NCERT Solutions for Class 8 Maths Chapter 4 Quadrilaterals

This worksheet provides complete and accurate NCERT Solutions for Class 8 Maths Chapter 4 Quadrilaterals from Ganita Prakash Part 1. Chapter 4 introduces students to the world of four-sided figures — helping them understand how different quadrilaterals are defined, constructed, and related to each other. From rectangles and squares to parallelograms, rhombuses, kites, and trapeziums, this chapter builds strong geometric reasoning and proof-writing skills. Parents and students can rely on this worksheet to find step-by-step solutions that are fully aligned with the NCERT Ganita Prakash curriculum, making exam preparation and homework much easier.

Chapter summary: shapes, properties & reasoning

Chapter 4 of Class 8 Maths Ganita Prakash Part 1 is entirely concept and reasoning-based. It does not contain stories or poems — instead, it uses a structured exploration approach where students observe figures, work through deductions, and develop geometric proofs step by step. The chapter begins with identifying what makes a figure a quadrilateral, then moves through the properties of rectangles, squares, parallelograms, rhombuses, kites, and trapeziums. It includes hands-on activities such as the Carpenter's Problem, Geoboard Activity, Joining Triangles, Puzzle Time paper folding, and Figure It Out exercises. The main theme is understanding how quadrilaterals are classified and how their properties — such as diagonal behaviour, angle relationships, and side lengths — can be proved using triangle congruence conditions like SAS, SSS, AAS, and RHS.

What this NCERT chapter covers?

The chapter builds understanding across several important geometric concepts:

- Identifying quadrilaterals based on sides, vertices, and angles
- Properties of rectangles: angles, opposite sides, diagonals
- Properties of squares: equal sides, perpendicular diagonals, angle bisectors
- Angle sum property of quadrilaterals (sum = 360°) with proof
- Properties of parallelograms: opposite sides, opposite angles, diagonals
- Properties of rhombuses: equal sides, perpendicular bisecting diagonals
- Properties of kites: diagonal bisection and perpendicularity
- Properties of trapeziums and isosceles trapeziums
- Congruence-based deductions (Deductions 1 through 9)
- Classification of quadrilaterals using a Venn diagram

How to use these NCERT solutions?

Students should first try to answer every Observe, Figure It Out, and Deduction question on their own before looking at the solutions in this worksheet. Attempting the question independently — even if the answer is incomplete — strengthens understanding and makes revision more effective.

Parents can use this worksheet to verify their child's answers at home and to explain reasoning when a student is stuck. Teachers can use it as a ready reference to plan discussions around key deductions and activity-based questions.

All solutions in this worksheet follow the exact section order of Chapter 4 in Ganita Prakash Part 1, making it easy to locate answers alongside the textbook. It is particularly useful during unit tests, half-yearly exams, and class assignments.

Student tips & learning tricks

Always write the reason alongside each step in geometry proofs. For example, when stating that two angles are equal, mention whether it is because of alternate angles, vertically opposite angles, or corresponding parts of congruent triangles (CPCT). Examiners expect reasoning to be explicitly stated.

A common mistake students make is writing triangle congruence in the wrong order — for example, writing △BAD ≅ △CDB instead of △BAD ≅ △DCB. In a congruence statement, the order of vertices must match the correspondence between the triangles exactly.

Remember the key fact: the sum of all angles in any quadrilateral is always 360°. This single rule helps solve many problems quickly — for instance, if three angles are known, the fourth is always 360° minus their sum.

When working with parallelograms, always remember that adjacent angles are supplementary (add up to 180°) and opposite angles are equal. These two properties together are very powerful for finding unknown angles.

For rhombus problems, recall that its diagonals always bisect each other at 90°, and they also bisect the angles of the rhombus — these are the two properties that distinguish a rhombus from a general parallelogram.

Why NCERT solutions are important?

NCERT Solutions for Class 8 Maths are built to match the exact learning objectives of the school curriculum. Working through correct, step-by-step solutions helps students understand not just what the answer is, but why it is correct — which is essential for building mathematical thinking at this stage.

For Chapter 4 Quadrilaterals, accurate solutions are especially important because the chapter is proof-heavy and logic-driven. Even a small error in reasoning — such as citing the wrong congruence condition — can lead to marks being deducted. Having NCERT-aligned answers gives students confidence and clarity before assessments.

Strong foundational knowledge of quadrilaterals at Class 8 level also prepares students for more advanced geometry in Classes 9 and 10, where these concepts reappear in greater depth.

Complete answer key – NCERT solutions

Observe (Page 82)

Figs. (i), (ii), and (iii) are quadrilaterals because they are closed figures bounded by exactly four line segments (sides), and their angles are formed between consecutive sides. Figs. (iv) and (v) are not quadrilaterals — fig. (iv) has a curved side, and fig. (v) has sides that cross each other (it is not a simple closed figure with four distinct vertices forming a proper quadrilateral).

Are there other ways to define a rectangle?

Yes. A rectangle can also be defined as: A quadrilateral whose diagonals are equal and bisect each other. (This is established through the Carpenter's Problem and Deductions 1–3.)

A Carpenter's Problem

1. What is the length of the other diagonal?
The other diagonal must also be 8 cm long. Since ABCD is a rectangle, by the SAS congruence condition, △ADC ≅ △DAB. Therefore AC = BD (corresponding parts of congruent triangles). So BD = 8 cm.

2. What is the point of intersection of the two diagonals?
The two diagonals intersect at their midpoints. That is, O is the midpoint of both AC and BD, so OA = OC and OB = OD. The diagonals bisect each other.

3. What should the angle be between the diagonals?
There is no restriction on the angle between the diagonals for a rectangle. Any angle x between the diagonals will give a rectangle, as long as the diagonals are equal and bisect each other. (This is proved in Deduction 3.)

Deduction 2 – Math Talk

Can the following equalities be used to establish that △AOD ≅ △COB?
Yes. These three conditions — two sides and the included angle (SAS) — can be used to establish △AOD ≅ △COB. Since AO = CO, ∠AOD = ∠COB (vertically opposite), and AD = CB (opposite sides of rectangle are equal), by the SAS condition, △AOD ≅ △COB.

Deduction 3

Can you find all the remaining angles when one diagonal angle is 60°?
The four angles at O are: 60°, 120°, 60°, 120° (using vertically opposite angles and linear pairs).
In △AOB, since OA = OB (diagonals bisect each other and are equal, so half-diagonals are equal):
a + a + 60 = 180
2a = 120
a = 60°
All angles of the quadrilateral = a + b = (90 – x/2) + x/2 = 90°.
So ABCD is a rectangle regardless of the value of x (the angle between the diagonals), as long as the diagonals are equal and bisect each other.

What is the value of a (in degrees) in terms of x?
a + a + x = 180
2a = 180 – x
a = (180 – x)/2 = 90 – x/2

What can we say about AB and CD, and AD and BC?
Triangle AOB is congruent to triangle COD, and triangle AOD is congruent to triangle COB.
Hence AB = CD and AD = CB (corresponding parts of congruent triangles).
So opposite sides are equal, and ABCD is a rectangle.

Deduction 4

Is it wrong to write △BAD ≅ △CDB? Why?
Yes, it is wrong to write △BAD ≅ △CDB. The correct correspondence is △BAD ≅ △DCB. In the congruence, vertex B corresponds to D, A corresponds to C, and D corresponds to B. Writing △CDB reverses the correspondence of the last two vertices, making the mapping incorrect.

Can you show that AB is parallel to DC (AB ∥ DC)?
BC acts as a transversal to AB and DC. Since ∠B = 90° and ∠C = 90°, we have ∠B + ∠C = 180°. As the sum of interior angles on the same side of the transversal is 180°, AB ∥ DC.

Properties of a Rectangle

Property 1: All the angles of a rectangle are 90°.
Property 2: The opposite sides of a rectangle are equal.
Property 3: The opposite sides of a rectangle are parallel to each other.
Property 4: The diagonals of a rectangle are of equal length and they bisect each other.

A Special Rectangle

Are there any non-rectangles among figures (i) to (iv)?
No. All four quadrilaterals are rectangles. Figure (iv) is a special rectangle with all sides equal — this is called a square.

Deduction 5

What more needs to be done to get equal side lengths (to form a square)?
The angle between the diagonals must be 90°.

Can △BOA ≅ △BOC be used to find the angles ∠BOA and ∠BOC formed by the diagonals of a square?
Yes. By the SSS condition, △BOA ≅ △BOC (OB is common, OA = OC since diagonals bisect each other, and BA = BC since all sides of a square are equal). Therefore ∠BOA = ∠BOC (corresponding parts of congruent triangles). Since ∠BOA + ∠BOC = 180° (linear pair), we get ∠BOA = ∠BOC = 90°. So the diagonals of a square bisect each other at right angles.

Using this fact, construct a square with a diagonal of length 8 cm:
1. Draw diagonal AC = 8 cm.
2. Find the midpoint O of AC.
3. Draw a line perpendicular to AC at O.
4. Mark points B and D on this perpendicular line, each at a distance of 4 cm from O (i.e., OB = OD = 4 cm).
5. Join A, B, C, D in order. ABCD is the required square.

Properties of a Square

Property 1: All the sides of a square are equal to each other.
Property 2: The opposite sides of a square are parallel to each other.
Property 3: The angles of a square are all 90°.
Property 4: The diagonals of a square are of equal length and they bisect each other at 90°.
Property 5: The diagonals of a square bisect the angles of the square.

What are the measures of ∠1, ∠2, ∠3, and ∠4?
In △ADC, AD = DC (sides of square), so ∠1 = ∠3.
∠1 + ∠3 + 90 = 180 → 2∠1 = 90 → ∠1 = ∠3 = 45°.
Similarly, ∠2 = ∠4 = 45°.
So all four angles formed by the diagonals at the corners of a square are 45° each.

Figure It Out (Rectangle section)

(i) Rectangle ABCD with one diagonal angle = 30°:
Since the diagonals of a rectangle bisect each other and are equal, the triangles formed are isosceles.
In △AOB: ∠OAB = 30° (given), OA = OB (diagonals bisect each other and are equal → half-diagonals equal).
∠OBA = 30° (base angles of isosceles triangle).
∠AOB = 180 – 30 – 30 = 120°.
The four angles at O: 120°, 60°, 120°, 60°.
∠DAB = 90° → ∠DAC = 90 – 30 = 60°.
Similarly by symmetry, all four triangles give angles 30°, 60°, 30°, 60° at the corners.
Angles at centre O: 120°, 60°, 120°, 60°.
Diagonal angles at vertices: 30° and 60° at each corner (each 90° angle split into 30° + 60°).

(ii) Rectangle PQRS with one diagonal angle = 110°:
∠QPR = 110° is the angle at O inside the rectangle (angle between diagonals).
Angles at O: 110°, 70°, 110°, 70°.
Using the same method: since OQ = OS and OP = OR (diagonals bisect each other):
In △QOR: ∠QOR = 70°, OQ = OR → base angles = (180 – 70)/2 = 55°.
In △POQ: ∠POQ = 110°, OP = OQ → base angles = (180 – 110)/2 = 35°.
So at each corner: ∠Q = 35° + 55° = 90° ✓
Diagonal angles at each vertex:
At P: 35° + 55° = 90° (split as 35° and 55°)
At Q: 55° + 35° = 90° (split as 55° and 35°)
At R: 35° + 55° = 90°
At S: 55° + 35° = 90°
Angles at centre O: 110°, 70°, 110°, 70°.

2. In each case, draw two line segments of total length 8 cm each (so each half = 4 cm), bisecting each other at the given angle. Join the four endpoints to get the quadrilateral. Since equal diagonals bisect each other, the result is always a rectangle (regardless of the angle between the diagonals, as proved in Deduction 3).
(i) 30° → Rectangle
(ii) 40° → Rectangle
(iii) 90° → Square (special case of rectangle where diagonals are perpendicular)
(iv) 140° → Rectangle

3. APML is a rectangle.
Justification: Since PL and AM are diameters, they are equal in length (both equal to the diameter of the circle). Since they are perpendicular to each other and both pass through the centre O of the circle, they bisect each other at O. Thus, the four endpoints A, P, M, L form a quadrilateral whose diagonals are equal and bisect each other — which is the definition of a rectangle. Hence APML is a rectangle.

4. Place the two sticks of equal length so that they cross each other at their midpoints (i.e., bisect each other). Stretch the thread through the four endpoints to form a quadrilateral. Since the sticks (diagonals) are equal and bisect each other, the quadrilateral formed is a rectangle. The angle between the two sticks can be any value — to get a square (with 90° diagonals), adjust the sticks until all four sides of the thread quadrilateral feel equal. At that point, the angle between the sticks is 90°. You have made an exact right angle.

5. No. A quadrilateral with opposite sides parallel and equal is a parallelogram, not necessarily a rectangle. A parallelogram has opposite sides parallel and equal, but its angles need not be 90°. For example, a parallelogram with angles 60° and 120° satisfies the condition but is not a rectangle. Hence, this property alone cannot be used as a definition of a rectangle.

Angles in a Quadrilateral

Is it possible to construct a quadrilateral with three angles equal to 90° and the fourth angle not equal to 90°?
No, it is not possible. The sum of all angles in any quadrilateral is 360°. If three angles are each 90°, their sum = 270°. The fourth angle = 360° – 270° = 90°. So the fourth angle must also be 90°.

The sum of all angles in any quadrilateral is 360°.
Proof (using quadrilateral SOME):
Draw diagonal SM, dividing SOME into △SEM and △SOM.
Sum of angles of △SEM: ∠1 + ∠2 + ∠3 = 180°
Sum of angles of △SOM: ∠4 + ∠5 + ∠6 = 180°
Total = (∠1 + ∠4) + ∠2 + (∠3 + ∠6) + ∠5 = 360°
These are the four angles of quadrilateral SOME.
Hence, the sum of angles of any quadrilateral = 360°.

Parallelograms

Are there quadrilaterals with parallel opposite sides that are not rectangles?
Yes. When two pairs of parallel lines are drawn that do not meet at right angles, the quadrilateral formed has parallel opposite sides but angles other than 90°. Such quadrilaterals are called parallelograms.

Construct using ruler and set-square or compass and ruler:
Draw line AB. Draw line DC parallel to AB. Draw AD connecting them at an angle other than 90°. Draw BC parallel to AD through B. ABCD is a parallelogram that is not a rectangle.

Is a rectangle a parallelogram?
Yes. A rectangle has both pairs of opposite sides parallel, satisfying the definition of a parallelogram. A rectangle is a special parallelogram with all angles equal to 90°.

Draw a parallelogram with adjacent sides 4 cm and 5 cm, angle 30°. What are the remaining angles and sides?
Since ABCD is a parallelogram:
∠A = 30° (given)
∠B = 180 – 30 = 150° (adjacent angles are supplementary)
∠C = 30° (opposite to ∠A)
∠D = 150° (opposite to ∠B)
AB = DC = 4 cm (opposite sides equal)
AD = BC = 5 cm (opposite sides equal)

Deduction 6

What are the other angles in parallelogram ABCD with ∠A = 30°?
∠D = 180 – 30 = 150°
∠B = 150°
∠C = 30°

What about opposite angles — will they be equal in all parallelograms?
Yes. Taking ∠P = x in parallelogram PEAR:
∠R = 180 – x (adjacent angles supplementary)
∠A = 180 – ∠R = 180 – (180 – x) = x
∠E = 180 – x
So ∠P = ∠A = x, and ∠R = ∠E = 180 – x.
The opposite angles of a parallelogram are always equal.

Deduction 7

Which two triangles to consider to show opposite sides of parallelogram are equal?
Consider △ABD and △CDB (formed by diagonal BD).
In △ABD and △CDB:
∠ABD = ∠CDB (alternate angles, since AB ∥ DC, BD is transversal)
∠ADB = ∠CBD (alternate angles, since AD ∥ BC, BD is transversal)
BD = BD (common side)
By AAS congruence: △ABD ≅ △CDB
Therefore AD = CB and AB = CD (corresponding parts of congruent triangles).
Thus, the opposite sides of a parallelogram are equal.

Is it wrong to write △ABD ≅ △CBD? Why?
Yes, it is wrong. In △ABD ≅ △CDB, vertex A corresponds to C, B to D, and D to B. Writing △CBD reverses the correspondence of the last two vertices (B↔B and D↔C), which is incorrect. The correct statement is △ABD ≅ △CDB.

Deduction 8

Is it wrong to write △AOE ≅ △SOY? Why?
Yes. In △AOE ≅ △YOS, vertex A corresponds to Y, O to O, and E to S. Writing △SOY changes the correspondence to S↔A, O↔O, Y↔E — this is a different (and incorrect) mapping. The correct statement is △AOE ≅ △YOS.

Do the diagonals of a parallelogram intersect at a particular angle?
No. The diagonals of a parallelogram do not necessarily intersect at any specific fixed angle. They can intersect at any angle depending on the shape of the parallelogram. (Unlike a rhombus, where diagonals always meet at 90°.)

Properties of a Parallelogram

Property 1: The opposite sides of a parallelogram are equal.
Property 2: The opposite sides of a parallelogram are parallel.
Property 3: In a parallelogram, the adjacent angles add up to 180°, and the opposite angles are equal.
Property 4: The diagonals of a parallelogram bisect each other.

Are the diagonals of a parallelogram always equal?
No. The diagonals of a parallelogram are not necessarily equal. They are equal only in the special case of a rectangle (or square).

Quadrilaterals with Equal Side Lengths

A quadrilateral in which all the sides have the same length is a rhombus.

Deduction 9

What are the other angles of rhombus ABCD with ∠A = 50°?
Let the four equal angles formed by the diagonal be a.
In △ADB: a + a + 50 = 180 → 2a = 130 → a = 65°.
So the diagonal bisects ∠A and ∠C into 65° each.
∠A = ∠C = 50° (opposite angles of rhombus are equal)
∠B = ∠D = 180 – 50 = 130°
The angles of rhombus ABCD are: 50°, 130°, 50°, 130°.

It can be seen that △GAE ≅ △MAE (How?):
In △GAE and △MAE:
GA = MA (sides of rhombus GAME), AE = AE (common side), GE = ME (sides of rhombus). By SSS condition, △GAE ≅ △MAE.

Every rhombus is a parallelogram — Verify that Deduction 6 applies:
In rhombus GAME, the four angles formed by the diagonal AE are all equal (a = b = c = d, as shown). Using alternate angle argument: since alternate angles formed by transversal AE on lines EM and GA are equal, EM ∥ GA. Similarly GE ∥ AM. So opposite sides are parallel → GAME is a parallelogram. Therefore all properties of a parallelogram (including adjacent angles summing to 180° and opposite angles being equal) apply to a rhombus.

In rhombus GAME, △GEO ≅ △MEO (Why?):
GO = MO (diagonals of a parallelogram bisect each other, so O is the midpoint of GM), EO = EO (common), GE = ME (all sides of rhombus are equal). By SSS condition, △GEO ≅ △MEO.
Therefore ∠GOE = ∠MOE.
Since ∠GOE + ∠MOE = 180° (linear pair), each angle = 90°.

Properties of a Rhombus

Property 1: All the sides of a rhombus are equal to each other.
Property 2: The opposite sides of a rhombus are parallel to each other.
Property 3: In a rhombus, the adjacent angles add up to 180°, and the opposite angles are equal.
Property 4: The diagonals of a rhombus bisect each other.
Property 5: The diagonals of a rhombus bisect its angles.
Property 6: Diagonals of a rhombus intersect each other at an angle of 90°.

Are the diagonals of a rhombus equal?
No. The diagonals of a rhombus are not necessarily equal. They are equal only when the rhombus is also a rectangle — i.e., when it is a square.

Figure It Out – Section 4.4

1.
(i) Parallelogram RAPE with ∠P = 40°:
∠P = 40°
∠A = 180 – 40 = 140° (adjacent angles supplementary)
∠R = 40° (opposite to ∠P)
∠E = 140° (opposite to ∠A)

(ii) Parallelogram SRQP with ∠P = 110°:
∠P = 110°
∠Q = 180 – 110 = 70°
∠R = 110° (opposite to ∠P)
∠S = 70° (opposite to ∠Q)

(iii) Rhombus XWVU with ∠U = 30° (given at vertex U):
∠U = 30° (given)
∠W = 30° (opposite angles of rhombus are equal)
∠X = 180 – 30 = 150°
∠V = 150° (opposite to ∠X)

(iv) Rhombus OAIE with ∠E = 20°:
∠E = 20° (given)
∠I = 20° (opposite angles equal)
∠O = 180 – 20 = 160°
∠A = 160° (opposite to ∠O)

2. 1. Draw diagonal AC = 7 cm. Mark midpoint O.
2. At O, draw a line at 140° to AC.
3. On this line, mark B and D such that OB = OD = 2.5 cm (since BD = 5 cm and diagonals bisect each other).
4. Join A, B, C, D in order.
ABCD is the required parallelogram with diagonals 7 cm and 5 cm intersecting at 140°.

3. 1. Draw diagonal AC = 5 cm. Mark midpoint O.
2. At O, draw a perpendicular line (since diagonals of rhombus are perpendicular).
3. On this perpendicular, mark B and D such that OB = OD = 2 cm (since BD = 4 cm).
4. Join A, B, C, D in order.
ABCD is the required rhombus with diagonals 4 cm and 5 cm.

Geoboard Activity

What quadrilateral do you get when the two rubber bands are perpendicular to each other with equal lengths?
You get a square. The diagonals are equal and bisect each other at 90° — which is the property unique to a square (among all rectangles). All four sides formed will be equal, and all angles will be 90°.

What quadrilateral do you get when one diagonal is extended on both sides by 2 cm (making it longer)?
Since the diagonals are perpendicular and bisect each other but are now unequal, the figure is a rhombus (not a square). The diagonals of a rhombus are perpendicular bisectors of each other. So extending one diagonal makes the quadrilateral a rhombus (not a rectangle and not a square).

Joining Triangles

What type of quadrilateral is formed?
A rhombus. When two equilateral triangles are joined along one of their sides (the common side of length 8 cm), all four sides of the resulting quadrilateral are 8 cm. Since all sides are equal, the figure is a rhombus. The angles are 60° and 120° alternately (since each angle of an equilateral triangle is 60°, and joining doubles one angle to 120°).

Two isosceles triangles with sides 8 cm, 8 cm, 6 cm joined in two different ways:
The quadrilateral formed has sides 8, 8, 8, 8 cm — all sides equal → it is a rhombus.

What are the different ways they can be joined?
They can be joined along the 6 cm side, the 9 cm side, or the 12 cm side.
- Joined along the 12 cm side: Results in a parallelogram (opposite sides equal: 6, 9, 6, 9 cm and equal opposite angles by symmetry of joining).
- Joined along the 9 cm side: Results in a parallelogram (sides 6, 12, 6, 12 cm).
- Joined along the 6 cm side: Results in a kite-like shape or a parallelogram with sides 9, 12, 9, 12 cm.
In general, when two congruent triangles are joined along a corresponding side (flipped), they form a parallelogram. When joined along a corresponding side (without flipping), they may form a kite or other quadrilateral.

Kite

A quadrilateral ABCD where AB = BC and CD = DA.

Property 1: In the kite ABCD, the diagonal BD:
(i) bisects ∠ABC and ∠ADC
(ii) bisects diagonal AC (AO = OC) and is perpendicular to it.

Proof using △AOB ≅ △COB:
In △AOB and △COB:
AB = CB (given)
OB = OB (common)
∠ABO = ∠CBO (proved above)
By SAS: △AOB ≅ △COB.
→ AO = CO (BD bisects AC) ✓
→ ∠AOB = ∠COB.
Since ∠AOB + ∠COB = 180° (linear pair) → ∠AOB = ∠COB = 90° (BD ⊥ AC) ✓

Trapezium

Construct a trapezium PQRS with PQ ∥ SR. Measure base angles at S and R.

Property 1: ∠S + ∠P = 180° and ∠R + ∠Q = 180°.
(Since PQ ∥ SR, interior angles on the same side of a transversal add up to 180°.)

Can you find the remaining angles without measuring?
Yes. Since PQ ∥ SR:
Property 1: angle S + angle P = 180° and angle R + angle Q = 180°.
(Co-interior angles on the same side of a transversal add up to 180°.)
Example: If angle S = 70°, then angle P = 110°.
If angle R = 65°, then angle Q = 115°.

How do we construct an isosceles trapezium?
Draw line UV. Mark X and W on a parallel line above such that the perpendicular distances are equal and UX = VW (non-parallel sides are equal). Join U to X and V to W. Construct isosceles trapezium UVWX with UV ∥ XW.

Isosceles Trapezium

What type of quadrilateral is XWZY?
XWZY is a rectangle. Since XW ∥ UV (given), the angles a and b at Y and Z respectively are 90° each (co-interior angles with the right angles at Y and Z). XW ∥ YZ, XY ∥ WZ (both perpendicular to UV). So XWZY has all angles 90° → it is a rectangle.

Now, △UXY ≅ △VWZ (How?):
In △UXY and △VWZ:
UX = VW (given, isosceles trapezium)
∠XYU = ∠WZV = 90° (by construction)
XY = WZ (opposite sides of rectangle XWZY are equal)
By RHS (or SAS): △UXY ≅ △VWZ.
Therefore ∠U = ∠V (corresponding parts of congruent triangles).

Property 2: In an isosceles trapezium, the angles opposite to the equal sides are equal. (∠U = ∠V and ∠X = ∠W)

Figure It Out – Section 4.6

1. All sides = 4 cm (equilateral triangles have all sides equal).
When two equilateral triangles (each with all angles 60°) are joined along one side:
The angles at the two shared vertices become 60° + 60° = 120°.
The angles at the other two vertices remain 60°.
So the quadrilateral (a rhombus) has:
Sides: 4 cm, 4 cm, 4 cm, 4 cm
Angles: 60°, 120°, 60°, 120°

2. 1. Draw AC = 6 cm.
2. Find the midpoint O of AC.
3. Draw a perpendicular to AC at O.
4. Mark B and D on this perpendicular such that OB + OD = 8 cm (choose any split, e.g., OB = 5 cm, OD = 3 cm, or OB = 4 cm, OD = 4 cm).
5. Join A, B, C, D in order.
ABCD is the required kite with diagonals AC = 6 cm and BD = 8 cm.

3.
(i) Trapezium with angles 135° and 105°:
Using co-interior angle property:
Third angle = 180° – 135° = 45°
Fourth angle = 180° – 105° = 75°
All angles: 135°, 105°, 45°, 75°.
(Verify: 135 + 105 + 45 + 75 = 360° ✓)

(ii) Trapezium with one angle = 100° (isosceles trapezium):
In an isosceles trapezium, base angles are equal.
Given angle = 100° → the angle on the same parallel side = 100°.
Co-interior angles:
Third angle = 180° – 100° = 80°
Fourth angle = 80°.
All angles: 100°, 100°, 80°, 80°.
(Verify: 100 + 100 + 80 + 80 = 360° ✓)

4. The Venn diagram (as shown in the book on page 110):
- Trapezium is the largest set (outermost oval).
- Inside Trapezium: Parallelogram and Kite (overlapping).
- Inside Parallelogram: Rectangle and Rhombus (overlapping).
- Inside the intersection of Rectangle and Rhombus: Square.
- Kite overlaps with Rhombus (their intersection = Square; Rhombus is a special kite too).

(i) What quadrilateral is both a kite and a parallelogram?
A rhombus. A rhombus has all sides equal and opposite sides parallel (parallelogram), and can be considered a kite where both pairs of adjacent sides are equal. More precisely, the only quadrilateral that is both a kite and a parallelogram is a rhombus.

(ii) Can there be a quadrilateral that is both a kite and a rectangle?
Yes, but only in the special case of a square. A square has all sides equal (so adjacent sides equal → kite condition satisfied) and all angles 90° (→ rectangle). No other quadrilateral can be both a kite and a rectangle.

(iii) Is every kite a rhombus?
No, not every kite is a rhombus. In a kite, only two pairs of adjacent sides are equal (AB = BC and CD = DA), but the two pairs need not be equal to each other (i.e., AB may ≠ CD). In a rhombus, all four sides are equal. So a kite is a rhombus only when AB = BC = CD = DA (i.e., all sides equal). The correct relationship: every rhombus is a kite, but not every kite is a rhombus.

5. ∠IOD = 60°.
(In △RIO: ∠IRO = 90°, ∠RIO = 45° → ∠ROI = 45°.
The angle ∠IOD: since O is on the diagonal of RODS, using properties:
∠IOD = 90° – 30° = 60°.)

6. 1. Draw line segment AC = 6 cm (one diagonal).
2. Find the midpoint O of AC using compass (draw arcs from A and C with radius > 3 cm; the arcs intersect above and below; join the intersection points — this gives the perpendicular bisector of AC).
3. On the perpendicular bisector, mark B and D such that OB = OD = 3 cm.
4. Join A, B, C, D in order.
ABCD is a square with diagonal 6 cm. (No protractor needed since the perpendicular is constructed using compass.)

7. UVWX is a square.
Geometric reasoning:
Let each side of square CASE = 2a. Then CU = UA = AX = XS = SE = EW = WC = CV = a (midpoints).
In each corner triangle (e.g., triangle CUV):
CU = CV = a, angle C = 90°.
By SAS: all four corner triangles are congruent.
So UV = UX = XW = WV (all sides of UVWX are equal).
angle VUX: In triangle CUV, angle CUV = angle CVU = 45°.
angle VUX = 180° – angle CUV – angle AUX = 180° – 45° – 45° = 90°.
All angles of UVWX = 90° and all sides equal → UVWX is a square.
For Figure (b) — other ways to construct a square within a square:
Mark points on each side of the outer square at a ratio other than 1:1 (e.g., 1:3) from each vertex, consistently going in the same direction around the square. Join these four points. The resulting inner quadrilateral will be a square tilted at a different angle, with vertices lying on the sides of the outer square.

8. Yes, it will be a square.
Geometric reasoning: A quadrilateral with four equal sides is a rhombus. A rhombus has opposite angles equal and adjacent angles supplementary. If one angle = 90°, then the opposite angle = 90°, and both adjacent angles = 90°. So all angles are 90° and all sides are equal → it is a square.

9. It is a parallelogram.
Justification: Draw diagonal BD in quadrilateral ABCD with AB = CD and AD = BC.
In △ABD and △CDB:
AB = CD (given), AD = BC (given), BD = BD (common).
By SSS: △ABD ≅ △CDB.
Therefore ∠ABD = ∠CDB (alternate angles) → AB ∥ DC.
And ∠ADB = ∠CBD (alternate angles) → AD ∥ BC.
So both pairs of opposite sides are parallel → ABCD is a parallelogram.

10. Yes, the sum of angles of any simple quadrilateral (including non-convex or re-entrant quadrilaterals) is 360°.
Geometric reasoning: Draw diagonal AC, splitting quadrilateral ABCD into △ABC and △ACD. The angle sum of each triangle is 180°. Total = 360°. The angle at D in such a quadrilateral is measured as the interior angle (which may be greater than 180° in a non-convex case, but the sum of all four interior angles still equals 360°).

11.
(i) A quadrilateral whose diagonals are equal and bisect each other must be a square.
FALSE.
Equal diagonals that bisect each other define a rectangle (as proved in Deductions 1 and 2). A rectangle need not have equal sides and is not necessarily a square. Example: a 3 cm × 5 cm rectangle.

(ii) A quadrilateral having three right angles must be a rectangle.
TRUE.
Sum of angles = 360°. Three angles of 90° sum to 270°. Fourth angle = 360° – 270° = 90°. All four angles are 90° → it is a rectangle.

(iii) A quadrilateral whose diagonals bisect each other must be a parallelogram.
TRUE.
If diagonals bisect each other at O: OA = OC and OB = OD.
In triangles AOB and COD: OA = OC, OB = OD, angle AOB = angle COD (vertically opposite). By SAS, the triangles are congruent.
angle OAB = angle OCD (alternate angles) → AB ∥ DC.
Similarly AD ∥ BC. So ABCD is a parallelogram.

(iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.
FALSE.
A kite has perpendicular diagonals but is not a rhombus (its sides are not all equal in general). Example: a kite with sides 3, 3, 5, 5 has perpendicular diagonals but is not a rhombus.

(v) A quadrilateral in which the opposite angles are equal must be a parallelogram.
TRUE.
Let angle A = angle C = x and angle B = angle D = y.
2x + 2y = 360° → x + y = 180°.
Adjacent angles are supplementary → AB ∥ DC and AD ∥ BC.
Both pairs of opposite sides parallel → parallelogram.

(vi) A quadrilateral in which all the angles are equal is a rectangle.
TRUE.
If all four angles are equal and sum = 360°, each angle = 90°.
A quadrilateral with all angles 90° is a rectangle by definition. (As proved in Deduction 4, opposite sides are automatically equal.)

(vii) Isosceles trapeziums are parallelograms.
FALSE.
A parallelogram has both pairs of opposite sides parallel. An isosceles trapezium has only one pair of parallel sides (the other pair of sides, though equal, is not parallel). So an isosceles trapezium is not a parallelogram. Example: A trapezium with parallel sides 4 cm and 8 cm, and equal non-parallel sides of 5 cm, is not a parallelogram.

Puzzle Time – Which Quad? (Paper Folding Activity)

Step 4: Open the sheet. What is the shape formed by the creases?
Answer: When the sheet is folded in half (step 1), then into a quarter (step 2), and a triangular crease is made at the corner that is at the middle of the paper (step 3), on opening the sheet, the creases form a rhombus (or a square if the triangular crease is at 45°). The four crease lines create a quadrilateral with equal sides, symmetric about both fold lines.

Step 5: How would you fold the quarter paper to get the multiple concentric diamond-shaped creases shown in the image?
Answer: Make multiple triangular creases of different sizes at the same central corner (the corner that is at the middle of the original sheet), each at the same angle, before opening the paper. Each triangular crease will unfold into a concentric rhombus/diamond shape. The result is a series of nested rhombuses.

Step 6: How would you fold the quarter paper such that a square is formed?
Answer: Make the triangular crease at exactly 45° to the folded edges at the central corner. This ensures that when opened, the crease forms a square (since the diagonal of the crease triangle is at equal angles to both fold lines, resulting in all four crease segments being equal and all angles being 90°).

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