Chapter 4 Quadrilaterals (Class 8 - Latest Maths NCERT (Ganita Prakash I) Solutions)

Solution:

---

Given:

Two rectangles with properties of their diagonals indicated by tick marks (showing diagonals are equal and bisect each other).

$\bullet$ In rectangle (i) ($ABCD$): Diagonal $AC$ and $BD$ intersect at point $O$, and $\angle CAB = 30^\circ$.

$\bullet$ In rectangle (ii) ($PQRS$): Diagonal $PR$ and $QS$ intersect at point $O$, and the angle between the intersecting diagonals is $110^\circ$ ($\angle POQ = 110^\circ$).

---

To Find:

All other interior and diagonal intersection angles in both rectangles.

---

Part (i): Analysis of Rectangle $ABCD$

Let the intersection of diagonals $AC$ and $BD$ be point $O$.

1. Since $ABCD$ is a rectangle, its diagonals are equal and bisect each other:

2. In $\triangle OAB$, since $OA = OB$, it is an isosceles triangle. Therefore, the base angles are equal:

3. Using the angle sum property in $\triangle OAB$:

4. Since adjacent angles on a straight line ($BD$) are supplementary ($\angle AOB + \angle AOD = 180^\circ$):

5. Similarly, using the isosceles triangles $\triangle OBC$, $\triangle OCD$, and $\triangle ODA$, all corresponding corner and central angles can be determined:

Visual Angle Breakdown for Rectangle (i):

---

Part (ii): Analysis of Rectangle $PQRS$

Let the intersection of diagonals $PR$ and $QS$ be point $O$, and given $\angle POQ = 110^\circ$.

1. Since vertically opposite angles are equal:

2. Since adjacent angles on line segment $PR$ are supplementary:

3. Consequently, its vertically opposite angle is also equal:

4. In the isosceles triangle $\triangle POQ$ ($OP = OQ$), the base angles are:

5. Similarly, in $\triangle QOR$ ($OQ = OR$), the base angles are:

Visual Angle Breakdown for Rectangle (ii):

---

Summary Table of Angles:

Rectangle	Central Angles at $O$	Corner Angles (Vertices)
(i) ($ABCD$)	$120^\circ, 60^\circ, 120^\circ, 60^\circ$	$30^\circ, 60^\circ, 30^\circ, 60^\circ$
(ii) ($PQRS$)	$110^\circ, 70^\circ, 110^\circ, 70^\circ$	$35^\circ, 55^\circ, 35^\circ, 55^\circ$

Solution:

---

Given:

A quadrilateral whose diagonals: 1. Have equal lengths of $8 \text{ cm}$. 2. Bisect each other. 3. Intersect at an angle of (i) $30^\circ$, (ii) $40^\circ$, (iii) $90^\circ$, and (iv) $140^\circ$.

---

To Find / Construct:

The geometric classification and step-by-step construction procedure for the quadrilateral in each of the four cases.

---

Geometric Property Note:

Any quadrilateral whose diagonals are equal in length and bisect each other is fundamentally a Rectangle (or specifically a square when the intersection angle is $90^\circ$). The shape of the rectangle changes depending on the angle between the intersecting diagonals.

---

General Steps of Construction (For all cases):

Step 1: Draw a line segment $AC = 8 \text{ cm}$ (representing the first diagonal).

Step 2: Find the midpoint $O$ of segment $AC$ by drawing its perpendicular bisector ($OA = OC = 4 \text{ cm}$).

Step 3: At midpoint $O$, construct the given intersecting angle using a protractor to draw a line intersecting the perpendicular bisector line.

Step 4: Cut off arcs of length $4 \text{ cm}$ on both sides of $O$ along this new line to get the second diagonal $BD = 8 \text{ cm}$ ($OB = OD = 4 \text{ cm}$).

Step 5: Join the outer vertex points $A, B, C,$ and $D$ in order to form the required quadrilateral ($ABCD$).

---

Case-by-Case Breakdown:

(i) Intersection Angle = $30^\circ$:

The diagonals intersect at $30^\circ$. The resulting quadrilateral is a narrow rectangle.

---

(ii) Intersection Angle = $40^\circ$:

The diagonals intersect at $40^\circ$. The resulting quadrilateral is a rectangle slightly broader than the $30^\circ$ case.

---

(iii) Intersection Angle = $90^\circ$:

The diagonals intersect at $90^\circ$. Since the diagonals are equal ($8 \text{ cm}$) and bisect each other at right angles ($90^\circ$), the resulting quadrilateral is a Square.

---

(iv) Intersection Angle = $140^\circ$:

The diagonals intersect at an obtuse angle of $140^\circ$ (which means the acute angle between them is $180^\circ - 140^\circ = 40^\circ$). The resulting quadrilateral is a rectangle identical in proportions to case (ii), oriented differently.

---

Summary Table:

Case	Diagonal Length	Intersection Angle	Type of Quadrilateral Formed
(i)	$8 \text{ cm}$	$30^\circ$	Rectangle
(ii)	$8 \text{ cm}$	$40^\circ$	Rectangle
(iii)	$8 \text{ cm}$	$90^\circ$	Square
(iv)	$8 \text{ cm}$	$140^\circ$	Rectangle

Solution:

---

Given:

1. A circle with centre $O$.

2. Line segments $PL$ and $AM$ are two perpendicular diameters of the circle ($PL \perp AM$).

---

To Find:

Identify the geometric figure formed by connecting the points $A, P, M,$ and $L$ in order ($APML$), and provide reasoning.

---

Solution / Identification:

The figure $APML$ is a Square.

---

Reasoning and Proof:

To prove that $APML$ is a square, let us examine the lengths of its sides and the angles between them using the properties of a circle:

1. Diagonals are Diameters: In quadrilateral $APML$, the line segments $PL$ and $AM$ are the diagonals. Since both are diameters of the same circle:

2. Diagonals are Equal in Length: Because $PL$ and $AM$ are both diameters, their lengths are equal.

3. Diagonals Bisect Each Other: Since $O$ is the centre of the circle, it is the midpoint of both diameters $PL$ and $AM$:

4. Diagonals are Perpendicular: We are given that the two diameters are perpendicular to each other ($PL \perp AM$), meaning they intersect at $90^\circ$ at centre $O$:

5. Conclusion on Quadrilateral Type: A quadrilateral whose diagonals are equal, bisect each other, and intersect at right angles ($90^\circ$) is always a Square.

---

Summary Table:

Property of $APML$	Mathematical Reason
Diagonals ($PL$ and $AM$)	Equal ($2 \times \text{radius}$)
Bisect each other at $O$	Both pass through the centre $O$
Intersection Angle	$90^\circ$ (Given as perpendicular diameters)
Final Shape	Square

Given:

Two sticks of equal length and a piece of thread. We need to construct an exact $90^\circ$ angle using only these tools (without paper folding).

---

To Construct:

An exact $90^\circ$ angle (perpendicular intersection) using two equal sticks and a thread.

---

Steps of Construction:

Step 1: Let $AB$ and $CD$ be two sticks of equal length.

Step 2: Find and mark the midpoints of both sticks using a measuring thread or ruler.

Step 3: Fix a pivot (like a pin or screw) at the midpoints of the two sticks so they can rotate freely like an adjustable cross ($X$-shape).

Step 4: Using a piece of thread, measure the distance between endpoint $A$ and endpoint $D$ ($AD$), and between endpoint $B$ and endpoint $D$ ($BD$).

Step 5: Pivot and adjust the sticks about the central screw until the distance $AD$ is exactly equal to the distance $BD$.

Step 6: When this condition is met, lock the screw to fix the sticks in place. The angle between the sticks at the centre is now exactly $90^\circ$.

---

Proof/Verification:

Let the intersection point (midpoint) be $M$. Since $M$ is the midpoint of both sticks:

Consider $\triangle AMD$ and $\triangle BMD$:

$\bullet$ $AM = BM$ (Midpoint of stick $AB$)

$\bullet$ $AD = BD$ (Constructed equal using thread measurement)

$\bullet$ $MD = MD$ (Common side)

Therefore, by Corresponding Parts of Congruent Triangles (CPCT):

Since angles $AMD$ and $BMD$ form a linear pair along the straight line $AB$:

Substituting $\angle AMD$ for $\angle BMD$:

Thus, $\angle AMD = \angle BMD = 90^\circ$, proving that the sticks intersect at an exact right angle.

Given:

A question asking whether a quadrilateral with opposite sides parallel and equal can be defined as a rectangle.

---

To Find / Prove:

Whether every quadrilateral that has opposite sides parallel and equal is necessarily a rectangle, or if another shape satisfies these conditions.

---

Solution:

Let $ABCD$ be a quadrilateral in which opposite sides are parallel and equal.

Any quadrilateral satisfying the condition of having opposite sides parallel and equal is defined as a Parallelogram, not necessarily a rectangle.

Why it cannot be chosen as a definition for a Rectangle:

1. By definition, a Rectangle is a quadrilateral in which all four interior angles are right angles ($90^\circ$).

2. The given information ($AB \parallel DC$ and $AD \parallel BC$ along with equal lengths) only guarantees that $ABCD$ is a parallelogram. It does not provide any information to prove that the interior angles are $90^\circ$.

3. A parallelogram can be slanted (like a rhombus or generic parallelogram) where interior angles are acute and obtuse (e.g., $60^\circ$ and $120^\circ$).

Conclusion:

$\therefore ABCD$ may not be a rectangle. $\therefore$ A rectangle cannot be defined as a quadrilateral with equal and parallel opposite sides.

---

Summary Table:

Quadrilateral Property	Parallelogram	Rectangle
Opposite sides parallel and equal	Yes	Yes
All interior angles equal to $90^\circ$	No (Can be any angle)	Yes (Mandatory)

(i) Parallelogram PEAR

Given: $PEAR$ is a parallelogram with $\angle P = 40^\circ$.

To Find: Remaining angles $\angle E, \angle A, \text{ and } \angle R$.

Solution:

In a parallelogram, opposite angles are equal and adjacent angles are supplementary.

$\angle E = 180^\circ - 40^\circ = 140^\circ$

Remaining angles of $PEAR$ are $140^\circ, 40^\circ, \text{ and } 140^\circ$.

---

(ii) Parallelogram PQRS

Given: $PQRS$ is a parallelogram with $\angle P = 110^\circ$.

To Find: Remaining angles $\angle Q, \angle R, \text{ and } \angle S$.

Solution:

Remaining angles of $PQRS$ are $70^\circ, 110^\circ, \text{ and } 70^\circ$.

---

(iii) Rhombus XUVW

Given: $XUVW$ is a rhombus. Side marks indicate all sides are equal. Diagonal $XV$ is drawn and $\angle UVX = 30^\circ$.

To Find: All internal angles of the rhombus.

Solution:

In a rhombus, diagonals bisect the vertex angles.

Total $\angle V = \angle WVX + \angle UVX = 30^\circ + 30^\circ = 60^\circ$

The angles of the rhombus are $60^\circ, 120^\circ, 60^\circ, \text{ and } 120^\circ$.

---

(iv) Rhombus OAIE

Given: $OAIE$ is a rhombus. Diagonal $OE$ is drawn and $\angle AEO = 20^\circ$.

To Find: All internal angles of the rhombus.

Solution:

Diagonal $OE$ bisects $\angle E$.

$\angle IEO = \angle AEO = 20^\circ$

Total vertex angle $\angle E = 20^\circ + 20^\circ = 40^\circ$

The angles of the rhombus are $140^\circ, 40^\circ, 140^\circ, \text{ and } 40^\circ$.

Given:

1. Diagonal $d_1 = 7 \text{ cm}$.

2. Diagonal $d_2 = 5 \text{ cm}$.

3. Angle of intersection between diagonals = $140^\circ$.

---

Properties Used:

The diagonals of a parallelogram bisect each other. Therefore, the halves of the diagonals are:

---

Steps of Construction for Parallelogram:

Step 1: Draw a line segment $AC = 7 \text{ cm}$ (the first diagonal).

Step 2: Locate the midpoint $O$ of $AC$ by drawing a perpendicular bisector ($AO = OC = 3.5 \text{ cm}$).

Step 3: At point $O$, construct an angle of $140^\circ$ (or its supplementary angle $180^\circ - 140^\circ = 40^\circ$) using a protractor, and draw a line passing through $O$.

Step 4: With centre $O$ and radius $2.5 \text{ cm}$ ($\frac{5}{2} \text{ cm}$), cut arcs on both sides of $O$ along this line to mark points $B$ and $D$ ($OB = OD = 2.5 \text{ cm}$).

Step 5: Join $AB$, $BC$, $CD$, and $DA$ to complete the required parallelogram $ABCD$.

Given:

1. Diagonal $d_1 = 4 \text{ cm}$.

2. Diagonal $d_2 = 5 \text{ cm}$.

---

Properties Used:

The diagonals of a rhombus bisect each other at right angles ($90^\circ$). Therefore:

---

Steps of Construction for Rhombus:

Step 1: Draw a line segment $AC = 5 \text{ cm}$ (the first diagonal).

Step 2: Construct the perpendicular bisector of $AC$ intersecting at midpoint $O$ ($AO = OC = 2.5 \text{ cm}$).

Step 3: Since diagonals intersect at $90^\circ$, draw a line perpendicular to $AC$ passing through $O$.

Step 4: With centre $O$ and radius $2 \text{ cm}$ ($\frac{4}{2} \text{ cm}$), cut arcs on the perpendicular line above and below $O$ to mark points $B$ and $D$ ($OB = OD = 2 \text{ cm}$).

Step 5: Join $AB$, $BC$, $CD$, and $DA$ to complete the required rhombus $ABCD$.

Given:

Two equilateral triangles, each having a side length of $4 \text{ cm}$, are joined together along a common side to form a quadrilateral.

---

To Find:

All the side lengths and interior angles of the resulting quadrilateral.

---

Geometric Analysis:

1. An equilateral triangle has all three sides equal and all three interior angles equal to $60^\circ$.

2. When two equilateral triangles are joined along one of their sides (base), they share that common side of $4 \text{ cm}$, forming a Rhombus (or a symmetric kite).

3. Finding the Interior Angles:

$\bullet$ Two angles of the quadrilateral are formed by single corner angles of the equilateral triangles. Therefore, each of these two opposite angles is equal to $60^\circ$.

$\bullet$ The other two opposite angles are formed by combining two adjacent $60^\circ$ angles from the two triangles ($60^\circ + 60^\circ = 120^\circ$).

---

Summary Table:

Feature	Values
Side Lengths	$4 \text{ cm}, 4 \text{ cm}, 4 \text{ cm}, 4 \text{ cm}$
Interior Angles	$60^\circ, 120^\circ, 60^\circ, 120^\circ$
Geometric Name	Rhombus

Given:

A kite whose diagonals measure $6 \text{ cm}$ and $8 \text{ cm}$.

---

To Construct:

A kite using the given diagonal lengths.

---

Properties of a Kite:

1. The diagonals of a kite are perpendicular to each other ($90^\circ$).

2. One diagonal (the main diagonal) bisects the other diagonal (the cross diagonal).

---

Steps of Construction:

Step 1: Draw a line segment $AC = 8 \text{ cm}$ (representing the longer diagonal).

Step 2: Construct the perpendicular bisector of $AC$. Let it intersect $AC$ at midpoint $O$ ($AO = OC = 4 \text{ cm}$).

Step 3: Since the second diagonal ($BD = 6 \text{ cm}$) must be bisected by the first diagonal, cut off arcs of length $3 \text{ cm}$ ($\frac{6}{2} = 3 \text{ cm}$) on both sides of $O$ along the perpendicular bisector line to mark points $B$ and $D$ ($OB = OD = 3 \text{ cm}$).

Step 4: Join $AB$, $BC$, $CD$, and $DA$ using a straightedge to form the required kite $ABCD$.

---

Summary Table:

Diagonal	Total Length	Bisected Segments
$AC$	$8 \text{ cm}$	$4 \text{ cm}$ and $4 \text{ cm}$
$BD$	$6 \text{ cm}$	$3 \text{ cm}$ and $3 \text{ cm}$

Property: In a trapezium, the angles on the same side of the non-parallel legs (consecutive interior angles) are supplementary, i.e., their sum is $180^\circ$.

(i) First Trapezium:

The top and bottom sides are parallel (indicated by arrows).

1. Angle at bottom left is $135^\circ$. Let the top left angle be $x$.

$x + 135^\circ = 180^\circ$

$x = 180^\circ - 135^\circ = 45^\circ$

2. Angle at bottom right is $105^\circ$. Let the top right angle be $y$.

$y + 105^\circ = 180^\circ$

$y = 180^\circ - 105^\circ = 75^\circ$

Result: The remaining angles are $45^\circ$ and $75^\circ$.

---

(ii) Second Trapezium:

This is an Isosceles Trapezium as indicated by the equal side markings on the non-parallel sides.

1. In an isosceles trapezium, base angles are equal and top angles are equal.

2. Given top left angle is $100^\circ$. Thus, the top right angle is also $100^\circ$.

3. Let the bottom angles be $z$. Since they are supplementary to the top angles:

$z + 100^\circ = 180^\circ$

Result: The remaining angles are $100^\circ, 80^\circ, \text{ and } 80^\circ$.

Venn Diagram Representation:

The hierarchical relationship among quadrilaterals (Parallelograms, Kites, Rhombuses, Rectangles, and Squares) can be visualized as follows:

---

Answers to Sub-Questions:

(i) What is the quadrilateral that is both a kite and a parallelogram?

A Rhombus (and by extension, a Square) is a quadrilateral that is both a kite and a parallelogram. A rhombus has adjacent sides equal (kite property) and opposite sides parallel and equal (parallelogram property).

---

(ii) Can there be a quadrilateral that is both a kite and a rectangle?

Yes. A Square is a quadrilateral that is both a kite and a rectangle. A square has all sides equal (making it a kite) and all angles equal to $90^\circ$ (making it a rectangle).

---

(iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals?

No, every kite is not a rhombus. In a kite, only two pairs of adjacent sides are equal, whereas in a rhombus, all four sides are equal.

Correct Relationship: Every rhombus is a kite, but every kite is not a rhombus.

---

Summary Table:

Intersection	Resulting Quadrilateral
Kite and Parallelogram	Rhombus / Square
Kite and Rectangle	Square
Rhombus to Kite	True (Every rhombus is a kite)

Given: $PAIR$ and $RODS$ are two rectangles. In rectangle $PAIR$, $PR = 5 \text{ cm}$. In rectangle $RODS$, $RS = 5 \text{ cm}$. The angle $\angle ORI = 30^\circ$. Point $O$ lies on the side $AI$ of rectangle $PAIR$.

To Find: The measure of $\angle IOD$.

---

Solution:

In rectangle $PAIR$, the side $RI$ is perpendicular to side $AI$.

$\angle RIA = 90^\circ$ (Vertex angle of a rectangle)

Since point $O$ lies on the segment $AI$, we consider the right-angled triangle $ROI$.

In $\triangle ROI$:

$\angle ORI = 30^\circ$ (Given)

$\angle RIO = 90^\circ$

Using the angle sum property of a triangle:

$\angle ROI = 180^\circ - (90^\circ + 30^\circ)$

$\angle ROI = 180^\circ - 120^\circ = 60^\circ$

Now, consider the rectangle $RODS$. In a rectangle, the angle at each vertex is $90^\circ$.

$\angle ROD = 90^\circ$ (Vertex angle of rectangle $RODS$)

From the figure, $\angle ROD$ is composed of $\angle ROI$ and $\angle IOD$.

$\angle IOD = \angle ROD - \angle ROI$

$\angle IOD = 90^\circ - 60^\circ$

$\angle IOD = 30^\circ$

Conclusion: The measure of $\angle IOD$ is $30^\circ$.

Given:

A square whose diagonal length is $6 \text{ cm}$.

---

To Construct:

A square using only a ruler and compass (without a protractor), utilizing diagonal properties.

---

Properties Used:

The diagonals of a square are equal in length ($6 \text{ cm}$), bisect each other, and intersect at right angles ($90^\circ$).

---

Steps of Construction:

Step 1: Draw a line segment $AC = 6 \text{ cm}$ (the first diagonal).

Step 2: Construct the perpendicular bisector of $AC$. Let it intersect $AC$ at midpoint $O$ ($AO = OC = 3 \text{ cm}$).

Step 3: Since diagonals of a square intersect at $90^\circ$, the perpendicular bisector line already passes through $O$ at $90^\circ$.

Step 4: With centre $O$ and radius $3 \text{ cm}$, cut arcs on the perpendicular bisector line above and below $O$ to mark points $B$ and $D$ ($OB = OD = 3 \text{ cm}$).

Step 5: Join $AB$, $BC$, $CD$, and $DA$ using a straightedge to complete the required square $ABCD$.

---

Summary Table:

Parameter	Value / Property
Diagonal Length	$6 \text{ cm}$
Intersection Angle	$90^\circ$ (Perpendicular bisector)
Segment Halves	$3 \text{ cm}$ each

Part (a): Geometric Reasoning for Quadrilateral $UVWX$

Given:

$CASE$ is a square where $U, V, W,$ and $X$ are the midpoints of sides $CA, CE, ES,$ and $SA$ respectively.

---

To Find:

The type of quadrilateral formed by $UVWX$.

---

Geometric Proof (Side Lengths):

Consider $\triangle VCU$ and $\triangle UAX$:

$\bullet$ $VC = UA$ (Halves of equal sides of square $CASE$)

$\bullet$ $\angle VCU = \angle UAX = 90^\circ$ (Corner angles of the square)

$\bullet$ $CU = AX$ (Halves of equal sides)

Therefore, by Corresponding Parts of Congruent Triangles (CPCT):

Similarly, by proving the other corner triangles congruent, we get $VU = XW = WV$. Thus, all sides of quadrilateral $UVWX$ are equal.

---

Geometric Proof (Interior Angles):

In the isosceles right-angled triangle $\triangle VCU$ ($VC = CU$):

Similarly, for triangle $\triangle UAX$, the base angles are also $45^\circ$.

Now, considering the straight line angle at vertex $U$:

Similarly, all other interior angles ($\angle VXW$, $\angle XWV$, and $\angle WVU$) are each equal to $90^\circ$.

Conclusion for Part (a): Since all sides are equal and all interior angles are $90^\circ$, quadrilateral $UVWX$ is a Square.

---

Part (b): Alternative Construction of a Square within a Square (Figure b)

Given:

A square $ABCD$ (or $CASE$) where vertices of the inner square lie on the sides of the outer square at unequal division points, as shown in Figure (b).

---

Steps of Construction:

Step 1: Let $ABCD$ be any given square.

Step 2: Take four points $P, Q, R,$ and $S$ on the four sides of the square such that they divide the sides at unequal but equal-offset lengths:

Step 3: Since the sides of the outer square are equal, the remaining segments are also equal:

Step 4: Join $PS$, $SR$, $RQ$, and $QP$ to form the inner quadrilateral $PQRS$.

---

Geometric Proof for Figure (b):

Consider $\triangle PAS$ and $\triangle SDR$:

$\bullet$ $PA = SD$

$\bullet$ $\angle PAS = \angle SDR = 90^\circ$

$\bullet$ $AS = DR$

Therefore, by CPCT, $PS = SR$. Similarly, $PS = RQ = QP$, meaning all sides are equal.

By congruent angle addition and straight-line supplementary angles ($180^\circ$), all interior angles of quadrilateral $PQRS$ are $90^\circ$.

Conclusion for Part (b): The inner quadrilateral $PQRS$ is a Square, demonstrating that a square can be constructed inside another square at any rotated angle as long as the offset distances from the corners are equal.

---

Summary Table:

Inner Square Type	Vertex Placement	Geometric Result
Figure (a) ($UVWX$)	Exact midpoints of outer sides	Square (Symmetric)
Figure (b) ($PQRS$)	Equal offset from corners ($AS=BP=\dots$)	Square (Rotated)

Solution:

---

Given:

A quadrilateral with: 1. Four equal sides. 2. Exactly one interior angle equal to $90^\circ$.

---

To Find:

Whether this quadrilateral is guaranteed to be a square.

---

Geometric Reasoning:

1. Properties of a Rhombus: Any quadrilateral with four equal sides is a rhombus. A rhombus also has its opposite sides parallel ($AB \parallel DC$ and $AD \parallel BC$).

2. Consecutive Angles in a Parallelogram / Rhombus: Since opposite sides are parallel, the consecutive angles are supplementary (their sum is $180^\circ$).

3. Similarly, the opposite angle will also be $90^\circ$ ($\angle C = 90^\circ$), and the fourth angle will be $90^\circ$ ($\angle D = 90^\circ$).

4. Therefore, having one interior angle equal to $90^\circ$ in a four-equal-sided quadrilateral forces all four interior angles to be $90^\circ$. Combined with four equal sides, the shape becomes a Square.

---

Construction and Measurement Method:

Step 1: Draw a base line segment $AB$ of any length (e.g., $5 \text{ cm}$).

Step 2: At vertex $A$, construct an angle of $90^\circ$ using a compass or set-square.

Step 3: Cut an arc of radius $5 \text{ cm}$ from $A$ along the perpendicular line to mark vertex $D$ ($AD = 5 \text{ cm}$).

Step 4: With centre $D$ and radius $5 \text{ cm}$, draw an arc. With centre $B$ and radius $5 \text{ cm}$, draw another arc intersecting the previous arc at point $C$.

Step 5: Join $DC$ and $BC$. Measure all interior angles ($\angle B, \angle C, \angle D$) using a protractor. You will find each of them measures exactly $90^\circ$, and all diagonals are equal in length, confirming it is a Square.

Solution:

---

Given:

A quadrilateral in which both pairs of opposite sides are equal ($AB = DC$ and $AD = BC$).

---

To Find / Prove:

Identify the type of quadrilateral and justify the answer using a diagonal and congruent triangles.

---

Identification:

A quadrilateral in which both pairs of opposite sides are equal is a Parallelogram.

---

Geometric Justification (Proof):

Let $ABCD$ be a quadrilateral where opposite sides are equal:

Step 1: Construct a Diagonal.

Draw the diagonal $AC$, which divides the quadrilateral $ABCD$ into two triangles: $\triangle ABC$ and $\triangle CDA$.

Step 2: Check for Congruent Triangles.

In $\triangle ABC$ and $\triangle CDA$:

$\bullet$ $AB = DC$ (Given opposite sides)

$\bullet$ $BC = DA$ (Given opposite sides)

$\bullet$ $AC = CA$ (Common diagonal shared by both triangles)

Step 3: Establish Parallelism.

Since the triangles are congruent, their corresponding parts (CPCT) are equal:

These alternate interior angles formed by the transversal line $AC$ with lines $AB, DC$ and $AD, BC$ are equal. Therefore:

Conclusion: Since both pairs of opposite sides are parallel, $ABCD$ is a Parallelogram.

Given:

A concave quadrilateral $ABCD$ (where one interior angle is reflex/pointing inwards).

---

To Find:

Whether the sum of the interior angles in this quadrilateral is also $360^\circ$, using geometric reasoning and construction/measurement.

---

Geometric Reasoning (Proof):

1. Construct a Diagonal: Join the non-adjacent vertices $B$ and $D$ with a line segment (diagonal $BD$).

2. Splitting into Triangles: The diagonal $BD$ divides the concave quadrilateral $ABCD$ into two separate triangles: $\triangle ABD$ and $\triangle CBD$.

3. Angle Sum Property of Triangles: We know that the sum of interior angles in any triangle is $180^\circ$:

4. Combining the Angles: Adding the angles of both triangles gives the total sum of interior angles of the quadrilateral $ABCD$:

Conclusion: Yes, the sum of the interior angles in a concave quadrilateral is also $360^\circ$.

(i) A quadrilateral whose diagonals are equal and bisect each other must be a square.

False. A quadrilateral whose diagonals are equal and bisect each other is a Rectangle. For it to be a square, the diagonals must also be perpendicular to each other.

---

(ii) A quadrilateral having three right angles must be a rectangle.

True. The sum of angles in a quadrilateral is $360^\circ$. If three angles are $90^\circ$, the fourth angle must be $360^\circ - (90^\circ + 90^\circ + 90^\circ) = 90^\circ$. A quadrilateral with all four angles as $90^\circ$ is a Rectangle.

---

(iii) A quadrilateral whose diagonals bisect each other must be a parallelogram.

True. By definition, if the diagonals of a quadrilateral bisect each other, it is a Parallelogram.

---

(iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.

False. A Kite also has perpendicular diagonals, but it is not necessarily a rhombus (since all four sides of a kite are not necessarily equal).

---

(v) A quadrilateral in which the opposite angles are equal must be a parallelogram.

True. This is a fundamental property. If opposite angles are equal, the adjacent angles become supplementary, implying the opposite sides are parallel.

---

(vi) A quadrilateral in which all the angles are equal is a rectangle.

True. If all angles are equal, each must be $360^\circ / 4 = 90^\circ$. An equiangular quadrilateral is a Rectangle.

---

(vii) Isosceles trapeziums are parallelograms.

False. A parallelogram must have two pairs of parallel sides. An isosceles trapezium has only one pair of parallel sides.