CBSE Class 9 Mathematics Chapter 8: Quadrilaterals NCERT Solutions

Solution:

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

The sum of the interior angles of any quadrilateral is always $360^\{\backslash circ\}$. This is known as the angle sum property of a quadrilateral.

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 each angle by substituting the value of $x$:

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

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

Answer: The angles of the quadrilateral are $36^\{\backslash circ\}$, $60^\{\backslash circ\}$, $108^\{\backslash circ\}$, and $156^\{\backslash circ\}$.

Solution:

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

To Prove: ABCD is a rectangle.

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

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, $\backslash triangle\; ABC\; \backslash cong\; \backslash triangle\; BAD$.

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

Therefore, $\backslash angle\; ABC\; =\; \backslash angle\; BAD$ (Equation i).

Now, since ABCD is a parallelogram, its consecutive interior angles are supplementary (sum up to $180^\{\backslash circ\}$).

So, $\backslash angle\; ABC\; +\; \backslash angle\; BAD\; =\; 180^\{\backslash circ\}$ (Equation ii).

Substitute $\backslash angle\; BAD$ with $\backslash angle\; ABC$ from Equation (i) into Equation (ii):

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

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

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

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

Since $\backslash angle\; ABC\; =\; \backslash angle\; BAD$, we also have $\backslash angle\; BAD\; =\; 90^\{\backslash circ\}$.

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

Hence, ABCD is a rectangle. Proved.

Solution:

Given: A quadrilateral ABCD, where the diagonals AC and BD bisect each other at point O, and $\backslash angle\; AOB\; =\; \backslash angle\; BOC\; =\; \backslash angle\; COD\; =\; \backslash angle\; DOA\; =\; 90^\{\backslash circ\}$.

To Prove: ABCD is a rhombus.

Proof: Since the diagonals AC and BD bisect each other, O is the midpoint of both AC and BD. This means $AO\; =\; OC$ and $BO\; =\; OD$. A quadrilateral whose diagonals bisect each other is a parallelogram.

Now, let's consider triangles $\backslash triangle\; AOB$ and $\backslash triangle\; BOC$.

1. $AO\; =\; OC$ (Diagonals bisect each other)
2. $\backslash angle\; AOB\; =\; \backslash angle\; COB$ (Given that they intersect at right angles, so each is $90^\{\backslash circ\}$)
3. $BO\; =\; BO$ (Common side)

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

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

Therefore, $AB\; =\; BC$ (Equation i).

Similarly, we can prove that $\backslash triangle\; BOC\; \backslash cong\; \backslash triangle\; COD$ (using $BO\; =\; OD$, $\backslash angle\; BOC\; =\; \backslash angle\; COD\; =\; 90^\{\backslash circ\}$, $OC\; =\; OC$) which gives $BC\; =\; CD$ (Equation ii).

And $\backslash triangle\; COD\; \backslash cong\; \backslash triangle\; DOA$ (using $OC\; =\; OA$, $\backslash angle\; COD\; =\; \backslash angle\; DOA\; =\; 90^\{\backslash circ\}$, $OD\; =\; OD$) which gives $CD\; =\; DA$ (Equation iii).

From equations (i), (ii), and (iii), 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 (i.e., $AO\; =\; OC$ and $BO\; =\; OD$), and $AC\; \backslash perp\; BD$ (i.e., angles at intersection are $90^\{\backslash circ\}$).

Proof:

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

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

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

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

Therefore, by CPCT (Corresponding Parts of Congruent Triangles), $AC\; =\; BD$.

Part 2: Proving diagonals bisect each other

Since ABCD is a square, it is also a parallelogram. In a parallelogram, diagonals bisect each other.

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

1. $AB\; =\; DC$ (Sides of a square)
2. $\backslash angle\; OAB\; =\; \backslash angle\; OCD$ (Alternate interior angles, since AB || DC)
3. $\backslash angle\; OBA\; =\; \backslash angle\; ODC$ (Alternate interior angles, since AB || DC)

By the ASA (Angle-Side-Angle) congruence criterion (or AAS using $\backslash angle\; AOB\; =\; \backslash angle\; COD$ as vertically opposite angles), $\backslash triangle\; AOB\; \backslash cong\; \backslash triangle\; COD$.

Therefore, by CPCT, $AO\; =\; OC$.

Similarly, consider triangles $\backslash triangle\; AOD$ and $\backslash triangle\; BOC$.

1. $AD\; =\; BC$ (Sides of a square)
2. $\backslash angle\; OAD\; =\; \backslash angle\; OCB$ (Alternate interior angles, since AD || BC)
3. $\backslash angle\; ODA\; =\; \backslash angle\; OBC$ (Alternate interior angles, since AD || BC)

By the ASA congruence criterion, $\backslash triangle\; AOD\; \backslash cong\; \backslash triangle\; BOC$.

Therefore, by CPCT, $OD\; =\; OB$.

Thus, the diagonals bisect each other.

Part 3: Proving diagonals intersect at right angles

Consider triangle $\backslash triangle\; ABC$. Since ABCD is a square, $\backslash angle\; ABC\; =\; 90^\{\backslash circ\}$.

Also, in a square, adjacent sides are equal ($AB\; =\; BC$). This makes $\backslash triangle\; ABC$ an isosceles right-angled triangle.

Therefore, the base angles are equal: $\backslash angle\; BAC\; =\; \backslash angle\; BCA$.

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

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

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

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

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

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

Now consider the intersection point O. We need to show that $\backslash angle\; AOB\; =\; 90^\{\backslash circ\}$.

In $\backslash triangle\; AOB$, we know $\backslash angle\; OAB\; =\; \backslash angle\; BAC\; =\; 45^\{\backslash circ\}$.

Since ABCD is a square, $AB\; =\; BC$. Also, we proved $AO\; =\; OC$ and $BO\; =\; OD$.

Consider $\backslash triangle\; BOC$. We know $\backslash angle\; OBC\; =\; \backslash angle\; ABC\; -\; \backslash angle\; OBA$. Since $\backslash triangle\; AOB\; \backslash cong\; \backslash triangle\; COD$, $\backslash angle\; OBA\; =\; \backslash angle\; ODC$. Since $\backslash triangle\; AOD\; \backslash cong\; \backslash triangle\; BOC$, $\backslash angle\; ODA\; =\; \backslash angle\; OBC$.

Let's use the property that $\backslash triangle\; ABC$ is isosceles with $AB=BC$. Also $\backslash triangle\; BCD$ is isosceles with $BC=CD$.

In $\backslash triangle\; BOC$, $BC$ is the hypotenuse if $\backslash angle\; BOC\; =\; 90^\{\backslash circ\}$. We know $\backslash angle\; BCA\; =\; 45^\{\backslash circ\}$.

Since $\backslash triangle\; AOB\; \backslash cong\; \backslash triangle\; BOC$ (as shown in Q3, if diagonals bisect at right angles), we have $AB\; =\; BC$. This is true for a square.

Let's reconsider $\backslash triangle\; BOC$. We know $\backslash angle\; OBC\; =\; \backslash angle\; OCB\; =\; 45^\{\backslash circ\}$ (since $\backslash triangle\; ABC$ is isosceles right triangle and $\backslash angle\; BAC\; =\; \backslash angle\; BCA\; =\; 45^\{\backslash circ\}$, and $\backslash angle\; OCB\; =\; \backslash angle\; BCA$).

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

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

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

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

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

Since $\backslash angle\; BOC\; =\; 90^\{\backslash circ\}$, the diagonals intersect at right angles.

Thus, we have proved that the diagonals of a square are equal ($AC\; =\; BD$), they bisect each other ($AO\; =\; OC,\; OB\; =\; OD$), and they intersect at right angles ($AC\; \backslash perp\; BD$). Proved.

Solution:

Given: A quadrilateral ABCD, where diagonals AC and BD are equal ($AC\; =\; BD$), bisect each other (i.e., $AO\; =\; OC$ and $BO\; =\; OD$), and intersect at right angles (i.e., $\backslash angle\; AOB\; =\; 90^\{\backslash circ\}$).

To Prove: ABCD is a square.

Proof:

Step 1: Prove ABCD is a parallelogram.

Since the diagonals AC and BD bisect each other, the quadrilateral ABCD is a parallelogram.

Step 2: Prove all sides are equal (ABCD is a rhombus).

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

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

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

Therefore, by CPCT, $AB\; =\; BC$.

Since ABCD is a parallelogram and one pair of adjacent sides is equal ($AB\; =\; BC$), all sides must be equal ($AB\; =\; BC\; =\; CD\; =\; DA$). Thus, ABCD is a rhombus.

Step 3: Prove one angle is 90° (ABCD is a rectangle).

We are given that the diagonals are equal, $AC\; =\; BD$.

We have already established that ABCD is a parallelogram (from Step 1).

A parallelogram with equal diagonals is a rectangle.

Alternatively, consider triangles $\backslash triangle\; ABC$ and $\backslash triangle\; BAD$.

1. $AB\; =\; BA$ (Common side)
2. $BC\; =\; AD$ (Opposite sides of parallelogram ABCD)
3. $AC\; =\; BD$ (Given)

By SSS congruence, $\backslash triangle\; ABC\; \backslash cong\; \backslash triangle\; BAD$.

Therefore, by CPCT, $\backslash angle\; ABC\; =\; \backslash angle\; BAD$.

Since ABCD is a parallelogram, consecutive angles are supplementary:

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

Substituting $\backslash angle\; BAD$ with $\backslash angle\; ABC$:

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

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

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

So, ABCD is a parallelogram with one angle equal to $90^\{\backslash circ\}$, which means it is a rectangle.

Step 4: Conclude ABCD is a square.

From Step 2, we proved that ABCD is a rhombus (all sides are equal).

From Step 3, we proved that ABCD is a rectangle (one angle is $90^\{\backslash circ\}$).

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

Hence, ABCD is a square. Proved.