CBSE Class 9 Maths Chapter 8 Quadrilaterals NCERT Solutions

Solution:

Let the measures of the four angles of the quadrilateral be $3x$, $5x$, $9x$, and $13x$, according to the given ratio.

The sum of the interior angles of any quadrilateral is $360^{\circ}$. Therefore, we can write the equation:

$$3x\; +\; 5x\; +\; 9x\; +\; 13x\; =\; 360^\{\backslash circ\}$$

Combining the terms on the left side:

$$30x\; =\; 360^\{\backslash circ\}$$

To find the value of $x$, divide both sides by 30:

$$x\; =\; \backslash frac\{360^\{\backslash circ\}\}\{30\}$$

$$x\; =\; 12^\{\backslash circ\}$$

Now, we can find the measure of each angle by substituting $x = 12^{\circ}$:

• First angle: $3x = 3 \times 12^{\circ} = 36^{\circ}
• Second angle: $5x = 5 \times 12^{\circ} = 60^{\circ}
• Third angle: $9x = 9 \times 12^{\circ} = 108^{\circ}
• Fourth angle: $13x = 13 \times 12^{\circ} = 156^{\circ}

Thus, the angles of the quadrilateral are $36^{\circ}$, $60^{\circ}$, $108^{\circ}$, and $156^{\circ}$.

Solution:

Given: ABCD is a parallelogram with diagonals AC and BD such that $AC = BD$.

To Prove: ABCD is a rectangle.

Proof: Consider triangles $\triangle ABC$ and $\triangle BAD$.

1. $AB = AB$ (Common side)
2. $BC = AD$ (Opposite sides of a parallelogram are equal)
3. $AC = BD$ (Given)

By the SSS (Side-Side-Side) congruence criterion, $\triangle ABC \cong \triangle BAD$.

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

$$\backslash angle\; ABC\; =\; \backslash angle\; BAD$$

This is equation (i).

Now, since ABCD is a parallelogram, its consecutive interior angles are supplementary. Thus,

$$\backslash angle\; ABC\; +\; \backslash angle\; BAD\; =\; 180^\{\backslash circ\}$$

This is equation (ii).

Substitute $\angle BAD$ from equation (i) into equation (ii):

$$\backslash angle\; ABC\; +\; \backslash angle\; ABC\; =\; 180^\{\backslash circ\}$$

$$2\backslash angle\; ABC\; =\; 180^\{\backslash circ\}$$

$$\backslash angle\; ABC\; =\; \backslash frac\{180^\{\backslash circ\}\}\{2\}\; =\; 90^\{\backslash circ\}$$

Since $\angle ABC = 90^{\circ}$ and $\angle ABC = \angle BAD$, we have $\angle BAD = 90^{\circ}$ as well.

A parallelogram with one angle equal to $90^{\circ}$ is a rectangle.

Hence, ABCD is a rectangle. Proved.

Solution:

Given: A quadrilateral ABCD, in which diagonals AC and BD bisect each other at right angles. Let the point of intersection be O. This means $AO = OC$, $BO = OD$, and $\angle AOB = \angle BOC = \angle COD = \angle DOA = 90^{\circ}$.

To Prove: ABCD is a rhombus.

Proof: Consider triangles $\triangle AOB$ and $\triangle BOC$.

1. $AO = OC$ (Diagonals bisect each other)
2. $\angle AOB = \angle COB$ (Each is $90^{\circ}$ as diagonals intersect at right angles)
3. $BO = BO$ (Common side)

By the SAS (Side-Angle-Side) congruence criterion, $\triangle AOB \cong \triangle BOC$.

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

$$AB\; =\; BC$$

This is equation (i).

Similarly, consider triangles $\triangle BOC$ and $\triangle DOC$:

1. $OC = OC$ (Common side)
2. $\angle BOC = \angle DOC = 90^{\circ}$
3. $BO = DO$ (Diagonals bisect each other)

By SAS congruence, $\triangle BOC \cong \triangle DOC$.

Therefore, $BC = CD$ (CPCT).

Now consider triangles $\triangle DOC$ and $\triangle AOD$:

1. $DO = DO$ (Common side)
2. $\angle DOC = \angle AOD = 90^{\circ}$
3. $OC = AO$ (Diagonals bisect each other)

By SAS congruence, $\triangle DOC \cong \triangle AOD$.

Therefore, $CD = DA$ (CPCT).

Combining the results, we have $AB = BC = CD = DA$.

A quadrilateral with all four sides equal is a rhombus.

Hence, ABCD is a rhombus. Proved.

Solution:

Given: ABCD is a square. AC and BD are its diagonals.

To Prove: $AC = BD$, AC and BD bisect each other, and $AC \perp BD$.

Proof:

Part 1: Proving diagonals are equal ($AC = BD$)

Consider triangles $\triangle ABC$ and $\triangle BAD$.

1. $AB = AB$ (Common side)
2. $BC = AD$ (Sides of a square are equal)
3. $\angle ABC = \angle BAD = 90^{\circ}$ (Angles of a square)

By the SAS (Side-Angle-Side) congruence criterion, $\triangle ABC \cong \triangle BAD$.

Therefore, $AC = BD$ (CPCT). The diagonals are equal.

Part 2: Proving diagonals bisect each other ($AO = OC$ and $BO = OD$)

Consider triangles $\triangle AOB$ and $\triangle COD$.

1. $AB = DC$ (Sides of a square are equal)
2. $\angle AOB = \angle COD$ (Vertically opposite angles)
3. $\angle OAB = \angle OCD$ (Alternate interior angles, since $AB \parallel DC$ and AC is a transversal)

By the AAS (Angle-Angle-Side) congruence criterion, $\triangle AOB \cong \triangle COD$.

Therefore, $AO = OC$ (CPCT). The diagonals bisect each other.

Similarly, by considering $\triangle AOD$ and $\triangle BOC$, we can prove that $DO = OB$.

Part 3: Proving diagonals bisect each other at right angles ($AC \perp BD$)

Consider triangle $\triangle AOB$. We know that $AO = OC$ and $BO = OD$. Since $AC = BD$, we have $AO = OC = BO = OD$.

In $\triangle ABC$, $\angle ABC = 90^{\circ}$. Since ABCD is a square, $AB = BC$. Thus, $\triangle ABC$ is an isosceles right-angled triangle.

The angles opposite to equal sides are equal: $\angle BAC = \angle BCA$.

The sum of angles in $\triangle ABC$ is $180^{\circ}$:

$$\backslash angle\; BAC\; +\; \backslash angle\; BCA\; +\; \backslash angle\; ABC\; =\; 180^\{\backslash circ\}$$

$$\backslash angle\; BAC\; +\; \backslash angle\; BAC\; +\; 90^\{\backslash circ\}\; =\; 180^\{\backslash circ\}$$

$$2\backslash angle\; BAC\; =\; 90^\{\backslash circ\}$$

$$\backslash angle\; BAC\; =\; 45^\{\backslash circ\}$$

So, $\angle BCA = 45^{\circ}$.

Now consider $\triangle BOC$. We know $BO = OC$ (since diagonals are equal and bisect each other). Thus, $\triangle BOC$ is an isosceles triangle.

The angle $\angle BOC$ is adjacent to $\angle AOB$. We need to show $\angle BOC = 90^{\circ}$.

In $\triangle BOC$, we have $\angle OBC$ and $\angle OCB$. We know $\angle OCB = \angle BCA = 45^{\circ}$.

The sum of angles in $\triangle BOC$ is $180^{\circ}$:

$$\backslash angle\; OBC\; +\; \backslash angle\; OCB\; +\; \backslash angle\; BOC\; =\; 180^\{\backslash circ\}$$

Since $BO = OC$, $\angle OBC = \angle OCB = 45^{\circ}$.

$$45^\{\backslash circ\}\; +\; 45^\{\backslash circ\}\; +\; \backslash angle\; BOC\; =\; 180^\{\backslash circ\}$$

$$90^\{\backslash circ\}\; +\; \backslash angle\; BOC\; =\; 180^\{\backslash circ\}$$

$$\backslash angle\; BOC\; =\; 90^\{\backslash circ\}$$

Since $\angle BOC = 90^{\circ}$, the diagonals AC and BD intersect at right angles.

Therefore, the diagonals of a square are equal, bisect each other, and intersect at right angles. Proved.

Solution:

Given: A quadrilateral ABCD, in which diagonals AC and BD are equal ($AC = BD$) and bisect each other at right angles. Let the point of intersection be O. This means $AO = OC$, $BO = OD$, and $\angle AOB = 90^{\circ}$.

To Prove: ABCD is a square.

Proof:

Step 1: Show that ABCD is a rhombus.

Consider triangles $\triangle AOB$ and $\triangle BOC$.

1. $AO = OC$ (Diagonals bisect each other)
2. $\angle AOB = \angle COB = 90^{\circ}$ (Diagonals intersect at right angles)
3. $BO = BO$ (Common side)

By SAS congruence, $\triangle AOB \cong \triangle BOC$.

Therefore, $AB = BC$ (CPCT).

Similarly, by considering other pairs of triangles (e.g., $\triangle BOC$ and $\triangle DOC$), we can show that $BC = CD$ and $CD = DA$.

Thus, $AB = BC = CD = DA$. A quadrilateral with all sides equal is a rhombus.

Step 2: Show that the rhombus is a rectangle (i.e., one angle is $90^{\circ}$).

Since the diagonals bisect each other, ABCD is a parallelogram.

We are given that the diagonals are equal: $AC = BD$.

In a parallelogram, if the diagonals are equal, then it is a rectangle (as proved in Question 2).

Therefore, ABCD is a rectangle.

Step 3: Conclude that ABCD is a square.

We have shown that ABCD is both a rhombus (all sides equal) and a rectangle (all angles $90^{\circ}$).

A quadrilateral that is both a rhombus and a rectangle is a square.

Hence, ABCD is a square. Proved.