Quadrilaterals CBSE Class 9 Mathematics Worksheets Set 06

Exercise I

Question 1. In a quadrilateral ABCD, AB is the smallest side and CD is the longest side. Show that ∠A > ∠C and ∠B > ∠D.
Answer:
Join AC. In △ABC, AB < BC, so ∠BCA < ∠BAC. In △ADC, AD < CD, so ∠ACD < ∠DAC.
Adding, ∠C < ∠A, that is ∠A > ∠C.
Join BD. In △ABD, AB < AD gives ∠ADB < ∠ABD. In △BCD, BC < CD gives ∠BDC < ∠CBD. Adding, ∠D < ∠B.
In simple words: The angle opposite the longer side is larger, and joining a diagonal lets us compare two triangles.

Teacher's Note:
1. Always join a diagonal to turn a quadrilateral problem into triangle problems.
2. Use: the larger side has the larger angle opposite it.

Question 2. The diagonals of a parallelogram ABCD meet at O. A line through O meets AB at P and DC at Q. Prove that OP = OQ.
Answer:
In △OPB and △OQD: OB = OD (diagonals bisect each other), ∠POB = ∠QOD (vertical angles) and ∠OBP = ∠ODQ (alternate angles, AB ∥ DC).
So △OPB ≅ △OQD (ASA), hence OP = OQ.
In simple words: Any line through the centre of a parallelogram is cut in half by the centre.

Teacher's Note:
1. The centre of a parallelogram is a centre of symmetry.
2. Mark the equal vertical and alternate angles on the figure.

Question 3. In a parallelogram ABCD, AX and CY bisect the angles A and C respectively (X on DC, Y on AB). Show that AX ∥ CY.
Answer:
AB ∥ DC, so ∠AXD = ∠XAB = ½∠A (alternate angles, AX bisects ∠A).
Also ∠YCD = ½∠C = ½∠A, because opposite angles of a parallelogram are equal.
So ∠AXD = ∠YCD. These are corresponding angles for the lines AX and CY cut by DC, hence AX ∥ CY.
In simple words: Opposite angles are equal, so their halves are equal and the bisectors are parallel.

Teacher's Note:
1. Halves of equal angles are equal.
2. Convert the alternate angle into a corresponding angle at C.

Question 4. ABCD is a parallelogram with ∠DAB = 60°. The bisectors of ∠A and ∠B meet at P on CD. Prove that P is the mid-point of CD.
Answer:
AB ∥ DC, so ∠DPA = ∠PAB = 30° (alternate angles). Also ∠DAP = 30°, so DP = DA (equal angles in △ADP).
Similarly ∠CPB = ∠PBA = ∠CBP (BP bisects ∠B), so CP = CB.
Since DA = CB, DP = CP, so P is the mid-point of CD.
In simple words: Each angle bisector makes an isosceles triangle, and the two equal sides are opposite sides of the parallelogram.

Teacher's Note:
1. An angle bisector in a parallelogram always cuts off an isosceles triangle.
2. Here ∠A + ∠B = 180°, so the bisectors meet at 90° at P.

Question 5. ABCD is a parallelogram. P lies on AD with AP = ⅓AD and Q lies on BC with CQ = ⅓BC. Prove that AQCP is a parallelogram.
Answer:
AD = BC and AD ∥ BC, so AP = ⅓AD = ⅓BC = CQ.
Also AP ∥ QC, because AD ∥ BC. In AQCP the opposite sides PA and QC are therefore equal and parallel.
Hence AQCP is a parallelogram.
In simple words: One pair of opposite sides that is equal and parallel is enough to prove a parallelogram.

Teacher's Note:
1. Check the third and fourth sides (AQ and PC) after proving this pair.
2. Here the pair AP and CQ are on opposite ends of the parallelogram.

Question 6. Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.
Answer:
Let P, Q, R, S be the mid-points of AB, BC, CD, DA of rectangle ABCD. By the mid-point theorem, PQ = ½AC = SR and QR = ½BD = PS.
The diagonals of a rectangle are equal, so AC = BD and PQ = QR = RS = SP.
A quadrilateral with all four sides equal is a rhombus.
In simple words: Each side of the inner figure is half a diagonal, and the diagonals of a rectangle are equal.

Teacher's Note:
1. Draw both diagonals in the figure.
2. Use the mid-point theorem in each corner triangle.

Question 7. Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.
Answer:
As above, PQRS has PQ = QR = RS = SP = ½ of a diagonal, so it is a rhombus.
PQ ∥ AC and QR ∥ BD, and AC ⊥ BD in a square, so ∠PQR = 90°.
A rhombus with a right angle is a square.
In simple words: Equal sides give a rhombus and perpendicular diagonals give a right angle.

Teacher's Note:
1. For a rectangle the same figure is a rhombus only.
2. Perpendicular diagonals are the key extra fact.

Question 8. In △ABC, M and N lie on AB and AC with AM = ¼AB and AN = ¼AC. Prove that MN = ¼BC.
Answer:
Let M′ and N′ be the mid-points of AB and AC. Then M′N′ = ½BC.
M is the mid-point of AM′ and N is the mid-point of AN′, so in △AM′N′, MN = ½M′N′.
Therefore MN = ½ × ½BC = ¼BC.
In simple words: Use the mid-point theorem twice.

Teacher's Note:
1. Introduce the mid-points of AB and AC as helpers.
2. The same method gives MN = BC/8 for one-eighth points.

Question 9. ABCD is a parallelogram and AP, CQ are perpendiculars from A and C on diagonal BD. Show that (i) △APB ≅ △CQD (ii) AP = CQ.
Answer:
(i) In △APB and △CQD: ∠APB = ∠CQD = 90°, AB = CD (opposite sides) and ∠ABP = ∠CDQ (alternate angles, AB ∥ CD). So △APB ≅ △CQD (AAS).
(ii) Corresponding parts give AP = CQ.
In simple words: The two right triangles at opposite corners are copies of each other.

Teacher's Note:
1. AAS needs two angles and a non-included side.
2. Equal distances from opposite vertices to a diagonal show the diagonal bisects the area.

Question 10. ABCD and PQRC are rectangles with P on DC and R on BC, and Q is the mid-point of AC. Prove that (i) DP = PC (ii) PR = ½AC.
Answer:
(i) PQ ⊥ DC and AD ⊥ DC, so PQ ∥ AD. In △ADC, Q is the mid-point of AC and PQ ∥ AD, so P is the mid-point of DC, i.e. DP = PC.
(ii) The diagonals of rectangle PQRC are equal, so PR = QC = ½AC.
In simple words: Use the converse of the mid-point theorem and the equal diagonals of a rectangle.

Teacher's Note:
1. PQ ∥ AD because both are perpendicular to DC.
2. QC is half of AC because Q is the mid-point.

Question 11. AD is a median of △ABC and DE ∥ BA, with E on AC. Show that BE is also a median of △ABC.
Answer:
D is the mid-point of BC and DE ∥ BA, so by the converse of the mid-point theorem in △ABC, E is the mid-point of AC.
Therefore BE joins B to the mid-point of AC, so BE is a median.
In simple words: A line through a mid-point parallel to a side bisects the other side.

Teacher's Note:
1. Always name the triangle in which you apply the theorem.
2. Medians join vertices to mid-points of opposite sides.

Question 12. ABCD is a parallelogram with ∠A = 72°. Find ∠B, ∠C and ∠D.
Answer:
Adjacent angles are supplementary: ∠B = 180° − 72° = 108°.
Opposite angles are equal: ∠C = 72° and ∠D = 108°.
In simple words: Adjacent angles add to 180° and opposite angles are equal.

Teacher's Note:
1. Check: 72 + 108 + 72 + 108 = 360.
2. Use the same two facts for any parallelogram angle problem.

Question 13. AB and CD are two parallel lines and a transversal l cuts AB at X and CD at Y. Prove that the bisectors of the interior angles form a rectangle.
Answer:
Interior angles on the same side of l: ∠BXY + ∠XYD = 180°, so half of them add to 90°. The bisectors meet at P with ∠XPY = 90°. The other two bisectors give the same at Q.
At X, the bisectors of the adjacent angles ∠AXY and ∠BXY are perpendicular (half of 180°), so the angle at X is 90°; likewise at Y.
The quadrilateral XPYQ has all four angles equal to 90°, so its opposite angles are equal and it is a parallelogram with right angles, that is a rectangle.
In simple words: All four corners are 90°, so the figure is a rectangle.

Teacher's Note:
1. Interior angles on the same side of a transversal are supplementary.
2. Half of 180° is 90°, which is the source of every right angle here.

Question 14. ABCD is a trapezium with AB ∥ DC. E is the mid-point of AD and F is a point on BC such that EF ∥ DC. Prove that EF = ½(AB + DC).
Answer:
Join AC to meet EF at G. In △ADC, E is the mid-point of AD and EG ∥ DC, so G is the mid-point of AC and EG = ½DC.
In △ABC, G is the mid-point of AC and GF ∥ AB, so F is the mid-point of BC and GF = ½AB.
EF = EG + GF = ½DC + ½AB = ½(AB + DC).
In simple words: The mid-line of a trapezium is the average of the two parallel sides.

Teacher's Note:
1. The diagonal AC is the helper that splits the trapezium into two triangles.
2. Use the mid-point theorem and its converse once each.

Exercise II

Question 15. Prove that in a parallelogram, the bisectors of any two consecutive angles intersect at a right angle.
Answer:
Let the bisectors of ∠A and ∠B meet at P. Consecutive angles are supplementary: ∠A + ∠B = 180°.
In △APB, ∠PAB + ∠PBA = ½(∠A + ∠B) = 90°.
So ∠APB = 180° − 90° = 90°.
In simple words: Half of 180° is 90°.

Teacher's Note:
1. Consecutive angles of a parallelogram are supplementary.
2. This is the quickest route to rectangle-formation results.

Question 16. ABCD is a parallelogram and AB is produced to E so that BE = AB. Prove that ED bisects BC.
Answer:
DC = AB = BE and DC ∥ BE, so DCEB is a parallelogram (one pair of opposite sides equal and parallel).
The diagonals of a parallelogram bisect each other, so DE bisects CB.
In simple words: The line ED is a diagonal of a second parallelogram.

Teacher's Note:
1. Spot the second parallelogram DCEB.
2. AB = DC and AB = BE together give DC = BE.

Question 17. Prove that the segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and equal to half their difference.
Answer:
Let AB ∥ DC with AB > DC, and let P, Q be the mid-points of AC and BD. Extend CQ to meet AB at E.
In △CDQ and △EBQ: DQ = BQ, ∠DQC = ∠BQE (vertical) and ∠DCQ = ∠BEQ (alternate). So △CDQ ≅ △EBQ and CQ = QE, BE = DC.
In △ACE, P and Q are mid-points of AC and CE, so PQ ∥ AE (that is, ∥ AB) and PQ = ½AE = ½(AB − BE) = ½(AB − DC).
In simple words: The mid-line of the triangle ACE is the required segment.

Teacher's Note:
1. The trick is to extend CQ and create a congruent pair of triangles.
2. AE = AB − BE because E lies on AB.

Question 18. ABCD is a quadrilateral with AB ∥ CD and AD = BC, where AD is not parallel to BC. Prove that ∠A = ∠B.
Answer:
Draw DE ⊥ AB and CF ⊥ AB. Then DE = CF (distance between parallels).
In right triangles AED and BFC: AD = BC (hypotenuse) and DE = CF, so △AED ≅ △BFC (RHS). Hence ∠A = ∠B.
In simple words: The base angles of an isosceles trapezium are equal.

Teacher's Note:
1. If AD ∥ BC the figure would be a parallelogram and the claim fails in general.
2. RHS needs a right angle, a hypotenuse and one side.

Question 19. ABCD is a parallelogram, P is the mid-point of DC and Q is a point on AC with CQ = ¼AC. PQ produced meets BC at R. Prove that R is the mid-point of BC.
Answer:
Let the diagonals meet at O, the mid-point of AC. CQ = ¼AC = ½CO, so Q is the mid-point of CO.
In △COD, P and Q are mid-points of CD and CO, so PQ ∥ DO (that is, PR ∥ DB).
Now consider △CDB: P is the mid-point of CD and PR ∥ DB, so R is the mid-point of CB (converse of the mid-point theorem).
In simple words: Two uses of the mid-point theorem pass the half-way property from P to R.

Teacher's Note:
1. Introduce O, the centre of the parallelogram.
2. The line PQ is parallel to the diagonal BD.

Question 20. The diagonals of a rectangle ABCD meet at O. If ∠BOC = 44°, find ∠OAD.
Answer:
∠AOD = ∠BOC = 44° (vertical angles).
OA = OD (half of equal diagonals), so ∠OAD = ∠ODA = (180° − 44°) ÷ 2 = 68°.
In simple words: Triangle AOD is isosceles with apex angle 44°.

Teacher's Note:
1. Diagonals of a rectangle are equal and bisect each other.
2. Check: 68 + 68 + 44 = 180.

Question 21. Each side of a rhombus is 10 cm and one diagonal is 16 cm. Find the other diagonal and the area of the rhombus.
Answer:
The diagonals bisect each other at right angles. Half of the given diagonal is 8 cm.
Half of the other diagonal = √(10² − 8²) = 6 cm, so the other diagonal is 12 cm.
Area = ½ × 16 × 12 = 96 cm².
In simple words: Use the right triangle formed by half-diagonals and a side.

Teacher's Note:
1. Area of a rhombus = half the product of its diagonals.
2. 6, 8, 10 is the 3, 4, 5 triplet doubled.

Question 22. Lines are drawn through A, B, C of △ABC parallel to BC, CA, AB respectively, forming a triangle PQR (P opposite A, Q opposite B, R opposite C). Prove that the perimeter of △PQR is double the perimeter of △ABC.
Answer:
Opposite sides of the three parallelograms formed are equal, and each side of △PQR is made of two sides of △ABC. For example the side of △PQR passing through C parallel to AB has length AB + AB = 2AB.
In the same way the other sides are 2BC and 2CA.
Perimeter of △PQR = 2(AB + BC + CA), double that of △ABC.
In simple words: △ABC is the mid-point triangle of △PQR.

Teacher's Note:
1. Each of the three parallelograms gives two equal pairs of sides.
2. A, B, C are the mid-points of the sides of △PQR.

Question 23. PQRS is a parallelogram. The bisectors of ∠P and ∠Q meet at O, and a line LOM is drawn parallel to PQ with L on PS and M on QR. Prove that (i) PL = QM (ii) OL = OM.
Answer:
(i) LM ∥ PQ and PS ∥ QR, so PQML is a parallelogram and PL = QM.
(ii) ∠LOP = ∠OPQ (alternate) = ∠OPL (bisector), so LP = LO. Similarly MQ = MO.
Since PL = QM, we get OL = OM.
In simple words: Each bisector makes an isosceles triangle, and the two equal legs are equal.

Teacher's Note:
1. Parallel lines with a bisector give isosceles triangles.
2. Combine with the parallelogram PQML.

Question 24. In △ABC, right-angled at B, AB = 9 cm and AC = 15 cm. D and E are the mid-points of AB and AC. Find (i) BC (ii) the area of △ADE.
Answer:
(i) BC = √(15² − 9²) = √144 = 12 cm.
(ii) Area of △ABC = ½ × 9 × 12 = 54 cm². DE ∥ BC and DE = 6 cm, so DE ⊥ AB, and area of △ADE = ½ × AD × DE = ½ × 4.5 × 6 = 13.5 cm².
In simple words: The small triangle is one-quarter of the large one.

Teacher's Note:
1. Check: 54 ÷ 4 = 13.5.
2. 9, 12, 15 is the 3, 4, 5 triplet tripled.

Question 25. △ABC is right-angled at B and D is the mid-point of AC. Show that DA = DB = DC.
Answer:
Draw DE ∥ BC meeting AB at E. D is the mid-point of AC, so E is the mid-point of AB.
DE ∥ BC and BC ⊥ AB give DE ⊥ AB, so DE is the perpendicular bisector of AB and DA = DB.
DA = DC since D is the mid-point of AC. So DA = DB = DC.
In simple words: The mid-point of the hypotenuse is the centre of the circle through all three vertices.

Teacher's Note:
1. A right-angled triangle has its circumcentre on the hypotenuse.
2. The median to the hypotenuse is half of it.

Question 26. A and B are on the same side of a line l. AD and BE are perpendicular to l with D, E on l, and C is the mid-point of AB. Prove that CD = CE.
Answer:
Draw CM ⊥ l with M on l. Then AD ∥ CM ∥ BE.
Three parallel lines make equal intercepts AC = CB on one transversal, so they make equal intercepts DM = ME on the line l.
CM ⊥ DE and M is the mid-point of DE, so CM is the perpendicular bisector of DE, hence CD = CE.
In simple words: Use the intercept theorem and the perpendicular bisector.

Teacher's Note:
1. Drop the helper perpendicular from C.
2. Every point of a perpendicular bisector is equidistant from the end points.

Question 27. One angle of a parallelogram is two-thirds of its adjacent angle. Find the angles of the parallelogram.
Answer:
Let the larger angle be x. Adjacent angles add to 180°: x + (2/3)x = 180°, so x = 108°.
The other angle is (2/3) × 108° = 72°. The angles are 108°, 72°, 108°, 72°.
In simple words: The two kinds of angle add to 180° and are in ratio 3 : 2.

Teacher's Note:
1. Treat 'two-thirds of' as a ratio 2 : 3.
2. Check: 108 + 72 = 180.

Question 28. D, E, F are the mid-points of BC, CA, AB of an isosceles triangle ABC with AB = AC. Prove that AD is perpendicular to EF and is bisected by it.
Answer:
AF = ½AB and AE = ½AC, so AF = AE. Also FD ∥ AC and DE ∥ AB, so AFDE is a parallelogram.
A parallelogram with a pair of equal adjacent sides is a rhombus. The diagonals of a rhombus, AD and FE, bisect each other at right angles.
So AD ⊥ EF and EF bisects AD.
In simple words: AFDE is a rhombus.

Teacher's Note:
1. The mid-point theorem gives the two parallel pairs.
2. Isosceles gives the equal adjacent sides.

Question 29. AD is a median of △ABC and E is the mid-point of AD. BE produced meets AC at F. Prove that AF = ⅓AC.
Answer:
Draw DG ∥ BF meeting AC at G. In △BCF, D is the mid-point of BC and DG ∥ BF, so G is the mid-point of CF: FG = GC.
In △ADG, E is the mid-point of AD and EF ∥ DG, so F is the mid-point of AG: AF = FG.
So AF = FG = GC and AF = ⅓AC.
In simple words: Two mid-point arguments split AC into three equal parts.

Teacher's Note:
1. Adding the parallel line DG is the key idea.
2. Use the converse of the mid-point theorem twice.