \).

Question 2. In quadrilateral \( ABCD \), \( CO \) and \( DO \) bisect \( \angle C \) and \( \angle D \). Prove \( \angle COD=\frac12(\angle A+\angle B) \).
Answer:
In \( \triangle COD \): \( \angle COD=180^\circ-\frac12(\angle C+\angle D) \).
\( \angle C+\angle D=360^\circ-(\angle A+\angle B) \), so \( \angle COD=180-180+\frac12(\angle A+\angle B)=\frac12(\angle A+\angle B) \).

Question 3. In parallelogram \( PQRS \), \( PL \) and \( RM \) are the bisectors of \( \angle P \) and \( \angle R \) (\( L \) on \( SR \), \( M \) on \( PQ \)). Prove that \( PMRL \) is a parallelogram.
Answer:
\( \angle P=\angle R \) (opposite angles of a parallelogram), so \( \angle LPQ=\frac12\angle P=\frac12\angle R=\angle SRM \).
\( \angle SRM=\angle RMQ \) (alternate angles, \( SR\parallel PQ \)), so \( \angle LPM=\angle RMQ \); these are corresponding angles for lines \( PL \), \( MR \), hence \( PL\parallel MR \).
Also \( LR\parallel PM \) (parts of \( SR\parallel PQ \)). Both pairs of opposite sides are parallel, so \( PMRL \) is a parallelogram.

Question 4. Prove that the line segment joining the mid-points of the non-parallel sides of a trapezium is parallel to the parallel sides and equal to half their sum.
Answer:
Let \( E,F \) be the mid-points of \( AD \) and \( BC \) in trapezium \( ABCD \) (\( AB\parallel DC \)). Join \( AF \) and produce it to meet \( DC \) produced at \( P \).
\( \triangle AFB\cong\triangle PFC \) (ASA: \( BF=CF \), vertically opposite angles, alternate angles), so \( AF=FP \) and \( CP=AB \).
In \( \triangle ADP \), \( E \) and \( F \) are mid-points of \( AD \) and \( AP \), so \( EF\parallel DP \) and \( EF=\frac12DP=\frac12(DC+CP)=\frac12(AB+CD) \).

Question 5. In a parallelogram \( ABCD \), \( P \) is on \( AD \) with \( AP=\frac13AD \) and \( Q \) is on \( BC \) with \( CQ=\frac13BC \). Prove that \( AQCP \) is a parallelogram.
Answer:
\( AD=BC \) gives \( AP=CQ \), and \( AD\parallel BC \) gives \( AP\parallel QC \).
One pair of opposite sides (\( AP \) and \( QC \)) is equal and parallel, so \( AQCP \) is a parallelogram.

Question 6. \( ABC \) is a triangle right-angled at \( C \). A line through the mid-point \( M \) of the hypotenuse \( AB \) and parallel to \( BC \) meets \( AC \) at \( D \). Show that (i) \( D \) is the mid-point of \( AC \), (ii) \( MD\perp AC \), (iii) \( CM=MA=\frac12AB \).
Answer:
(i) \( M \) is the mid-point of \( AB \) and \( MD\parallel BC \), so by the converse of the mid-point theorem \( D \) is the mid-point of \( AC \).
(ii) \( MD\parallel BC \) and \( BC\perp AC \), so \( MD\perp AC \).
(iii) \( MD \) is the perpendicular bisector of \( AC \), so \( MC=MA \); and \( MA=\frac12AB \) since \( M \) is the mid-point.

Question 7. Show that the quadrilateral formed by joining the mid-points of the sides of a rectangle is a rhombus.
Answer:
In rectangle \( ABCD \) with mid-points \( P,Q,R,S \) of \( AB,BC,CD,DA \): \( PQ\parallel AC \), \( PQ=\frac12AC \) and \( SR\parallel AC \), \( SR=\frac12AC \), so \( PQRS \) is a parallelogram.
\( \triangle APS\cong\triangle BPQ \) (SAS: \( AP=BP \), right angles, \( AS=BQ \)), so \( PS=PQ \).
A parallelogram with two adjacent equal sides is a rhombus.

Question 8. A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle in common. Show that the vertex of the square opposite the common angle bisects the hypotenuse.
Answer:
Let the right angle be at \( C \) and the square \( CDME \) with \( M \) on the hypotenuse \( AB \).
\( CM \) is a diagonal of the square, so it bisects the right angle at \( C \).
In the isosceles right triangle the bisector of the right angle is also the median to the hypotenuse, so \( M \) is the mid-point of \( AB \).