Quadrilaterals ⇒⇒ Exercise 4

Question 1

Observe the following figures.

Solution :

From observation, we find that Figure (iv) and (v) are not a quadrilateral because one of its “sides” is curved, which does not fulfill the condition that all sides of a polygon (and thus a quadrilateral) must be straight line segments.

Question 2

Is it wrong to write ∆BAD ≅ ∆CDB? Why?

Solution :

The statement Δ𝐵𝐴D≅ Δ𝐶𝐷𝐵 is potentially incorrect, and here's why:

For two triangles to be congruent, they must satisfy one of the congruence criteria (such as SSS, SAS, ASA, AAS, or HL). This means the corresponding sides and angles must be equal.

The congruence symbol ≅ implies that these two triangles have identical shape and size, meaning their corresponding sides and angles must be equal.

Possible issues:

Orientation: Without a clearer context, the triangles may not even be in the same orientation or relative position in the plane.

Just saying Δ𝐵𝐴D≅ Δ𝐶𝐷𝐵 doesn't indicate whether the triangles are flipped or rotated relative to each other.

Side Lengths and Angles:

For instance, the side 𝐵𝐷 is common to both triangles, but the other sides AB,AD and CD,BC might not be equal.

Question 1 (i)

Find all the other angles inside the following rectangles.

Solution :

(i) Given $ABCD$ is a rectangle :

All interior angles are $90^\circ$ ($\angle DAB = \angle ABC$ $= \angle BCD = \angle CDA = 90^\circ$).

The diagonals are equal in length ($AC = BD$) and bisect each other.

Because the diagonals bisect each other, and they are equal, all four segments created at the intersection point $O$ are equal:

$$AO = OC = BO = OD$$

This means the four triangles formed at the center ($\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$) are all isosceles triangles.

In $\triangle AOB$

We are given $\angle CAB = 30^\circ$. Note that $\angle CAB$ is the same as $\angle OAB$ in $\triangle AOB$.

Triangle $\triangle AOB$ is isosceles because $AO = BO$.

$$\angle OBA = \angle OAB = 30^\circ$$

( The angles opposite the equal sides are equal: )

We can now find $\angle AOB$:

(Angle sum property of triangles )

$$\angle AOB = 180^\circ - (\angle OAB + \angle OBA)$$

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

$$\angle AOB = 180^\circ - 60^\circ$$

$$\mathbf{\angle AOB = 120^\circ}$$

Now Find $\angle AOD$:

$$\angle AOD = 180^\circ - \angle AOB \quad \text{(Linear Pair)}$$

(The angles $\angle AOB$ and $\angle AOD$ form a linear pair on the diagonal $AC$, so their sum is $180^\circ$ )

$$\angle AOD = 180^\circ - 120^\circ$$

$$\mathbf{\angle AOD = 60^\circ}$$

Other Adjacent Central Angles : $$\angle BOC = \angle AOD $$

$$\angle COD = \angle AOB $$

( Vertically Opposite Angles )

In $\triangle AOD$

Triangle $\triangle AOD$ is isosceles because $AO = OD$.

$$\angle ODA = \angle OAD $$

( The angles opposite the equal sides are equal: )

We know $\angle AOD = 60^\circ$

$$\angle OAD = \angle ODA = \frac{180^\circ - \angle AOD}{2}$$

$$\angle OAD = \angle ODA = \frac{180^\circ - 60^\circ}{2} = \frac{120^\circ}{2}$$

$$\mathbf{\angle OAD = \angle ODA = 60^\circ}$$

Therefore

$$\angle OAD = \angle OCB = 60^\circ$$

$$\angle OAB = \angle OCD = 30^\circ$$

( Alternate interior angles )

$$\angle OBC = \angle ODA = 60^\circ$$

$$\angle OBA = \angle ODC = 30^\circ$$

( Alternate interior angles )

Question 1 (ii)

Find all the other angles inside the following rectangles.

Solution :

(ii) Given $PQRS$ is a rectangle :

The diagonals $PR$ and $QS$ are equal and bisect each other.

Because the diagonals bisect each other, and they are equal, all four segments created at the intersection point $O$ are equal:

$$PO = OR = QO = OS$$

This means the four triangles formed at the center ($\triangle POQ, \triangle QOR, \triangle ROS, \triangle SOP$) are all isosceles triangles.

( The angles opposite the equal sides are equal: )

In $\triangle QOR$

Triangle $\triangle QOR$ is isosceles because $QO = RO$.

$\angle OQR$ and $\angle ORQ$ are equal.

We know $\angle QOR = 110^\circ$

$$\angle OQR = \angle ORQ = \frac{180^\circ - \angle QOR}{2}$$

(Angle sum property of triangles )

$$\angle OQR = \angle ORQ = \frac{180^\circ - 110^\circ}{2} = \frac{70^\circ}{2}$$

$$\mathbf{\angle OQR = \angle ORQ = 35^\circ}$$

Now Find $\angle POQ$:

$$\angle POQ = 180^\circ - \angle QOR \quad \text{(Linear Pair)}$$

(The angles $\angle POQ$ and $\angle QOR$ form a linear pair on the diagonal $PR$, so their sum is $180^\circ$ )

$$\angle POQ = 180^\circ - 110^\circ$$

$$\mathbf{\angle POQ = 70^\circ}$$

Now Find Other Adjacent Central Angles :

$$\angle ROS = \angle POQ = {70^\circ}$$

$$\angle QOR= \angle POS = {110^\circ}$$

( (Vertically Opposite Angles) )

In $\triangle POQ$

Triangle $\triangle POQ$ is isosceles because $PO = QO$.

$$\angle OPQ = \angle OQP $$

( The angles opposite the equal sides are equal: )

We know $\angle POQ = 70^\circ$

$$\angle OPQ = \angle OQP = \frac{180^\circ - \angle POQ}{2}$$

$$\angle OPQ = \angle OQP = \frac{180^\circ - 70^\circ}{2} = \frac{110^\circ}{2}$$

$$\mathbf{\angle OPQ = \angle OQP = 55^\circ}$$

Therefore

$$\angle OPQ = \angle ORS = 55^\circ$$

$$\angle OPS = \angle ORQ = 35^\circ$$

( Alternate interior angles )

$$\angle OSR = \angle OQP = 55^\circ$$

$$\angle OSP = \angle OQR = 35^\circ$$

( Alternate interior angles )

Question 2

Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of
(i) 30° (ii) 40°
(iii) 90° (iv) 140°

Solution :

(i) Steps for Construction (Intersection Angle $\mathbf{30^\circ}$)

• 1. Draw a line segment $AC$ of length 8 cm.
• 2. Locate the midpoint, $O$, of $AC$ by measuring 4 cm from $A$ (or $C$) and marking the point $O$.


• Mark a point that corresponds to a $30^\circ$ angle with $OA$.
• Draw a long straight line (a ray) passing through $O$ and the $30^\circ$ mark. This line will contain the second diagonal, $BD$.







• Set your compass to a radius of 4 cm (half the length of the diagonal).
• With the compass point at $O$, mark a point $B$ on the $30^\circ$ line on one side of $AC$.


• Extend the $30^\circ$ line through $O$. With the compass point still at $O$, mark a point $D$ on the opposite side of $AC$ such that $OD = 4$ cm.
• The length of $BD$ should now be $4 \text{ cm} + 4 \text{ cm} = 8 \text{ cm}$.








• join the consecutive points : $A$ to $B$ , $B$ to $C$, $C$ to $D$, $D$ to $A$

The resulting figure $ABCD$ is the required quadrilateral.

Now, Draw yourself images for (ii) and (iii), (iv) as it is solved in above (i) part of question.

Question 3

Consider a circle with centre O. Line segments PL and AM are two perpendicular diameters of the circle. What is the figure APML? Reason and/or experiment to figure this out.

Solution :

The sides $PL$ and $AM$ are given as line segments. In the context of the quadrilateral $APML$, these segments are the diagonals.

$PL$ and $AM$ are both diameters of the same circle.

$$PL = AM$$

( All diameters of a circle are equal in length )

The diagonals $PL$ and $AM$ both pass through the centre $O$.

Since $O$ is the centre, it divides each diameter into two equal radii.

$PO = OL$ and $AO = OM$

Thus, the diagonals bisect each other.

The line segments $PL$ and $AM$ are given as perpendicular diameters.

$\angle AOL = \angle LOM$ $= \angle MOP = \angle POA = 90^\circ$

Conclusion from Diagonals: A quadrilateral whose diagonals are equal, bisect each other, and are perpendicular is a square.

Question 4

We have seen how to get 90° using paper folding. Now, suppose we do not have any paper but two sticks of equal length and a thread. How do we make an exact 90° using these?

Solution :

You can make an exact $90^\circ$ angle (a right angle) using the two equal sticks and the thread by creating a specific right-angled triangle.

This method is based on the Pythagorean triple $(3, 4, 5)$. If a triangle's sides have lengths that are in the ratio $3:4:5$, the angle opposite the longest side (the hypotenuse) is exactly $90^\circ$.

Since your sticks are of equal length, let's call this length $L$. We'll use the thread to form a loop that measures $3L + 4L + 5L = 12L$ in total circumference.

Take the equal length two sticks and place them on the ground to form a corner .

• Use the first stick to measure and form the side corresponding to 3 units on the ground.
• Use the second stick to measure and form the side corresponding to 4 units on the ground, making sure the ends meet at a single point (the potential $90^\circ$ corner).


• Stretch the 5-unit segment of the thread between the free ends of the 3-unit side and the 4-unit side.


• If the thread is pulled taut and the 3-unit, 4-unit, and 5-unit segments meet exactly at their respective marks, the angle between the 3-unit side and the 4-unit side will be exactly $90^\circ$.

Since $3^2 + 4^2 = 5^2$, the triangle is guaranteed to have a right angle.

Question 5

We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal a rectangle?

Solution :

The answer is No, the properties "opposite sides are parallel" and "opposite sides are equal" cannot be chosen as the definition of a rectangle.

These two properties define a broader category of quadrilaterals: the Parallelogram.

The Missing Defining Property of a Rectangle

A parallelogram becomes a rectangle if you add just one of the following conditions:

1. At least one of its interior angles is a right angle ($90^\circ$).

2. Its diagonals are equal in length.

Therefore, A Valid Definition of a Rectangle

A quadrilateral with four right angles and A parallelogram with equal diagonals.

Question 1 (i)

Find the remaining angles in the following quadrilaterals.

Solution :

A quadrilateral PEAR in which both pairs of opposite sides are parallel is defined as a Parallelogram

The given information:

$PR \parallel EA$

$PE \parallel RA$

$\angle P = 40^\circ$

In a parallelogram,

$$\angle A = \angle P = {40^\circ}$$

$$\angle E = \angle R$$

( All Opposite Angles are Equal )

In a parallelogram, consecutive (adjacent) angles are supplementary (they add up to $180^\circ$).

$\angle E$ is consecutive to $\angle P$.

$$\angle P + \angle E = 180^\circ$$

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

$$\mathbf{\angle E = 140^\circ}$$

Since opposite angles are equal, and we found

$$\angle R = \angle E = {140^\circ}$$

$$\angle A = \angle P = {40^\circ}$$

Question 1 (ii)

Find the remaining angles in the following quadrilaterals.

Solution :

A quadrilateral PQRS in which both pairs of opposite sides are parallel is defined as a Parallelogram

The given information:

$PQ \parallel SR$

$PS \parallel QR$

$\angle P = 110^\circ$

In a parallelogram,

$$\angle P = \angle R = {110^\circ}$$

$$\angle S = \angle Q$$

( All Opposite Angles are Equal )

In a parallelogram, consecutive (adjacent) angles are supplementary (they add up to $180^\circ$).

$\angle Q$ is consecutive to $\angle P$.

$$\angle P + \angle Q = 180^\circ$$

$$\angle Q = 180^\circ - 110^\circ$$

$$\mathbf{\angle Q = 70^\circ}$$

Since opposite angles are equal, and we found

$$\angle Q = \angle S = {70^\circ}$$

$$\angle R = \angle P = {110^\circ}$$

Question 1 (iii)

Find the remaining angles in the following quadrilaterals.

Solution :

The quadrilateral $XWUV$ where all sides are equal is a Rhombus.

In a rhombus, the diagonals bisect the corner angles. If the quadrilateral's vertices are $X, W, U, V$, the diagonals would be $XV$ and $WU$.

The given information:

$XW = WV = VU = UX$ .

$\angle UVX = 30^\circ$

In $\triangle VUX$

Triangle $\triangle VUX$ is isosceles because $UV = UX$.

$$\angle UXV = \angle UVX = 30^\circ$$

( The angles opposite the equal sides are equal: )

The third angle is: $\angle X U V = \angle U$ $$\angle U = 180^\circ - (30^\circ + 30^\circ) $$

$$\angle U = 120^\circ$$

(Angle sum property of triangles )

Now, Find $\angle V$ :

In a rhombus, consecutive (adjacent) angles are supplementary (they add up to $180^\circ$).

$$\angle V + \angle U = 180^\circ$$

$$\angle V + 120^\circ = 180^\circ$$

$$\mathbf{\angle V = 60^\circ}$$

Since Opposite angles of a rhombus are equal, and we found

$$\angle X = \angle V = {60^\circ}$$

$$\angle W = \angle U = {120^\circ}$$

Question 1 (iv)

Find the remaining angles in the following quadrilaterals.

Solution :

The quadrilateral $AEIO$ where all sides are equal is a Rhombus.

In a rhombus, the diagonals bisect the corner angles. If the quadrilateral's vertices are $A, E, I, O$, the diagonals would be $OE$ and $AI$.

The given information:

$AE = AO = OI = EI$ .

$\angle AEO = 20^\circ$

In $\triangle OAE$

Triangle $\triangle OAE$ is isosceles because $AE = AO$.

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

( The angles opposite the equal sides are equal: )

The third angle is: $\angle OAE = \angle A$ $$\angle A = 180^\circ - (20^\circ + 20^\circ) $$

$$\angle A = 140^\circ$$

(Angle sum property of triangles )

Now, Find $\angle E$ :

In a rhombus, consecutive (adjacent) angles are supplementary (they add up to $180^\circ$).

$$\angle E + \angle A = 180^\circ$$

$$\angle E + 140^\circ = 180^\circ$$

$$\mathbf{\angle E = 40^\circ}$$

Since Opposite angles of a rhombus are equal, and we found

$$\angle E = \angle O = {40^\circ}$$

$$\angle A = \angle I = {140^\circ}$$

Question 2

Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140°.

Solution :

(i) Steps for Construction (Intersection Angle $\mathbf{140^\circ}$)

• 1. Draw a line segment $AC$ of length 7 cm.
• 2. Locate the midpoint, $O$, of $AC$ by measuring 3.5 cm from $A$ (or $C$) and marking the point $O$.


• Mark a point that corresponds to a $140^\circ$ angle with $OA$.
• Draw a long straight line (a ray) passing through $O$ and the $140^\circ$ mark. This line will contain the second diagonal, $BD$.







• Set your compass to a radius of 2.5 cm (half the length of the diagonal).
• With the compass point at $O$, mark a point $B$ on the $140^\circ$ line on one side of $AC$.


• Extend the $140^\circ$ line through $O$. With the compass point still at $O$, mark a point $D$ on the opposite side of $AC$ such that $OD = 2.5$ cm.
• The length of $BD$ should now be $2.5 \text{ cm} + 2.5 \text{ cm} = 5 \text{ cm}$.








• join the consecutive points : $A$ to $B$ , $B$ to $C$, $C$ to $D$, $D$ to $A$

The resulting figure $ABCD$ is the required quadrilateral.

Question 3

Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm.

Solution :

(i) Steps for Construction (Intersection Angle $\mathbf{90^\circ}$)

• 1. Draw a line segment $AC$ of length 4 cm.
• 2. Locate the midpoint, $O$, of $AC$ by measuring 2 cm from $A$ (or $C$) and marking the point $O$.


• Mark a point that corresponds to a $90^\circ$ angle with $OA$.
• Draw the perpendicular bisector of AC passing through $O$ . This line will contain the second diagonal, $BD$.


• Set your compass to a radius of 2.5 cm (half the length of the diagonal).
• With the compass point at $O$, mark a point $B$ on the $90^\circ$ line on one side of $AC$.


• Extend the $90^\circ$ line through $O$. With the compass point still at $O$, mark a point $D$ on the opposite side of $AC$ such that $OD = 2.5$ cm.
• The length of $BD$ should now be $2.5 \text{ cm} + 2.5 \text{ cm} = 5 \text{ cm}$.








• join the consecutive points : $A$ to $B$ , $B$ to $C$, $C$ to $D$, $D$ to $A$

The resulting figure $ABCD$ is the required rhombus.

Question 1

Take two cardboard cutouts of an equilateral triangle of side length 8 cm. Can you join them to get a quadrilateral? What type of quadrilateral is this? Justify your answer.

Solution :

Let the two equilateral triangles be $\triangle ABC$ and $\triangle ADC$. When joined along the common side $AC$, the resulting quadrilateral is $ABCD$.

Since both triangles are equilateral with a side length of 8 cm, all their sides are 8 cm long.

A quadrilateral with all four sides equal is defined as a Rhombus.

Opposite Angles $\angle B$ and $\angle D$:

Since both triangles are equilateral, all their interior angles are $60^\circ$.

$\angle BAC = \angle BCA = 60^\circ$

$\angle DAC = \angle DCA = 60^\circ$

$$\angle B = \angle ABC = \mathbf{60^\circ}$$

$$\angle D = \angle ADC = \mathbf{60^\circ}$$

Opposite Angles $\angle A$ and $\angle C$:

$\angle A = \angle BAC + \angle DAC = 60^\circ + 60^\circ = \mathbf{120^\circ}$

$\angle C = \angle BCA + \angle DCA = 60^\circ + 60^\circ = \mathbf{120^\circ}$

Since the figure has four equal sides (8 cm) and opposite angles are equal ($60^\circ$ and $120^\circ$), it is confirmed to be a Rhombus.

Question 2

Take two cardboard cutouts of an isosceles triangle with sidelengths 8 cm, 8 cm, and 6 cm.

Solution :

For an isosceles triangle with side lengths 8 cm, 8 cm, and 6 cm, there are two possibilities for the common side:

Case 1: Joining Along the Base (6 cm side)

The resulting quadrilateral has two pairs of equal adjacent sides

The two $8 \text{ cm}$ sides of the first triangle.

The two $8 \text{ cm}$ sides of the second triangle.

The common side (6 cm) is one diagonal. The other diagonal is formed by joining the vertices opposite the 6 cm side.

Since $8 = 8$ and $8 = 8$, the sides are grouped into two distinct pairs of equal-length sides that are next to each other. This is the definition of a Kite.

Case 2: Joining Along one of the Equal Sides (8 cm side)

The resulting quadrilateral has opposite sides that are equal:

One pair of sides is 8 cm.

The other pair of sides is 6 cm.

A quadrilateral whose opposite sides are equal is a Parallelogram.

The specific internal angles of the $8, 8, 6$ triangle (which are not $60^\circ$) will determine the precise angles of the parallelogram, but the figure will always be a parallelogram.

Question 1

Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.

Solution :

Let the two equilateral triangles be $\triangle ABC$ and $\triangle ADC$. When joined along the common side $AC$, the resulting quadrilateral is $ABCD$.

Since both triangles are equilateral with a side length of 4 cm, all their sides are 4cm long.

A quadrilateral with all four sides equal is defined as a Rhombus.

Opposite Angles $\angle B$ and $\angle D$:

Since both triangles are equilateral, all their interior angles are $60^\circ$.

$\angle BAC = \angle BCA = 60^\circ$

$\angle DAC = \angle DCA = 60^\circ$

$$\angle B = \angle ABC = \mathbf{60^\circ}$$

$$\angle D = \angle ADC = \mathbf{60^\circ}$$

Opposite Angles $\angle A$ and $\angle C$:

$\angle A = \angle BAC + \angle DAC = 60^\circ + 60^\circ = \mathbf{120^\circ}$

$\angle C = \angle BCA + \angle DCA = 60^\circ + 60^\circ = \mathbf{120^\circ}$

Since the figure has four equal sides (4 cm) and opposite angles are equal ($60^\circ$ and $120^\circ$), it is confirmed to be a Rhombus.

Question 2

Construct a kite whose diagonals are of lengths 6 cm and 8 cm.

Solution :

(i) Steps for Construction (Intersection Angle $\mathbf{90^\circ}$)

• 1. Draw a line segment $AC$ of length 6 cm.
• 2. Locate the midpoint, $O$, of $AC$ by measuring 3 cm from $A$ (or $C$) and marking the point $O$.


• Mark a point that corresponds to a $90^\circ$ angle with $OA$.
• Draw the perpendicular bisector of AC passing through $O$ . This line will contain the second diagonal, $BD$.


• Set your compass to a radius of 4 cm (half the length of the diagonal).
• With the compass point at $O$, mark a point $B$ on the $90^\circ$ line on one side of $AC$.


• Extend the $90^\circ$ line through $O$. With the compass point still at $O$, mark a point $D$ on the opposite side of $AC$ such that $OD = 4$ cm.
• The length of $BD$ should now be $4 \text{ cm} + 4 \text{ cm} = 8 \text{ cm}$.








• join the consecutive points : $A$ to $B$ , $B$ to $C$, $C$ to $D$, $D$ to $A$

The resulting figure $ABCD$ is the required rhombus.

Question 3

Find the remaining angles in the following trapeziums.

Solution :

A trapezium is defined as a quadrilateral with at least one pair of parallel sides.

(i) Finding the Remaining Angles

Let's assume the given angles, $\angle P = 135^\circ$ and $\angle Q = 105^\circ$, belong to two different pairs of supplementary angles (i.e., they are on different non-parallel sides).

Let $PQ \parallel SR$ be the parallel sides of the trapezium $PQRS$.

Let's assume the given angles are:

$\angle P = 135^\circ$ (which means $\angle S$ is its supplement)

$$\angle S = 180^\circ - \angle P$$

$$\angle S = 180^\circ - 135^\circ$$

$$\mathbf{\angle S = 45^\circ}$$

$\angle Q = 105^\circ$ (which means $\angle R$ is its supplement)

$$\angle R = 180^\circ - \angle Q$$

$$\angle R = 180^\circ - 105^\circ$$

$$\mathbf{\angle R = 75^\circ}$$

The remaining angles are $\mathbf{45^\circ}$ and $\mathbf{75^\circ}$.

(ii) Finding the Remaining Angles Isosceles Trapezium

Let the Isosceles trapezium be $ABCD$ with $AB \parallel DC$.

Side $AD = BC$, then the two non-parallel sides ($AD$ and $BC$) act as transversals.

1. Find the angle on the same transversal ($\angle D$)

Since $AB \parallel DC$ and $AD$ is the transversal:

$$\angle A + \angle D = 180^\circ$$

When two parallel lines are cut by a transversal, the sum of the interior angles on the same side of the transversal is $180^\circ$

$$100^\circ + \angle D = 180^\circ$$

$$\mathbf{\angle D = 80^\circ}$$

2. Use symmetry of the isosceles trapezium

Since the trapezium is isosceles: We can use this property, the base angles are equal

$\angle D = \angle C$ and $\angle A = \angle B$.

(The shape is symmetric, so angles at the same base are equal.)

(Because the non-parallel sides are equal, the angles adjacent to each base are equal: )

Therefore

$\angle D = 80^\circ$ = $\angle C$

$\angle A = 100^\circ$ = $\angle B$

Question 4

Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles, and squares. Then, answer the following questions —

(i) What is the quadrilateral that is both a kite and a parallelogram?
(ii) Can there be a quadrilateral that is both a kite and a rectangle?
(iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals?

Solution :

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

Rhombus

• A kite has two pairs of adjacent equal sides.
• A parallelogram has opposite sides parallel.
• A rhombus satisfies both conditions.

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

No

• A rectangle has opposite sides equal and all angles 90°.
• A kite requires two pairs of adjacent equal sides.
• A rectangle does not generally have adjacent sides equal (unless it is a square).

Only a square has properties of both, but a square is not considered a kite in the strict school definition, so the answer is No.

(iii) Is every kite a rhombus ?

No, not every kite is a rhombus.

• A kite needs only two pairs of adjacent equal sides.
• A rhombus needs all four sides equal and opposite sides parallel.

However, a "general" kite (like the classic toy shape) has two short sides next to each other and two long sides next to each other. Because not all four sides are equal, it is not a rhombus.

Question 5

If PAIR and RODS are two rectangles, find ∠IOD.

Solution :

Given : the two rectangles $PAIR$ and $RODS$

$\angle IRO = 30^\circ$:

In Triangle $ROI$

$\angle RIO = 90^\circ$

(Because $I$ is a vertex of the rectangle $PAIR$, and all interior angles of a rectangle are $90^\circ$.)

$\angle IRO = 30^\circ$

$$\angle ROI + \angle RIO + \angle IRO = 180^\circ$$

( The sum of angles in any triangle is $180^\circ$ )

$$\angle ROI + 90^\circ + 30^\circ = 180^\circ$$

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

$$\angle ROI = \mathbf{60^\circ}$$

Now, Find $\angle IOD$

In Rectangle $RODS$

$\angle ROD = 90^\circ$

( Because $O$ is a vertex of the rectangle $RODS$ )

From the diagram, $\angle ROD$ is made up of two smaller angles: $\angle ROI$ and $\angle IOD$.

$$\angle ROI + \angle IOD = 90^\circ$$

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

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

$$\mathbf{\angle IOD = 30^\circ}$$

The value of $\angle IOD$ is $30^\circ$.

Question 6

Construct a square with a diagonal 6 cm without using a protractor.

Solution :

To construct a square with a diagonal of 6 cm, follow these steps using a ruler and a compass:

(i) Steps for Construction (Intersection Angle $\mathbf{90^\circ}$)

• 1. Draw a line segment $AC$ of length 6 cm.
• 2. Locate the midpoint, $O$, of $AC$ by measuring 3 cm from $A$ (or $C$) and marking the point $O$.
• 3. Open your compass to a radius slightly more than half of AC (more than 3 cm).
• 4. Place the compass pointer at point A and draw arcs above and below the line AC.
• 5. Without changing the radius, place the pointer at point C and draw arcs that intersect the first two arcs.
• 6. Use your ruler to draw a straight line through these two intersection points. Label the point where this vertical line crosses AC as O (the midpoint).


• 7. Using your ruler, set your compass width to exactly 3 cm (half of the diagonal).
• 8.Place the compass pointer at the midpoint O.







• 9. Draw an arc on the perpendicular bisector line above AC and label it point B.
• 10. Draw an arc on the same line below AC and label it point D.
• 11. Now that you have all four vertices (A, B, C, and D), simply use your ruler to connect them

The resulting figure $ABCD$ is the required Square.

Question 7

CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b)

Solution :

(a) Type of Quadrilateral $UVWX$

Let the side length of the large square $CASE$ be $2s$.

Since the points are midpoints, segments like $CU, CV, AX, AU$, etc., all have length $'s'$.

In $\triangle CUV$

By the Pythagorean theorem

$$UV^2 = CU^2 + CV^2 $$

$$UV^2 = s^2 + s^2 = 2s^2 $$

$$UV = s\sqrt{2}$$

Since all four corner triangles ($\triangle CUV, \triangle AUX, \triangle SXW, \triangle EWV$) are congruent (SAS property), all sides of $UVWX$ are equal to $s\sqrt{2}$.

In the isosceles right triangle $\triangle CUV$

$UC = VC$.

$\angle CUV = \angle CVU$.

( The angles opposite the equal sides are equal: )

$$\angle CUV + \angle CVU + \angle VCU = 180^\circ$$

( The sum of angles in any triangle is $180^\circ$ )

$$\angle CUV + \angle CVU + 90^\circ = 180^\circ$$

$$2\angle CUV + 90^\circ = 180^\circ$$

$$2\angle CUV = 180^\circ - 90^\circ$$

$$\angle CUV = 45^\circ $$

$\angle CUV = \angle CVU$ = 45°

Similarly, in $\triangle AUX$,

$\angle AUX = 45^\circ$.

Since $CUA$ is a straight line ($180^\circ$):

$$\angle VUX = 180^\circ - (\angle CUV + \angle AUX) $$

$$\angle VUX = 180^\circ - (45^\circ + 45^\circ) = 90^\circ$$

This applies to all four interior angles, proving $UVWX$ is a square.

Therefore $UVWX$ is a square, we must show that all sides are equal and all angles are 90°.

(b) To construct a square like Figure (b)

Figure (b) shows that the inner shape remains a square even if the points are not midpoints,

provided they are "equidistant from the vertices."

Steps to construct a square like Figure (b)

• 1. Pick a distance $d$ that is less than the side of the outer square.
• 2. On each side of the outer square, mark a point that is distance $d$ from the vertex, moving in a clockwise (or counter-clockwise) direction.
• 3. This creates four congruent right triangles at the corners. Because the triangles are congruent, the hypotenuses (the sides of the inner quadrilateral) are all equal.
• 4. Because the two acute angles of a right triangle add up to $90^\circ$, the angle remaining on the straight line of the outer square side will always be:
• $$180^\circ - 90^\circ = 90^\circ$$

Any time you rotate an inner square such that its vertices touch the sides of an outer square symmetrically, you maintain the properties of a square.

Question 8

If a quadrilateral has four equal sides and one angle of 90°, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement.

Solution :

(A) Geometric Reasoning

To prove this, we can look at the properties of quadrilaterals step-by-step:

A quadrilateral with all four sides equal is a rhombus.

In a rhombus:

Opposite angles are equal and adjacent angles are supplementary (sum to 180°)

If one angle = 90°, then:

Its adjacent angle = 180°−90° = 90°

Opposite angles are also equal

So all four angles become 90°.

Conclusion

If all four angles are $90^\circ$ and all four sides are equal, the shape fits the mathematical definition of a square.

2. Construction and Measurement

• 1. Draw a horizontal line segment $AB$ of $5\text{ cm}$.
• 2. At point $A$, use a protractor to mark a $90^\circ$ angle. Draw a vertical line segment $AD$ that is also exactly $5\text{ cm}$.
• 3. Set a compass to a width of $5\text{ cm}$ ,
• 4. Place the point on $B$ and draw an arc in the empty corner space, and Place the point on $D$ and draw another arc.
• 5. Mark the point where the two arcs cross as $C$. Connect $B$ to $C$ and $D$ to $C$.

Measurement Results:

Use a ruler to measure $BC$ and $DC$; they will both be $5\text{ cm}$.

Use a protractor to measure the angles at $B$, $C$, and $D$; they will all measure exactly $90^\circ$.

Question 9

What type of quadrilateral is one in which the opposite sides are equal? Justify your answer.
Hint: Draw a diagonal and check for congruent triangles.

Solution :

A quadrilateral with opposite sides equal is a parallelogram.

Let quadrilateral $PQRS$ has $PQ = RS$ and $QR = SP$, we can prove it is a parallelogram by showing the opposite sides are parallel.

(A) Geometric Reasoning

First we Draw quadrilateral $PQRS$ and draw the diagonal $PR$

let’s use a diagonal to split the shape into triangles

This gives us two triangles: $\triangle PQR$ and $\triangle RSP$.

In $\triangle PQR$ and $\triangle RSP$

$PQ = RS$ (Given)

$QR = SP$ (Given)

$PR = PR$ (Common side shared by both triangles)

By the Side-Side-Side (SSS) postulate,

$\triangle PQR \cong \triangle RSP$

Because the triangles are congruent, their corresponding parts are equal (CPCTC)

$\angle QPR$ = $\angle SRP$

These two angles are alternate interior angles for the lines $PQ$ and $SR$ with transversal $PR$.

Since the alternate interior angles are equal, the lines $PQ$ and $SR$ must be parallel.

Following the same logic for the other set of angles, $QR$ must be parallel to $SP$.

∴ PQ || RS and QR || SP.

Hence, $PQRS$ is a parallelogram.

Question 10

Will the sum of the angles in a quadrilateral such as the following one also be 360°? Find the answer using geometric reasoning as well as by constructing this figure and measuring.

Solution :

This specific shape is known as a concave (or re-entrant) quadrilateral.

(A) Geometric Reasoning

The most reliable way to prove the angle sum of any quadrilateral is to divide it into triangles.

Let’s draw a straight line segment connecting vertex $B$ to vertex $D$.

This gives us two triangles: $\triangle ABD$ and $\triangle CBD$.

We know that the sum of interior angles in any triangle is $180^\circ$

Sum of angles in $\triangle ABD = 180^\circ$

Sum of angles in $\triangle CBD = 180^\circ$

Total sum for the quadrilateral = $180^\circ + 180^\circ = 360^\circ$

$m\angle A + m\angle B + m\angle C + m\angle D = 360^\circ$.

This confirms that the $360^\circ$ rule applies to all quadrilaterals, regardless of their orientation .

Question 11 (i)

State whether the following statements are true or false. Justify your answers.

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

Solution :

The statement is false. A quadrilateral whose diagonals are equal and bisect each other need not be a square.

Geometric Reasoning

To identify a quadrilateral based on its diagonals, we look at three specific properties:

• 1. Bisecting each other: This property confirms the quadrilateral is a parallelogram.
• 2. Equal in length: When the diagonals of a parallelogram are equal, the figure is a rectangle.
• 3. Perpendicular bisectors: For the shape to be a square (or a rhombus), the diagonals must also cross at a $90^\circ$ angle.

Because the statement only specifies that the diagonals are equal and bisect each other, it describes all rectangles.

A square is a type of rectangle, but a standard rectangle (where adjacent sides are unequal) also fits this description.

Therefore, it is not necessarily a square.

Question 11 (ii)

State whether the following statements are true or false. Justify your answers.

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

Solution :

The statement is true. A quadrilateral having three right angles must be a rectangle.

(A) Geometric Reasoning

The most direct way to justify this is by looking at the Angle Sum Property of a quadrilateral.

• 1. Total Interior Angles: The sum of the interior angles of any quadrilateral is always $360^\circ$.
• 2. If three of the angles are each $90^\circ$, we can find the measure of the fourth angle ($x$) as follows:
• $$90^\circ + 90^\circ + 90^\circ + x = 360^\circ$$
• $$x = 360^\circ - 270^\circ = 90^\circ$$

By definition, a rectangle is a quadrilateral in which all four angles are right angles.

Since the fourth angle is mathematically forced to be $90^\circ$, the shape must be a rectangle.

A rectangle only requires all angles to be $90^\circ$ (which this shape has), but its adjacent sides can be different lengths.

Question 11 (iii)

State whether the following statements are true or false. Justify your answers.

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

Solution :

The statement is true. A quadrilateral whose diagonals bisect each other is always a parallelogram.

(A) Geometric Reasoning

let us consider a quadrilateral $PQRS$ where the diagonals $PR$ and $QS$ intersect at point $M$.

If the diagonals bisect each other, then $PM = MR$ and $QM = MS$.

In $\triangle PMQ$ and $\triangle RMS$.

• 1. $PM = RM$ (Given: diagonals bisect each other).
• 2. $QM = SM$ (Given: diagonals bisect each other).
• 3. $\angle PMQ = \angle RMS$ (Vertical angles are always equal).

By the Side-Angle-Side (SAS) property, $\triangle PMQ \cong \triangle RMS$

Because the triangles are congruent, their corresponding parts are equal (CPCTC).

Therefore, $\angle MPQ = \angle MRS$.

These are alternate interior angles for lines $PQ$ and $RS$ with $PR$ acting as a transversal.

If alternate interior angles are equal, then $PQ$ is parallel to $RS$.

By repeating this logic for the other pair of triangles ($\triangle PMS$ and $\triangle RMQ$), we find that $PS$ is parallel to $RQ$

Since both pairs of opposite sides are parallel, the quadrilateral is a parallelogram.

Question 11 (iv)

State whether the following statements are true or false. Justify your answers.

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

Solution :

The statement is false. A quadrilateral whose diagonals are perpendicular to each other is not necessarily a rhombus.

(A) Geometric Reasoning

For a quadrilateral to be a rhombus, the diagonals must satisfy two conditions:

they must be perpendicular and they must bisect each other.

If the diagonals are only perpendicular but do not bisect each other, the resulting shape could be a kite or even an irregular quadrilateral.

• 1. In a rhombus, all four sides must be equal.
• 2. In a kite, only two pairs of adjacent sides are equal.
• 3. In an irregular quadrilateral with perpendicular diagonals, no sides may be equal at all.

This quadrilateral whose diagonals are perpendicular to each other may not be a rhombus, because the diagonals AC and BD may not bisect each other.

∴ The given statement is false.

Question 11 (v)

State whether the following statements are true or false. Justify your answers.

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

Solution :

The statement is true.

A quadrilateral in which the opposite angles are equal is always a parallelogram.

(A) Geometric Reasoning

Let the angles of the quadrilateral be $A, B, C,$ and $D$.

According to the statement, $\angle A = \angle C$ and $\angle B = \angle D$.

We know that $\angle A + \angle B + \angle C + \angle D = 360^\circ$.

( The sum of angles in any quadrilateral is $360^\circ$ )

$$\angle A + \angle B + \angle A + \angle B = 360^\circ$$

$$2(\angle A + \angle B) = 360^\circ$$

$$\angle A + \angle B = 180^\circ$$

When the sum of adjacent (consecutive) interior angles is $180^\circ$, the lines are parallel.

Since $\angle A + \angle B = 180^\circ$, side $AD$ is parallel to side $BC$.

Similarly, since $\angle A + \angle D = 180^\circ$, side $AB$ is parallel to side $CD$.

Conclusion: A quadrilateral with two pairs of parallel opposite sides is, by definition, a parallelogram.

Question 11 (vi)

State whether the following statements are true or false. Justify your answers.

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

Solution :

The statement is true.

If all four angles are equal, each angle must be 360°/4 = 90°. A quadrilateral with four right angles is a rectangle.

Question 11 (vii)

State whether the following statements are true or false. Justify your answers.

(vii) Isosceles trapeziums are parallelograms.

Solution :

The statement is false. An isosceles trapezium is not a parallelogram.

Since isosceles trapeziums have one pair of parallel sides and equal non-parallel sides, whereas parallelograms have two pairs of parallel sides. So, an isosceles trapezium is not a parallelogram.

To understand why, we must compare the both quadrilaterals:

Because an isosceles trapezium specifically has only one pair of parallel sides, it lacks the second pair of parallel sides required to be a parallelogram.

Furthermore, in a parallelogram, opposite angles must be equal;

in an isosceles trapezium, base angles are equal (adjacent to the same parallel side), but opposite angles are usually supplementary (adding to 180°).