Quadrilaterals

This chapter rigorously examines the fundamental properties and classifications of quadrilaterals, ranging from basic forms like trapeziums to more complex structures such as cyclic quadrilaterals. A thorough understanding of these geometric figures is crucial for mastering foundational concepts in geometry and is frequently assessed in CMI examinations, requiring precise application of theorems and definitions for problem-solving.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Trapezium | | 2 | Parallelogram properties | | 3 | Rhombus | | 4 | Rectangle and square | | 5 | Cyclic quadrilateral |

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We begin with Trapezium.

Part 1: Trapezium

Trapezium

Overview

A trapezium is a quadrilateral with one pair of opposite sides parallel. Problems on trapeziums often mix parallel-line angle facts, proportionality, midpoint ideas, and special properties of the isosceles trapezium. In exam problems, the most useful skill is to separate what is true for every trapezium from what is true only for special ones. ---

Learning Objectives

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Definition

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Angle Relations

These are often the first equations to write in angle problems. ---

Midpoint Segment

This is one of the most important quantitative properties of a trapezium. ---

Isosceles Trapezium

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What Is Not True in General

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Minimal Worked Examples

Example 1 In trapezium $ABCD$, suppose $\qquad AB \parallel CD$ and $\qquad \angle A = 68^\circ$. Find $\angle D$. Since $AD$ is a transversal to the parallel lines, $\qquad \angle A + \angle D = 180^\circ$. So $\qquad \angle D = 180^\circ - 68^\circ = 112^\circ$. --- Example 2 In a trapezium, the bases are of lengths $10$ and $16$. Find the length of the segment joining the midpoints of the non-parallel sides. Using the midpoint theorem, $\qquad MN = \dfrac{10+16}{2} = 13$. So the required length is $\boxed{13}$. ---

How to Recognize an Isosceles Trapezium

In proof-based settings, these facts are often used in reverse. ---

Common Patterns in Questions

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="In trapezium $ABCD$, if $AB\parallel CD$ and $\angle A=74^\circ$, then $\angle D$ equals" options=["$74^\circ$","$96^\circ$","$106^\circ$","$116^\circ$"] answer="C" hint="Use the fact that angles on the same leg are supplementary." solution="Since $AB\parallel CD$, the side $AD$ is a transversal. Hence $\qquad \angle A + \angle D = 180^\circ$. So $\qquad \angle D = 180^\circ - 74^\circ = 106^\circ$. Therefore the correct option is $\boxed{C}$." ::: :::question type="NAT" question="The bases of a trapezium have lengths $8$ and $14$. Find the length of the segment joining the midpoints of the non-parallel sides." answer="11" hint="Use the average of the base lengths." solution="If $M$ and $N$ are the midpoints of the non-parallel sides, then $\qquad MN = \dfrac{8+14}{2} = \dfrac{22}{2} = 11$. Hence the answer is $\boxed{11}$." ::: :::question type="MSQ" question="Which of the following statements are always true in an isosceles trapezium?" options=["The non-parallel sides are equal","The diagonals are equal","The base angles are equal in pairs","Both pairs of opposite sides are parallel"] answer="A,B,C" hint="Remember the difference between trapezium and parallelogram." solution="1. True. This is the definition of an isosceles trapezium.

True. Diagonals are equal in an isosceles trapezium.

True. Base angles are equal in pairs.

False. If both pairs of opposite sides were parallel, it would be a parallelogram.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="State and prove the theorem for the segment joining the midpoints of the non-parallel sides of a trapezium." answer="It is parallel to the bases and has length equal to half the sum of the base lengths." hint="Use a standard auxiliary construction or triangle midpoint arguments." solution="Let $ABCD$ be a trapezium with $\qquad AB \parallel CD$, and let $M$ and $N$ be the midpoints of the non-parallel sides $AD$ and $BC$ respectively. The standard theorem states that the segment $MN$ joining these midpoints is parallel to the two bases and satisfies $\qquad MN = \dfrac{AB+CD}{2}$. A clean proof can be obtained by drawing a diagonal and applying the midpoint theorem in the two triangles formed. Since $M$ and $N$ are midpoints of the legs, the line through them is forced to be parallel to the bases. The lengths contributed from the two triangular midpoint segments add, giving $\qquad MN = \dfrac{AB}{2} + \dfrac{CD}{2} = \dfrac{AB+CD}{2}$. Hence the segment joining the midpoints of the non-parallel sides is parallel to the bases and has length equal to half the sum of the base lengths." ::: ---

Summary

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Part 2: Parallelogram properties

Parallelogram Properties

Overview

A parallelogram is one of the most important quadrilaterals in Euclidean geometry because several side, angle, and diagonal properties interact at once. In exam problems, the main skill is not memorizing isolated facts, but knowing which property is the fastest entry point: parallel sides, equal opposite sides, angle relations, diagonal bisection, or a converse test. ---

Learning Objectives

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Definition

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Basic Properties

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Why Adjacent Angles Are Supplementary

This helps convert one known angle into all the others. ---

Diagonal Properties

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Triangle Structure Inside a Parallelogram

So area and length relations often come from triangle congruence. ---

Converse Tests

These are very important in proof questions. ---

One Powerful Converse

This theorem appears often because it is shorter to verify than checking both pairs. ---

Minimal Worked Examples

Example 1 In a parallelogram, one angle is $68^\circ$. Find all four angles. Opposite angles are equal and adjacent angles are supplementary. So: $\qquad \angle A = 68^\circ$ $\qquad \angle B = 180^\circ - 68^\circ = 112^\circ$ Hence: $\qquad \angle C = 68^\circ,\quad \angle D = 112^\circ$ --- Example 2 In parallelogram $ABCD$, diagonals intersect at $O$. If $AO=7$ and $BO=5$, find $AC$ and $BD$. Since diagonals bisect each other: $\qquad AO = OC = 7$ $\qquad BO = OD = 5$ So: $\qquad AC = AO + OC = 14$ $\qquad BD = BO + OD = 10$ ---

Common Patterns in Questions

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="In parallelogram $ABCD$, if $\angle A=72^\circ$, then $\angle B$ equals" options=["$72^\circ$","$108^\circ$","$118^\circ$","$144^\circ$"] answer="B" hint="Adjacent angles in a parallelogram are supplementary." solution="In a parallelogram, adjacent angles sum to $180^\circ$. Hence $\qquad \angle B = 180^\circ - 72^\circ = 108^\circ$. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="The diagonals of a parallelogram intersect at $O$. If $AO=9$, find $AC$." answer="18" hint="Diagonals bisect each other." solution="In a parallelogram, diagonals bisect each other, so $\qquad AO = OC = 9$. Hence $\qquad AC = AO + OC = 18$. Therefore the answer is $\boxed{18}$." ::: :::question type="MSQ" question="Which of the following statements are always true for a parallelogram?" options=["Opposite sides are equal","Opposite angles are equal","Diagonals bisect each other","Diagonals are equal"] answer="A,B,C" hint="Separate general parallelogram facts from rectangle-specific facts." solution="1. True. Opposite sides are equal.

True. Opposite angles are equal.

True. Diagonals bisect each other.

False. Diagonals need not be equal in a general parallelogram.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Prove that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram." answer="Use congruent triangles formed by the diagonals." hint="Let the diagonals intersect at $O$ and compare suitable triangles." solution="Let the diagonals of quadrilateral $ABCD$ intersect at $O$, and suppose $\qquad AO=OC \quad \text{and} \quad BO=OD$. Now consider triangles $\triangle AOB$ and $\triangle COD$. We have:

• $\qquad AO = OC$

• $\qquad BO = OD$

• $\qquad \angle AOB = \angle COD$ because they are vertically opposite angles

So $\qquad \triangle AOB \cong \triangle COD$ by SAS. Hence $\qquad \angle ABO = \angle CDO$ and $\qquad \angle BAO = \angle DCO$. These equal angles imply $\qquad AB \parallel CD$. Similarly, by comparing triangles $\triangle AOD$ and $\triangle COB$, we get $\qquad AD \parallel BC$. Thus both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram." ::: ---

Summary

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Part 3: Rhombus

Rhombus

Overview

A rhombus is one of the most important special quadrilaterals in Euclidean geometry because it combines the structure of a parallelogram with the extra condition that all four sides are equal. In exam problems, rhombus questions often involve diagonals, angle bisectors, right triangles, area formulas, and sometimes a larger surrounding figure such as a rectangle. The PYQ for this topic also shows that rhombus questions can lead to nontrivial similarity and ratio arguments. ---

Learning Objectives

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Definition and Basic Structure

So in rhombus $ABCD$, $\qquad AB\parallel CD,\quad BC\parallel AD$ $\qquad \angle A=\angle C,\quad \angle B=\angle D$ $\qquad \angle A+\angle B=180^\circ$ ::: ---

Diagonal Properties

So the diagonals of a rhombus are not just bisectors of each other; they are also perpendicular and act as angle bisectors. ---

Why the Diagonals Are So Useful

This gives one of the most useful formulas: ---

Area of a Rhombus

The diagonal formula is especially efficient in geometry problems. ---

Angle Bisector Role of the Diagonals

This often creates similar triangles. ---

Minimal Worked Examples

Example 1 The diagonals of a rhombus are $10$ and $24$. Find its side length. Each diagonal is bisected at the intersection point, so the half-diagonals are $\qquad 5\quad \text{and}\quad 12$ Thus the side is the hypotenuse of a right triangle: $\qquad s^2=5^2+12^2=25+144=169$ So $\qquad s=13$ Hence the side length is $\boxed{13}$. --- Example 2 The diagonals of a rhombus are $8$ and $6$. Find its area. Using the diagonal formula, $\qquad \text{Area}=\dfrac{1}{2}\cdot 8\cdot 6=24$ So the area is $\boxed{24}$. ---

Similarity Inside a Rhombus

In particular, if diagonals meet at $G$, then the four triangles around $G$ are right triangles, and many of them are congruent. ---

Rhombus Inside a Rectangle

The PYQ for this topic uses a rectangle around the rhombus. This is an important advanced configuration. Suppose rectangle $EBFD$ is given and rhombus $ABCD$ is placed so that:

• $B$ and $D$ are opposite corners of the rectangle,

• $A$ lies on side $ED$,

• $C$ lies on side $BF$,

• diagonals of the rhombus meet at $G$.

Then the following facts become very useful. ---

Coordinate Model for Advanced Problems

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Common Question Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which of the following is always true for a rhombus?" options=["Its diagonals are always equal","Its diagonals bisect each other at right angles","Each angle is $90^\circ$","Its adjacent sides are unequal"] answer="B" hint="Use the standard diagonal properties of a rhombus." solution="A rhombus is a parallelogram with all sides equal. Its diagonals bisect each other and are perpendicular. They are not always equal, and all angles need not be right angles. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="The diagonals of a rhombus are $10$ and $24$. Find its side length." answer="13" hint="Half the diagonals form a right triangle with the side." solution="The half-diagonals are $5$ and $12$. Since the diagonals of a rhombus are perpendicular bisectors, the side is $\qquad \sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$ Hence the answer is $\boxed{13}$." ::: :::question type="MSQ" question="Which of the following statements are true for every rhombus?" options=["Opposite sides are parallel","Diagonals bisect each other","Diagonals are always equal","Each diagonal bisects a pair of opposite angles"] answer="A,B,D" hint="Remember that a rhombus is a special parallelogram." solution="A rhombus is a parallelogram, so opposite sides are parallel and diagonals bisect each other. Also, in a rhombus each diagonal bisects a pair of opposite angles. But the diagonals are not always equal; that happens only in a square. Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Prove that the diagonals of a rhombus are perpendicular." answer="Using equal side lengths and diagonal bisection, the triangles formed are congruent, which forces the diagonals to meet at right angles" hint="Use the midpoint of the diagonals and compare the triangles around the intersection." solution="Let rhombus $ABCD$ have diagonals $AC$ and $BD$ intersecting at $G$. Since a rhombus is a parallelogram, its diagonals bisect each other, so $\qquad AG=GC$ and $\qquad BG=GD$. Also all sides of a rhombus are equal, so $\qquad AB=BC=CD=DA$. Now compare triangles $\triangle AGB$ and $\triangle CGB$:

• $AG=CG$

• $BG$ is common

• $AB=CB$

So the triangles are congruent. Hence $\qquad \angle AGB=\angle BGC$. But $A,G,C$ are collinear, so $\qquad \angle AGB+\angle BGC=180^\circ$. Equal supplementary angles are each $90^\circ$, so $\qquad \angle AGB=\angle BGC=90^\circ$. Therefore $\qquad AC\perp BD$. Hence the diagonals of a rhombus are perpendicular." ::: ---

Summary

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Part 4: Rectangle and square

Rectangle and Square

Overview

Rectangle and square are among the most important special quadrilaterals in Euclidean geometry. In exam problems, they are rarely tested only through definitions; instead, they appear through diagonals, angle conditions, symmetry, cyclicity, coordinate geometry, and characterization inside a larger figure such as a parallelogram or rhombus. ---

Learning Objectives

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Definitions

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Core Properties of a Rectangle

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Core Properties of a Square

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Area, Perimeter, and Diagonal

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Relation With Other Quadrilaterals

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Cyclic Nature

A square is therefore also cyclic. ---

High-Value Geometry Observations

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Coordinate View

This makes it easy to compute:

• diagonal lengths

• midpoint of diagonals

• slopes for perpendicularity or parallelism

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which of the following is always true for a rectangle?" options=["Its diagonals are perpendicular","Its diagonals are equal","All its sides are equal","It has exactly one pair of parallel sides"] answer="B" hint="Recall the defining and derived properties of a rectangle." solution="A rectangle is a parallelogram with all angles equal to $90^\circ$. Its diagonals are always equal and bisect each other. They are not always perpendicular, and all sides need not be equal. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="The diagonal of a square is $10\sqrt{2}$. Find its side length." answer="10" hint="Use $d=s\sqrt{2}$." solution="For a square, $\qquad d=s\sqrt{2}$. Given $\qquad d=10\sqrt{2}$, we get $\qquad s=10$. Hence the answer is $\boxed{10}$." ::: :::question type="MSQ" question="Which of the following conditions are sufficient to conclude that a quadrilateral is a square?" options=["It is a rectangle with two adjacent equal sides","It is a rhombus with one right angle","It is a parallelogram with equal diagonals and perpendicular diagonals","It is a quadrilateral with equal diagonals"] answer="A,B,C" hint="A square is both a rectangle and a rhombus." solution="1. True. A rectangle with adjacent equal sides has all four sides equal, so it is a square.

True. A rhombus with one right angle has all right angles, hence is a square.

True. In a parallelogram, equal diagonals imply rectangle and perpendicular diagonals imply rhombus, so both together imply square.

False. Equal diagonals alone do not force a square; a non-square rectangle also has equal diagonals.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Prove that a parallelogram with equal diagonals is a rectangle." answer="True" hint="Use triangle congruence or angle comparison." solution="Let $ABCD$ be a parallelogram with $AC=BD$. In a parallelogram, opposite sides are equal, so $\qquad AD=BC$, and $AB$ is common to triangles $ABD$ and $BAC$. Thus in triangles $ABD$ and $BAC$:

• $AB=BA$

• $AD=BC$

• $BD=AC$

So the triangles are congruent by SSS. Hence $\qquad \angle DAB = \angle ABC$. But adjacent angles of a parallelogram are supplementary. Therefore two equal supplementary angles must each be $90^\circ$. So the parallelogram has a right angle, and hence it is a rectangle. Therefore the answer is $\boxed{\text{True}}$." ::: ---

Summary

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Part 5: Cyclic quadrilateral

Cyclic Quadrilateral

Overview

A cyclic quadrilateral is a quadrilateral whose four vertices lie on a single circle. This is one of the most important structures in Euclidean geometry because it converts lengths, chords, arcs, and angles into a tightly connected system. In exam-level questions, cyclic quadrilaterals are often hidden inside bigger figures and used through angle-chasing, diameter-based right angles, and strong identities such as Ptolemy’s theorem. The PYQ pattern for this topic strongly suggests that you should be comfortable not only with standard angle facts, but also with:

• identifying right angles from diameters

• using equal angles in the same segment

• handling variable points on an arc

• applying Ptolemy’s theorem in a clean algebraic way

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Learning Objectives

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Core Definition

When this happens, the circle is called the circumcircle of the quadrilateral. ---

First Main Property

This is often the fastest way to use cyclicity in an angle-chasing problem. ---

Converse of the First Property

This is one of the standard converse tests for cyclicity. ---

Exterior Angle Property

This is just another form of the supplementary-angle property and is extremely useful in short proofs. ---

Angles in the Same Segment

This fact is one of the deepest angle tools inside cyclic-quadrilateral problems. ---

Diameter Gives Right Angle

This is especially important when a cyclic quadrilateral contains one or more diameters. Similarly, if $BD$ is a diameter, then the angles subtending $BD$ are right angles. ::: ---

Ptolemy’s Theorem

This is the strongest standard length relation for cyclic quadrilaterals and is often the key in medium-to-hard exam problems. ::: ---

When Ptolemy Becomes Especially Useful

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Diameter + Ptolemy Combination

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Minimal Worked Examples

Example 1 If $ABCD$ is cyclic and $\angle A = 68^\circ$, find $\angle C$. Since opposite angles of a cyclic quadrilateral are supplementary, $\qquad \angle C = 180^\circ - 68^\circ = 112^\circ$ So the answer is $\boxed{112^\circ}$. --- Example 2 Suppose $ABCD$ is cyclic with $\qquad AB=1,\ BC=2,\ CD=3,\ DA=6$ and $\qquad AC=5$. Find $BD$. Using Ptolemy’s theorem, $\qquad AC\cdot BD = AB\cdot CD + BC\cdot AD$ So $\qquad 5\cdot BD = 1\cdot 3 + 2\cdot 6 = 15$ Thus $\qquad BD = 3$ So the answer is $\boxed{3}$. ---

Common Cyclicity Tests

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $ABCD$ is a cyclic quadrilateral and $\angle A=72^\circ$, then $\angle C$ equals" options=["$72^\circ$","$98^\circ$","$108^\circ$","$118^\circ$"] answer="C" hint="Use the opposite-angle property." solution="Opposite angles of a cyclic quadrilateral are supplementary. So $\qquad \angle C = 180^\circ - 72^\circ = 108^\circ$. Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="In a cyclic quadrilateral, one exterior angle is $47^\circ$. Find the interior opposite angle." answer="47" hint="Use the exterior-angle property." solution="In a cyclic quadrilateral, an exterior angle equals the interior opposite angle. Therefore the required angle is $\boxed{47^\circ}$, so the numerical answer is $\boxed{47}$." ::: :::question type="MSQ" question="Which of the following statements are true for a cyclic quadrilateral?" options=["Opposite angles are supplementary","If one pair of opposite angles is supplementary, the quadrilateral is cyclic","An exterior angle equals the adjacent interior angle","Ptolemy's theorem holds"] answer="A,B,D" hint="Check which are special to cyclic quadrilaterals." solution="1. True.

True, by the converse test for cyclicity.

False. An exterior angle equals the interior opposite angle, not the adjacent interior angle.

True.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Suppose $ABCD$ is cyclic and $AC$ is a diameter. Prove that $\angle ABC=90^\circ$." answer="$\angle ABC=90^\circ$" hint="Use the angle-in-a-semicircle theorem." solution="Since $AC$ is a diameter, the arc $AC$ is a semicircle. The angle $\angle ABC$ subtends the diameter $AC$. By the angle-in-a-semicircle theorem, any angle subtending a diameter is a right angle. Therefore, $\qquad \boxed{\angle ABC = 90^\circ}$." ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="If the diagonals of a parallelogram are equal in length, which of the following types of parallelogram must it be?" options=["Rhombus","Square","Rectangle","Trapezium"] answer="Rectangle" hint="Consider the properties of diagonals for each special parallelogram." solution="A parallelogram with equal diagonals is a rectangle. While a square also has equal diagonals, it is a specific type of rectangle, and 'Rectangle' is the most general correct answer here."
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:::question type="NAT" question="In a cyclic quadrilateral $ABCD$, if $\angle A = 2x^\circ$, $\angle B = (3x-10)^\circ$, and $\angle C = (x+50)^\circ$, what is the measure of $\angle D$ in degrees?" answer="60" hint="In a cyclic quadrilateral, opposite angles sum to $180^\circ$. Use this property to find $x$ and then $\angle B$ and $\angle D$." solution="For a cyclic quadrilateral, $\angle A + \angle C = 180^\circ$.
So, $2x + (x+50) = 180 \implies 3x + 50 = 180 \implies 3x = 130 \implies x = \frac{130}{3}$.
Now, $\angle B = (3x-10)^\circ = \left(3\left(\frac{130}{3}\right)-10\right)^\circ = (130-10)^\circ = 120^\circ$.
Since $\angle B + \angle D = 180^\circ$, we have $\angle D = 180^\circ - 120^\circ = 60^\circ$."
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:::question type="MCQ" question="The midpoints of the sides of a quadrilateral $PQRS$ are joined in order to form a new quadrilateral $ABCD$. If $PQRS$ is a kite, which of the following best describes quadrilateral $ABCD$?" options=["Rhombus","Rectangle","Square","Trapezium"] answer="Rectangle" hint="Recall the property that the quadrilateral formed by joining the midpoints of a general quadrilateral has sides parallel to the diagonals of the original quadrilateral. Consider the properties of diagonals in a kite." solution="The quadrilateral formed by joining the midpoints of the sides of any quadrilateral is a parallelogram. For a kite, the diagonals are perpendicular. The sides of the midpoint quadrilateral are parallel to the diagonals of the original quadrilateral. If the diagonals of $PQRS$ are perpendicular, then the sides of $ABCD$ will also be perpendicular, making $ABCD$ a rectangle."
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:::question type="NAT" question="A trapezium has parallel sides of length $10 \text{ cm}$ and $18 \text{ cm}$. If its height is $6 \text{ cm}$, what is its area in $\text{cm}^2$?" answer="84" hint="The area of a trapezium is given by the formula $\frac{1}{2}(a+b)h$, where $a$ and $b$ are the lengths of the parallel sides and $h$ is the height." solution="The area of a trapezium is given by $\operatorname{Area} = \frac{1}{2}(a+b)h$.
Given $a=10 \text{ cm}$, $b=18 \text{ cm}$, and $h=6 \text{ cm}$.
$\operatorname{Area} = \frac{1}{2}(10+18)(6) = \frac{1}{2}(28)(6) = 14 \times 6 = 84 \text{ cm}^2$."
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What's Next?