Quadrilaterals Class 8 Mathematics NCERT Solutions & Question Answers

Q1Figure it Out (Section 4.1)

Find all the other angles inside the following rectangles.

Solution

Given: Two rectangles with diagonals drawn.

(i)

A rectangle where one angle between a diagonal and a side is

30^\circ

.

(ii)

A rectangle where one angle between the two diagonals is

110^\circ

.

To Find: All the other angles inside the rectangles.

Solution:

For rectangle (i): Let the rectangle be ABCD and the angle given be

\angle BAC = 30^\circ

. Let the diagonals AC and BD intersect at O.

In a rectangle, all angles are $90^\circ$ . So, $\angle ABC = \angle BCD = \angle CDA = \angle DAB = 90^\circ$ .
In $\triangle ABC$ , $\angle BCA = 90^\circ - \angle BAC = 90^\circ - 30^\circ = 60^\circ$ .
Diagonals of a rectangle are equal and bisect each other. So, $OA = OB = OC = OD$ .
In $\triangle AOB$ , since $OA = OB$ , it is an isosceles triangle. Therefore, $\angle OBA = \angle OAB = 30^\circ$ .
The angle between the diagonals at O is $\angle AOB = 180^\circ - (\angle OAB + \angle OBA) = 180^\circ - (30^\circ + 30^\circ) = 120^\circ$ .
The other angle between diagonals is $\angle BOC = 180^\circ - \angle AOB = 180^\circ - 120^\circ = 60^\circ$ (linear pair).
Also, $\angle COD = \angle AOB = 120^\circ$ and $\angle DOA = \angle BOC = 60^\circ$ (vertically opposite angles).
Similarly, we can find all other angles:
- In $\triangle BOC$ , $OB=OC$ , so $\angle OBC = \angle OCB = (180^\circ - 60^\circ)/2 = 60^\circ$ . So $\triangle BOC$ is equilateral.
- In $\triangle COD$ , $OC=OD$ , so $\angle OCD = \angle ODC = (180^\circ - 120^\circ)/2 = 30^\circ$ .
- In $\triangle DOA$ , $OD=OA$ , so $\angle ODA = \angle OAD = (180^\circ - 60^\circ)/2 = 60^\circ$ .

Summary for (i):

Angles at vertices, split by diagonals: $(30^\circ, 60^\circ), (30^\circ, 60^\circ), (30^\circ, 60^\circ), (30^\circ, 60^\circ)$ .
Angles at the intersection of diagonals: $120^\circ, 60^\circ, 120^\circ, 60^\circ$ .

For rectangle (ii): Let the rectangle be PQRS and diagonals PR and QS intersect at O. Let

\angle POQ = 110^\circ

.

Diagonals are equal and bisect each other, so $OP = OQ = OR = OS$ .
The other angle between diagonals is $\angle QOR = 180^\circ - 110^\circ = 70^\circ$ (linear pair).
Also, $\angle ROS = \angle POQ = 110^\circ$ and $\angle SOP = \angle QOR = 70^\circ$ (vertically opposite angles).
In $\triangle POQ$ , $OP=OQ$ , so $\angle OPQ = \angle OQP = (180^\circ - 110^\circ)/2 = 35^\circ$ .
In $\triangle QOR$ , $OQ=OR$ , so $\angle OQR = \angle ORQ = (180^\circ - 70^\circ)/2 = 55^\circ$ .
The angles of the rectangle are $90^\circ$ . For example, $\angle PQR = \angle OQP + \angle OQR = 35^\circ + 55^\circ = 90^\circ$ .
Similarly, we can find all other angles:
- In $\triangle ROS$ , $OR=OS$ , so $\angle ORS = \angle OSR = (180^\circ - 110^\circ)/2 = 35^\circ$ .
- In $\triangle SOP$ , $OS=OP$ , so $\angle OSP = \angle OPS = (180^\circ - 70^\circ)/2 = 55^\circ$ .

Summary for (ii):

Angles at vertices, split by diagonals: $(35^\circ, 55^\circ), (35^\circ, 55^\circ), (35^\circ, 55^\circ), (35^\circ, 55^\circ)$ .
Angles at the intersection of diagonals: $110^\circ, 70^\circ, 110^\circ, 70^\circ$ .

Final Answer: The angles are calculated as shown above for both cases.

Q2Figure it Out (Section 4.1)

Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of

(i)

$30^{\circ}$

(ii)

$40^{\circ}$

(iii)

$90^{\circ}$

(iv)

$140^{\circ}$

Solution

To Construct: A quadrilateral with diagonals of 8 cm that are equal, bisect each other, and intersect at a given angle.

Property: A quadrilateral whose diagonals are equal and bisect each other is always a rectangle.

Steps of Construction (General):

Draw a line segment AC of length 8 cm.
Find the midpoint of AC by constructing its perpendicular bisector. Let the midpoint be O.
At point O, use a protractor to draw a line XY making the required angle with AC (e.g., $30^\circ$ for part i).
Since the other diagonal BD is also 8 cm and is bisected at O, we have $OB = OD = 4$ cm.
With O as the center, draw arcs of radius 4 cm on the line XY on both sides of O. Mark these points as B and D.
Join the points A, B, C, and D to form the quadrilateral ABCD.

Resulting Quadrilaterals: In all four cases, the resulting quadrilateral ABCD will be a rectangle, as its diagonals are equal (8 cm) and bisect each other.

(i)

Angle = $30^\circ$ : A rectangle where the acute angle between the diagonals is

30^\circ

.

(ii)

Angle = $40^\circ$ : A rectangle where the acute angle between the diagonals is

40^\circ

.

(iii)

Angle = $90^\circ$ : A rectangle where the diagonals intersect at

90^\circ

. A rectangle with perpendicular diagonals is a square. So, this construction will result in a square with diagonals of 8 cm.

(iv)

Angle = $140^\circ$ : A rectangle where the obtuse angle between the diagonals is

140^\circ

. The acute angle will be

180^\circ - 140^\circ = 40^\circ

. This is the same rectangle as in part (ii), just oriented differently.

Final Answer: The construction steps provided will produce a rectangle for all cases, and specifically a square for case (iii).

Q3Figure it Out (Section 4.1)

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

Given: A circle with center O. PL and AM are two diameters. The diameters are perpendicular to each other, so

\angle POL = \angle LOM = \angle MOA = \angle AOP = 90^\circ

.

To Find: The type of quadrilateral APML.

Reasoning: The figure APML is a quadrilateral formed by joining the endpoints of the two diameters.

Properties of Diagonals: The diagonals of the quadrilateral APML are the line segments PL and AM.
Length of Diagonals: Since PL and AM are both diameters of the same circle, their lengths are equal. $PL = AM = \text{Diameter of the circle}$
Intersection of Diagonals: The diameters of a circle always pass through the center O. Therefore, the diagonals PL and AM intersect at the center O.
Bisection of Diagonals: The center O is the midpoint of any diameter. Thus, O is the midpoint of both PL ( $PO = OL = \text{radius}$ ) and AM ( $AO = OM = \text{radius}$ ). This means the diagonals bisect each other.
Angle between Diagonals: It is given that the diameters are perpendicular. Therefore, the angle between the diagonals is $90^\circ$ .

Conclusion: Let's analyze the properties of the quadrilateral APML based on its diagonals:

Since the diagonals are equal and bisect each other, the quadrilateral is a rectangle.
Since the diagonals are perpendicular and bisect each other, the quadrilateral is a rhombus. A quadrilateral that is both a rectangle and a rhombus must be a square.

Final Answer: The figure APML is a square.

Q4Figure it Out (Section 4.1)

We have seen how to get $90^{\circ}$ 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 $90^{\circ}$ using these?

Solution

Objective: To create a

90^\circ

angle using two equal length sticks and a thread.

Concept: The diagonals of a square are equal in length and they bisect each other at a right angle (

90^\circ

). We can use this property.

Method:

Represent Diagonals: Use the two sticks of equal length as the diagonals of a quadrilateral.
Bisect Each Other: Find the midpoint of each stick. You can do this by using the thread. Take the thread, mark the length of a stick on it. Then fold the marked length of the thread in half to find the midpoint.
Position the Sticks: Place one stick on the ground. Place the second stick over the first one such that their midpoints coincide. Now the sticks bisect each other.
Form a Square: The endpoints of the sticks will form the vertices of a quadrilateral. Use the thread to connect the four endpoints of the sticks. The quadrilateral formed will be a rectangle because its diagonals are equal and bisect each other. To make the angle between the sticks exactly $90^\circ$ , we need this rectangle to be a square. A square is a rectangle with all four sides equal.
Ensure Right Angle: Adjust the angle between the two sticks until all four sides of the quadrilateral formed by the thread are equal in length. You can check this by measuring one side with the thread and then comparing that length to the other three sides.

Result: When the four sides formed by the thread are equal, the quadrilateral is a rhombus. Since we already established it's a rectangle (equal diagonals that bisect), a figure that is both a rectangle and a rhombus is a square. The diagonals of a square are perpendicular.

Final Answer: By arranging the two equal sticks to bisect each other and adjusting them until the quadrilateral formed by connecting their endpoints with a thread has four equal sides, the angle at the intersection of the sticks will be exactly

90^\circ

.

Q5Figure it Out (Section 4.1)

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

Question: Is every quadrilateral that has opposite sides parallel and equal, a rectangle?

Answer: No.

Reasoning:

A quadrilateral in which opposite sides are parallel is the definition of a parallelogram.
In a parallelogram, it is a property that opposite sides are also equal in length.
Therefore, the condition given ("a quadrilateral that has opposite sides parallel and equal") describes a parallelogram.
A rectangle is a special type of parallelogram where all interior angles are $90^\circ$ .
However, not all parallelograms are rectangles. For example, a parallelogram can have angles of $60^\circ$ and $120^\circ$ . Such a figure satisfies the condition of having opposite sides parallel and equal, but it is not a rectangle.

Conclusion: The definition "a quadrilateral that has opposite sides parallel and equal" is a definition for a parallelogram, not specifically for a rectangle. A rectangle requires the additional condition that its angles must be right angles.

Final Answer: No, because this is the definition of a parallelogram. A parallelogram is only a rectangle if its angles are all

90^\circ

.

Q1Figure it Out (Section 4.4)

Find the remaining angles in the following quadrilaterals.

Solution

Given: Two quadrilaterals with some angles given.

(i)

A parallelogram with one angle of

40^\circ

.

(ii)

A rhombus with one angle of

120^\circ

.

To Find: The remaining angles in each quadrilateral.

Solution:

For quadrilateral (i) - Parallelogram: Let the parallelogram be ABCD with

\angle A = 40^\circ

.

Opposite angles are equal: In a parallelogram, opposite angles are equal. So, $\angle C = \angle A = 40^\circ$ .
Adjacent angles are supplementary: Adjacent angles in a parallelogram add up to $180^\circ$ $18 0^{\circ}$ .
- $\angle A + \angle B = 180^\circ \implies 40^\circ + \angle B = 180^\circ \implies \angle B = 140^\circ$ .
- $\angle A + \angle D = 180^\circ \implies 40^\circ + \angle D = 180^\circ \implies \angle D = 140^\circ$ .
Check: Opposite angle $\angle D$ is indeed equal to $\angle B$ .

Remaining angles for the parallelogram are: $140^\circ, 40^\circ, 140^\circ$ .

For quadrilateral (ii) - Rhombus: A rhombus is a type of parallelogram, so it has the same angle properties. Let the rhombus be PQRS with

\angle P = 120^\circ

.

Opposite angles are equal: $\angle R = \angle P = 120^\circ$ .
Adjacent angles are supplementary:
- $\angle P + \angle Q = 180^\circ \implies 120^\circ + \angle Q = 180^\circ \implies \angle Q = 60^\circ$ .
- $\angle P + \angle S = 180^\circ \implies 120^\circ + \angle S = 180^\circ \implies \angle S = 60^\circ$ .
Check: Opposite angle $\angle S$ is indeed equal to $\angle Q$ .

Remaining angles for the rhombus are: $60^\circ, 120^\circ, 60^\circ$ .

Final Answer:

(i)

The angles of the parallelogram are

40^\circ, 140^\circ, 40^\circ, 140^\circ

.

(ii)

The angles of the rhombus are

120^\circ, 60^\circ, 120^\circ, 60^\circ

.

Q2Figure it Out (Section 4.4)

Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of $140^{\circ}$ .

Solution

To Construct: A parallelogram with diagonals 7 cm and 5 cm, intersecting at

140^\circ

.

Property: The diagonals of a parallelogram bisect each other.

Steps of Construction:

Draw the first diagonal, let's say AC, of length 7 cm.
Find the midpoint of AC. Let this point be O. This can be done by drawing the perpendicular bisector or by measuring 3.5 cm from A (or C).
At the midpoint O, use a protractor to draw a line XY that makes an angle of $140^\circ$ with the line segment AC.
The second diagonal is BD, with length 5 cm. Since it is bisected at O, the lengths $OB$ and $OD$ must both be half of 5 cm, which is 2.5 cm.
With O as the center and a radius of 2.5 cm, draw two arcs cutting the line XY on either side of O. Mark these intersection points as B and D.
Join the points A, B, C, and D in order to form the quadrilateral.

Result: ABCD is the required parallelogram. Its diagonals AC = 7 cm and BD = 5 cm, they bisect each other at O, and the angle of intersection

\angle AOD

(or

\angle BOC

) is

140^\circ

. The other angle of intersection,

\angle AOB

(or

\angle DOC

), will be

180^\circ - 140^\circ = 40^\circ

.

Q3Figure it Out (Section 4.4)

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

Solution

To Construct: A rhombus with diagonals 4 cm and 5 cm.

Properties:

The diagonals of a rhombus are perpendicular to each other.
The diagonals of a rhombus bisect each other.

Steps of Construction:

Draw the first diagonal, let's say AC, of length 5 cm.
Construct the perpendicular bisector of AC. Let the midpoint be O.
The second diagonal, BD, has a length of 4 cm and is bisected by O. So, OB = OD = 2 cm.
With O as the center, draw arcs of radius 2 cm on the perpendicular bisector on both sides of AC. Mark these points as B and D.
Join the points A, B, C, and D.

Result: ABCD is the required rhombus because its diagonals are perpendicular bisectors of each other.

QuadrilateralsClass 8 Mathematics NCERT Solutions

Find all the other angles inside the following rectangles.

Solution

Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of

30∘30^{\circ}30∘

40∘40^{\circ}40∘

90∘90^{\circ}90∘

140∘140^{\circ}140∘

Solution

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

We have seen how to get 90∘90^{\circ}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∘90^{\circ}90∘ using these?

Solution

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

Find the remaining angles in the following quadrilaterals.

Solution

Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140∘140^{\circ}140∘.

Solution

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

Solution

$30^{\circ}$

$40^{\circ}$

$90^{\circ}$

$140^{\circ}$

We have seen how to get $90^{\circ}$ 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 $90^{\circ}$ using these?

Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of $140^{\circ}$ .