How to Use This Page
1. Solve each exercise problem on your own first. 2. Compare your method and answer with the step-by-step solution below. 3. Focus on understanding why each property is applied, not just the final answer. 4. Revisit the Common Mistakes section before exams.
Key Concepts (Quick Recap)
Types of Quadrilaterals and Defining Properties
| Quadrilateral | Defining Properties |
|---|---|
| Parallelogram | Opposite sides equal and parallel; opposite angles equal; diagonals bisect each other |
| Rectangle | All parallelogram properties + all angles 90; diagonals equal |
| Rhombus | All sides equal; opposite sides parallel; opposite angles equal; diagonals perpendicular bisectors |
| Square | All rectangle + rhombus properties; all sides equal; all angles 90; diagonals equal and perpendicular |
| Trapezium | One pair of opposite sides parallel |
| Kite | Two pairs of adjacent sides equal; one diagonal perpendicular bisector of other |
Angle Sum Property
The sum of interior angles of any quadrilateral is 360 degrees. Proof: draw diagonal AC, creating triangles ABC and ADC. Each has angle sum 180, so total = 180 + 180 = 360.
Midpoint Theorem
| Statement: The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half its length. If D and E are midpoints of AB and AC, then DE | BC and DE = BC/2. |
|---|
Converse: If a line passes through the midpoint of one side and is parallel to a second side, it bisects the third side.
Diagonal Properties Summary
| Quadrilateral | Diagonals Equal | Diagonals Bisect | Diagonals Perpendicular |
|---|---|---|---|
| Parallelogram | No | Yes | No |
| Rectangle | Yes | Yes | No |
| Rhombus | No | Yes | Yes |
| Square | Yes | Yes | Yes |
| Kite | No | One bisects other | Yes |
Exercise 5.1 — Properties of Quadrilaterals
Q1. Three angles of a quadrilateral are 75, 85 and 110 degrees. Find the fourth angle.
Property: Sum of interior angles = 360 degrees.
Let fourth angle = x
75 + 85 + 110 + x = 360
270 + x = 360
x = 90 degrees
Q2. Angles of a quadrilateral are in ratio 1 : 2 : 3 : 4. Find each angle.
x + 2x + 3x + 4x = 360
10x = 360, x = 36
Angles: 36, 72, 108, 144 degrees
Q3. In parallelogram ABCD, angle A = 65. Find all angles.
Properties: Opposite angles equal; adjacent angles supplementary.
angle C = angle A = 65
angle B = 180 − 65 = 115
angle D = angle B = 115
Q4. In a parallelogram, one angle is 30 more than its adjacent angle. Find all angles.
x + (x + 30) = 180 (adjacent supplementary)
2x + 30 = 180, x = 75
Angles: 75, 105, 75, 105 degrees
Q5. Diagonals of parallelogram ABCD intersect at O. If AO = 5 cm and BO = 4 cm, find AC and BD.
Property: Diagonals bisect each other.
AC = AO + OC = 5 + 5 = 10 cm
BD = BO + OD = 4 + 4 = 8 cm
Q6. Diagonals of a rhombus are 10 cm and 24 cm. Find its side.
Property: Diagonals bisect each other at right angles.
Half-diagonals: 5 cm and 12 cm
Side = sqrt(5^2 + 12^2) = sqrt(25 + 144) = sqrt(169) = 13 cm
Q7. Rectangle ABCD has AB = 8 cm, BC = 6 cm. Find diagonal AC.
AC = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10 cm
Q8. Perimeter of parallelogram is 60 cm. One side is 18 cm. Find the other side.
2(18 + x) = 60
18 + x = 30
x = 12 cm
Q9. In quadrilateral ABCD, angle A = angle C and angle B = angle D. Is ABCD a parallelogram?
Both pairs of opposite angles are equal.
angle A + angle B = 360/2 = 180 (supplementary)
A quadrilateral with both pairs of opposite angles equal
is a parallelogram.
Q10. In kite PQRS, PQ = PS = 7 cm and QR = SR = 5 cm. Find perimeter.
Perimeter = 7 + 5 + 5 + 7 = 24 cm
Q11. Find angles of a parallelogram if one angle is twice its adjacent angle.
x + 2x = 180, x = 60
Angles: 60 and 120 degrees
Q12. Rhombus with diagonals 16 cm and 12 cm. Find area and side.
Area = (1/2) x 16 x 12 = 96 sq cm
Half-diagonals: 8 cm, 6 cm
Side = sqrt(64 + 36) = sqrt(100) = 10 cm
Q13. Trapezium ABCD with AB || DC. angle A = 70, angle D = 80. Find angle B and angle C.
Property: Co-interior angles between parallel lines are supplementary.
angle B = 180 − 70 = 110 (same side as angle A)
angle C = 180 − 80 = 100 (same side as angle D)
Q14. Diagonals of rectangle ABCD meet at O. If angle AOD = 120, find angle ABO.
Property: Diagonals equal and bisect each other, so AO = BO.
angle AOB = 180 − 120 = 60
Triangle AOB is isosceles (AO = BO)
angle OAB = angle OBA = (180 − 60)/2 = 60
Q15. True or False: A quadrilateral can have all angles obtuse.
If all angles > 90, sum > 360. But sum must be exactly 360.
Impossible.
Exercise 5.2 — Midpoint Theorem
Q1. In triangle ABC, D and E are midpoints of AB and AC. If BC = 12 cm, find DE.
DE = (1/2) x BC = (1/2) x 12 = 6 cm
Q2. In triangle PQR, S and T are midpoints of PQ and PR. If ST = 4.5 cm, find QR.
ST = (1/2) x QR
QR = 2 x 4.5 = 9 cm
Q3. D, E, F are midpoints of BC, CA, AB. If BC = 8, CA = 6, AB = 10, find perimeter of triangle DEF.
DE = AB/2 = 5, EF = BC/2 = 4, FD = CA/2 = 3
Perimeter = 5 + 4 + 3 = 12 cm
Q4. Prove: The line segment joining midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
| To prove: DE | BC and DE = BC/2. |
|---|
Construction: Extend DE to F such that EF = DE. Join CF.
Proof:
In triangles AED and CEF:
AE = CE (E is midpoint)
angle AED = angle CEF (vertically opposite)
DE = EF (construction)
Triangles AED and CEF are congruent (SAS).
So AD = CF, angle ADE = angle CFE.
But AD = DB (D is midpoint), so CF = DB.
angle ADE = angle CFE means CF || DB (alternate angles equal).
Therefore DBCF is a parallelogram.
Hence DF || BC and DF = BC.
Since DF = DE + EF = 2DE:
2DE = BC, so DE = BC/2 and DE || BC.
Hence proved.
Q5. In triangle ABC, D is midpoint of AB. A line through D parallel to BC meets AC at E. Prove E is midpoint of AC.
| By converse of midpoint theorem: since D is midpoint of AB and DE | BC, E must be the midpoint of AC. |
|---|
Q6. Triangle ADE has perimeter 15 cm. D, E are midpoints of AB, AC. BC = 12 cm. Find perimeter of triangle ABC.
By midpoint theorem:
AB = 2AD, AC = 2AE, BC = 2DE
Perimeter of ABC = 2(AD + AE + DE) = 2 x 15 = 30 cm
Q7. P, Q, R, S are midpoints of sides AB, BC, CD, DA of quadrilateral ABCD. Prove PQRS is a parallelogram.
Construction: Join AC.
In triangle ABC: PQ || AC and PQ = AC/2 (midpoint theorem)
In triangle ADC: SR || AC and SR = AC/2 (midpoint theorem)
Therefore PQ = SR and PQ || SR.
Hence PQRS is a parallelogram.
Q8. In triangle LMN, P is midpoint of LM. PQ || MN. If MN = 14 cm, find PQ.
By converse of midpoint theorem, Q is midpoint of LN. By midpoint theorem, PQ = MN/2 = 7 cm.
Q9. Midpoints of sides of a triangle are joined. The medial triangle has what fraction of the original area?
The medial triangle divides the original triangle into 4 congruent triangles. Area of medial triangle = 1/4 of original.
Q10. In triangle ABC, D, E, F are midpoints of BC, CA, AB. Show triangle DEF and ABC are similar. Find the ratio of perimeters.
DE = AB/2, EF = BC/2, FD = CA/2
Ratio of corresponding sides = 1/2
Therefore DEF ~ ABC (SSS similarity)
Ratio of perimeters = 1:2
Exercise 5.3 — Constructions
Q1. Construct quadrilateral ABCD with AB = 4.5 cm, BC = 3.5 cm, CD = 4 cm, AD = 3 cm, angle A = 90 degrees.
1. Draw AB = 4.5 cm
2. At A, construct 90-degree angle; draw ray AX
3. From A, cut arc of radius 3 cm on AX; mark D
4. From B, draw arc of radius 3.5 cm
5. From D, draw arc of radius 4 cm
6. Arcs intersect at C
7. Join BC and CD
Verify: AB = 4.5, BC = 3.5, CD = 4, DA = 3, angle A = 90.
Q2. Construct parallelogram ABCD with AB = 5 cm, BC = 3.5 cm, angle A = 60 degrees.
1. Draw AB = 5 cm
2. At A, construct 60-degree angle; draw ray AX
3. From A, cut arc of radius 3.5 cm on AX; mark D
4. From B, draw arc of radius 3.5 cm (BC)
5. From D, draw arc of radius 5 cm (DC = AB)
6. Arcs intersect at C; join BC and CD
Q3. Construct rhombus ABCD with side 4 cm, angle A = 70 degrees.
1. Draw AB = 4 cm
2. At A, construct 70-degree angle; draw ray AX
3. From A, cut arc of radius 4 cm on AX; mark D
4. From B, draw arc of radius 4 cm
5. From D, draw arc of radius 4 cm
6. Arcs intersect at C; join BC and CD
Q4. Construct rectangle ABCD with AB = 6 cm, BC = 4 cm.
1. Draw AB = 6 cm
2. At A, construct 90-degree angle; draw ray AX
3. From A, cut arc of radius 4 cm on AX; mark D
4. At B, construct 90-degree angle; draw ray BY
5. From B, cut arc of radius 4 cm on BY; mark C
6. Join CD; verify CD = 6 cm and all angles 90
Q5. Construct a square with side 4.5 cm.
1. Draw AB = 4.5 cm
2. At A, construct 90-degree angle; draw ray AX
3. From A, cut arc of radius 4.5 cm on AX; mark D
4. At B, construct 90-degree angle; draw ray BY
5. From B, cut arc of radius 4.5 cm on BY; mark C
6. Join CD; verify CD = 4.5 cm
Q6. Construct quadrilateral PQRS with PQ = 4 cm, QR = 3 cm, RS = 4.5 cm, PS = 3.5 cm, diagonal PR = 5 cm.
1. Draw PQ = 4 cm
2. From P, arc of radius 5 cm (diagonal PR)
3. From Q, arc of radius 3 cm (QR)
4. Arcs intersect at R
5. From R, arc of radius 4.5 cm (RS)
6. From P, arc of radius 3.5 cm (PS)
7. Arcs intersect at S; join RS and PS
Q7. Construct kite ABCD with AB = AD = 5 cm, BC = DC = 3 cm, diagonal AC = 6 cm.
1. Draw diagonal AC = 6 cm
2. From A, arc of radius 5 cm (AB = AD)
3. From C, arc of radius 3 cm (CB = CD)
4. On one side, arcs intersect at B
5. On other side, arcs intersect at D
6. Join AB, BC, CD, DA
Q8. Construct trapezium ABCD with AB || DC, AB = 6 cm, DC = 4 cm, AD = 3 cm, BC = 3.5 cm.
1. Draw AB = 6 cm
2. From A, arc of radius 3 cm (AD)
3. From B, arc of radius 3.5 cm (BC)
4. On same side of AB, locate D (3 cm from A) and C (3.5 cm from B)
with DC = 4 cm and DC || AB
5. Adjust: mark D at 3 cm from A, then C at 3.5 cm from B
such that DC = 4 cm
6. Join AD, DC, CB
Q9. Construct parallelogram with sides 5 cm and 3.5 cm, one diagonal 6 cm.
1. Draw AB = 5 cm
2. From A, arc of radius 3.5 cm (AD)
3. From B, arc of radius 6 cm (diagonal BD)
4. Arcs intersect at D
5. From D, arc of radius 5 cm (DC = AB)
6. From B, arc of radius 3.5 cm (BC = AD)
7. Arcs intersect at C; join AD, DC, BC
Q10. Construct rhombus with diagonals 8 cm and 6 cm.
1. Draw AC = 8 cm (longer diagonal)
2. Find midpoint O (OA = OC = 4 cm)
3. At O, draw perpendicular bisector of AC
4. From O, mark B and D on perpendicular bisector
with OB = OD = 3 cm (half of BD = 6 cm)
5. Join AB, BC, CD, DA
Exercise 5.4 — Word Problems and Proofs
Q1. Angles of a quadrilateral are (3x+10), (2x+30), (x+40), (4x). Find all angles.
3x + 10 + 2x + 30 + x + 40 + 4x = 360
10x + 80 = 360
10x = 280, x = 28
Angles: 94, 86, 68, 112 degrees
Check: 94 + 86 + 68 + 112 = 360
Q2. Prove: Diagonals of a parallelogram bisect each other.
To prove: AO = OC and BO = OD.
In triangles AOB and COD:
AB = CD (opposite sides)
angle OAB = angle OCD (alternate angles, AB || CD)
angle OBA = angle ODC (alternate angles, AB || CD)
Triangles AOB and COD are congruent (ASA).
Hence AO = OC and BO = OD.
Q3. Prove: If diagonals of a quadrilateral bisect each other, it is a parallelogram.
In triangles AOB and COD:
AO = CO, BO = DO, angle AOB = angle COD (vertically opposite)
Triangles AOB and COD congruent (SAS).
So AB = CD and angle OAB = angle OCD (alternate angles).
AB || CD and AB = CD, therefore ABCD is a parallelogram.
Q4. Prove: A rectangle is a parallelogram.
All angles are 90 degrees.
angle A + angle B = 180 (supplementary) implies AD || BC
angle A + angle D = 180 (supplementary) implies AB || DC
Both pairs of opposite sides parallel, so ABCD is a parallelogram.
(A special one with all right angles and equal diagonals.)
Q5. Perimeter of a rhombus is 52 cm. One diagonal is 24 cm. Find side and other diagonal.
Side = 52/4 = 13 cm
Half of d1 = 12 cm
13^2 = 12^2 + (d2/2)^2
169 = 144 + (d2/2)^2
(d2/2)^2 = 25, d2/2 = 5, d2 = 10 cm
Q6. In parallelogram ABCD, angle A : angle B = 3 : 2. Find all angles.
3x + 2x = 180 (adjacent supplementary)
5x = 180, x = 36
angle A = 108, angle B = 72
angle C = 108, angle D = 72
Q7. In triangle ABC, D and E are midpoints of AB and AC. BC is extended to F such that CE = EF. Prove DBFE is a parallelogram.
By midpoint theorem: DE || BC and DE = BC/2.
Since F is on BC extended, DE || BF.
Also DE = BC/2.
Since CE = EF, E is midpoint of CF.
In triangle with parallel lines: DB || EF and DB = EF.
Both pairs of opposite sides parallel and equal,
so DBFE is a parallelogram.
Q8. A farmer's rhombus-shaped field has perimeter 160 m and one diagonal 48 m. Find the area.
Side = 160/4 = 40 m
Half of d1 = 24 m
40^2 = 24^2 + (d2/2)^2
1600 = 576 + (d2/2)^2
(d2/2)^2 = 1024, d2/2 = 32, d2 = 64 m
Area = (1/2) x 48 x 64 = 1536 sq m
Q9. Prove: The segment joining midpoints of diagonals of a trapezium is parallel to the parallel sides and equals half their difference.
| Given: Trapezium ABCD with AB | DC. E and F are midpoints of AC and BD. |
|---|
In triangle ABC, with G midpoint of BC:
EG || AB and EG = AB/2 (midpoint theorem)
In triangle BCD, with G midpoint of BC:
FG || DC and FG = DC/2 (midpoint theorem)
E, G, F are collinear (on same line parallel to both AB and DC).
EF = EG − FG = AB/2 − DC/2 = (AB − DC)/2
Q10. In parallelogram ABCD, bisector of angle A meets DC at E. AB = 8 cm, AD = 5 cm. Find AE.
AE bisects angle A. Let angle DAE = angle EAB = x.
Since AB || DC: angle AEB = angle DAE = x (alternate angles).
In triangle AEB: angle AEB = angle EAB = x.
Triangle AEB is isosceles, so AE = AB = 8 cm.
Common Mistakes
1. Using 180 instead of 360 for quadrilateral angle sum. Four sides means two triangles means 2 x 180 = 360. Think "quad = four = 360."
2. Confusing quadrilateral diagonal properties. Rhombus does NOT have equal diagonals. Rectangle does NOT have perpendicular diagonals. Only a square has both. Draw and memorize the properties table.
3. Midpoint theorem direction errors. DE is parallel to BC (the THIRD side), not AB or AC. DE = BC/2, not AB/2 or AC/2.
4. Rhombus area formula mistake. Area = (1/2) x d1 x d2, NOT d1 x d2. The half is essential.
5. Mixing up adjacent vs opposite angle properties in parallelograms. Adjacent angles are supplementary (180); opposite angles are equal. These are different facts.
6. Assuming all trapeziums are isosceles. A general trapezium has only one pair of parallel sides. Only an isosceles trapezium has equal non-parallel sides and equal base angles.
7. Skipping verification in constructions. Always measure all sides and angles after constructing a quadrilateral to confirm accuracy.
8. Confusing kite and rhombus. Kite has two pairs of adjacent equal sides (AB = AD and CB = CD). Rhombus has all four sides equal.
Exam Checklist
- [ ] State and prove the angle sum property of a quadrilateral (360 degrees via diagonal division)
- [ ] List defining properties of parallelogram, rectangle, rhombus, square, trapezium, kite
- [ ] Identify quadrilateral type from conditions (e.g., "diagonals bisect and are equal" = rectangle)
- [ ] Find unknown angles using 360-degree sum
- [ ] Apply parallelogram properties: opposite sides equal, opposite angles equal, adjacent supplementary
- [ ] Apply midpoint theorem to find unknown lengths (segment = half of third side)
- [ ] Apply converse of midpoint theorem to prove a point is a midpoint
- [ ] Prove the midpoints of any quadrilateral form a parallelogram
- [ ] Construct quadrilaterals from 5 independent measurements
- [ ] Construct special quadrilaterals (parallelogram, rhombus, rectangle, square)
- [ ] Calculate rhombus area from diagonals: (1/2) x d1 x d2
- [ ] Prove key theorems: parallelogram diagonal bisection, midpoint theorem and converse
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Frequently Asked Questions
The sum of interior angles of any quadrilateral is 360 degrees. This can be proved by dividing the quadrilateral into two triangles using a diagonal. Each triangle has angle sum 180 degrees, so the quadrilateral has 180 + 180 = 360 degrees.