--- title: "NCERT Solutions for Class 8 Maths Chapter 4 Quadrilaterals" url: https://www.swavid.com/maths/class/8/chapter/quadrilaterals/ncert-solutions dateModified: 2026-10-07T15:26:38+00:00 --- # NCERT Solutions for Class 8 Maths Chapter 4 Quadrilaterals This chapter covers questions related to the properties, definitions, and geometric reasoning of various quadrilaterals, including rectangles, squares, parallelograms, rhombuses, kites, and trapeziums. Students also explore the sum of angles in a quadrilateral and how to construct these shapes. Free PDF (30 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-8/swavid-ncert-solutions-class-8-maths-chapter-4-quadrilaterals-d7069d64f0.pdf ## Observe the following figures ### Question 1 *3 marks · Short answer* Figs. (i), (ii), and (iii) are quadrilaterals, and the others are not. Why? **Solution** 1. A quadrilateral is a closed figure bounded by four line segments. 2. Figures (i), (ii), and (iii) are formed by four straight line segments intersecting only at their endpoints. 3. Figures (iv) and (v) are not quadrilaterals because figure (iv) contains a curved boundary and figure (v) has intersecting lines forming a self-intersecting polygon. **Answer:** Figures (i), (ii), and (iii) are closed figures made of four line segments, whereas the others are not. > Common mistake: Confusing open or curved boundaries with line segments. ## Are there other ways to define a rectangle? ### Question 1 *2 marks · Very short answer* What is the length of the other diagonal? **Solution** 1. In a rectangle, the diagonals are of equal length. 2. Since the given strip is $8\text{ cm}$ long, the length of the other diagonal must also be $8\text{ cm}$. **Answer:** $8\text{ cm}$ > Common mistake: Assuming diagonals of a rectangle are not equal. ### Question 2 *2 marks · Very short answer* What is the point of intersection of the two diagonals? **Solution** 1. The diagonals of a rectangle bisect each other. 2. Therefore, the point of intersection is the midpoint of both diagonals. **Answer:** The midpoint of both diagonals > Common mistake: Stating they only intersect without mentioning they bisect each other. ### Question 3 *2 marks · Very short answer* What should the angle be between the diagonals? **Solution** 1. In a rectangle, the diagonals are of equal length and bisect each other. 2. However, for a general rectangle, the angle between the diagonals is not necessarily $90^\circ$ (it is $90^\circ$ only in a square or rhombus). **Answer:** The angle between the diagonals of a rectangle can be any arbitrary angle (say, $x$) and need not be $90^\circ$. > Common mistake: Assuming that the diagonals of every rectangle intersect at right angles like a square. ## Can the following equalities be used to establish that $\Delta \text{AOD} \cong \Delta \text{COB}$? AO = CO (proved above) $\angle \text{AOB} = \angle \text{COD}$ (vertically opposite angles) AD = CB ### Question 1 *3 marks · Short answer* Can the following equalities be used to establish that $\Delta \text{AOD} \cong \Delta \text{COB}$? AO = CO (proved above) $\angle \text{AOB} = \angle \text{COD}$ (vertically opposite angles) AD = CB **Solution** 1. Examine the given conditions: $AO = CO$, $\angle AOB = \angle COD$, and $AD = CB$. 2. Note that the given angle $\angle AOB$ is between sides $AO$ and $OB$, whereas the given sides $AD$ and $CB$ are not included between the given angle and $AO$ or $CO$. 3. Since the side-angle-side (SAS) condition requires the angle to be included between the two sides, these equalities cannot be used to prove $\Delta AOD \cong \Delta COB$. **Answer:** No, these equalities cannot be used because the angle is not the included angle between the given sides. > Common mistake: Assuming any three corresponding parts are sufficient for congruence without checking the congruence criteria like SAS. ## Can you find all the remaining angles? ### Question 1 *3 marks · Short answer* Can you find all the remaining angles? **Solution** 1. In the rectangle with the angle between the diagonals given as $60^\circ$, the vertically opposite angle is also $60^\circ$, and the linear pair angles are $180^\circ - 60^\circ = 120^\circ$ each. 2. Each triangle formed at the corners by the diagonals and sides is isosceles since the halves of diagonals are equal, making the base angles equal to $(180^\circ - 60^\circ)/2 = 60^\circ$ and $(180^\circ - 120^\circ)/2 = 30^\circ$. 3. Thus, the remaining interior angles formed around the diagonal intersections and corners are $60^\circ$ and $30^\circ$. **Answer:** The remaining angles are $60^\circ$ and $30^\circ$. > Common mistake: Confusing the base angles of the isosceles triangles formed by the diagonals. ## Can you find the value of a? ### Question 1 *3 marks · Short answer* Can you find the value of a? **Solution** 1. Consider $\triangle \text{AOB}$ where the two base angles are equal to $a$ and the vertex angle is $60^\circ$. 2. Using the sum of interior angles of a triangle, we have $a + a + 60^\circ = 180^\circ$. 3. Solving this, $2a = 120^\circ$, which gives $a = 60^\circ$. **Answer:** $a = 60^\circ$ > Common mistake: Forgetting that the base angles of an isosceles triangle are equal. ## Can we now identify what type of quadrilateral ABCD is? ### Question 1 *3 marks · Short answer* Can we now identify what type of quadrilateral ABCD is? Notice that its angles all add up to 90° ($30° + 60°$). What can we say about its sides? **Solution** 1. We are given a quadrilateral ABCD formed by diagonals that are equal and bisect each other at angles of $30^\circ$ and $60^\circ$. 2. From the congruence of triangles $\triangle AOB \cong \triangle COD$ and $\triangle AOD \cong \triangle COB$, we find that their corresponding sides are equal, so $AB = CD$ and $AD = CB$. 3. Since all four angles are $90^\circ$ and opposite sides are equal, ABCD is a rectangle. **Answer:** The quadrilateral ABCD is a rectangle, and its opposite sides are equal ($AB = CD$ and $AD = CB$). > Common mistake: Confusing a rectangle with a general parallelogram without checking the interior angles. ## Will ABCD remain a rectangle if the angles between the diagonals are changed? Can we generalise this? Take one of the angles between the diagonals as x. ### Question 1 *3 marks · Short answer* Will ABCD remain a rectangle if the angles between the diagonals are changed? Can we generalise this? Take one of the angles between the diagonals as x. **Solution** 1. Let the diagonals of quadrilateral ABCD be of equal length and bisect each other at angle $x$. 2. In the isosceles triangle formed by the sides and diagonals, the base angles work out to $\frac{180^\circ - x}{2}$ and $\frac{x}{2}$. 3. The interior angle of the quadrilateral formed at each vertex is the sum of these base angles, which equals $90^\circ$. 4. Thus, no matter what the angle $x$ between the diagonals is, the resulting quadrilateral is always a rectangle as long as the diagonals are equal and bisect each other. **Answer:** Yes, ABCD remains a rectangle for any angle $x$ between the diagonals, because the four angles are always $90^\circ$ and opposite sides are equal. > Common mistake: Thinking that changing the angle between the diagonals alters the $90^\circ$ corner angles of the quadrilateral. ## Can you find the other angles? ### Question 1 *3 marks · Short answer* Can you find the other angles? **Solution** 1. Let the two intersecting angles between the diagonals be $x$ and $180^\circ - x$ vertically opposite and linear pairs. 2. In the isosceles triangle formed by the sides and diagonals, the base angles are $a = 90^\circ - \frac{x}{2}$ and $b = \frac{x}{2}$. 3. The four interior angles of the quadrilateral are each $a + b = \left(90^\circ - \frac{x}{2}\right) + \frac{x}{2} = 90^\circ$. **Answer:** The four angles between the diagonals are $x$, $x$, $180^\circ - x$, and $180^\circ - x$, and the four interior angles of the rectangle are all $90^\circ$. > Common mistake: Forgetting that the sum of angles in a triangle is $180^\circ$ when expressing base angles in terms of $x$. ## What is the value of a (in degrees) in terms of x? ### Question 1 *3 marks · Short answer* What is the value of a (in degrees) in terms of x? **Solution** 1. In $\triangle \text{AOB}$, the measures of the base angles are both $a$ since $\text{OA} = \text{OB}$. 2. The sum of the interior angles of a triangle is $180^\circ$, so $a + a + x = 180$. 3. Simplifying the equation gives $2a = 180 - x$, so $a = \frac{180 - x}{2} = 90 - \frac{x}{2}$. **Answer:** $a = 90^\circ - \frac{x}{2}$ > Common mistake: Forgetting that the triangle is isosceles and not dividing the remaining angle equally between the two base angles. ## What can we say about AB and CD, and AD and BC? ### Question 1 *3 marks · Short answer* What can we say about AB and CD, and AD and BC? **Solution** 1. Consider triangles $\Delta AOB$ and $\Delta COD$. 2. We have $\Delta AOB \cong \Delta COD$ and $\Delta AOD \cong \Delta COB$. 3. Hence, $AB = CD$ and $AD = BC$ since they are corresponding parts of congruent triangles. **Answer:** $AB = CD$ and $AD = BC$ > Common mistake: Stating sides are equal without mentioning the corresponding congruent triangles. ## In the quadrilaterals below, are there any non-rectangles? ### Question 1 *2 marks · Very short answer* In the quadrilaterals below, are there any non-rectangles? **Solution** 1. Quadrilateral (iv) is a non-rectangle in terms of traditional perception, but geometrically all four figures (i), (ii), (iii), and (iv) are rectangles because their angles are all right angles ($90^\circ$). 2. Quadrilateral (iv) is a special kind of rectangle known as a square, where all sides are of equal length. **Answer:** No, all the given quadrilaterals, including (iv), are rectangles. > Common mistake: Thinking that a square is not a rectangle. ## Are the opposite sides of a rectangle parallel? ### Question 1 *3 marks · Short answer* Can you similarly show that AB is parallel to DC ($AB||DC$)? **Solution** 1. Consider AD as a transversal intersecting the parallel lines AB and DC. 2. Since ABCD is a rectangle, the angles at A and D are right angles, so $\angle A + \angle D = 90^\circ + 90^\circ = 180^\circ$. 3. When the sum of the interior angles on the same side of the transversal is $180^\circ$, the lines are parallel, therefore $AB \parallel DC$. **Answer:** Yes, $AB \parallel DC$ can be shown using the transversal property where the sum of interior angles on the same side is $180^\circ$. > Common mistake: Forgetting that consecutive interior angles supplementary condition proves lines are parallel. ## Let us consider the Carpenter’s Problem again. If the wooden strips have to be placed such that the thread passing through their endpoints forms a square, what must be done? ### Question 1 *3 marks · Short answer* What more needs to be done to get equal sidelengths as well? Can this be achieved by properly choosing the angle between the diagonals? See if you can reason and/or experiment to figure this out! **Solution** 1. For a quadrilateral to be a square, its diagonals must be equal in length. 2. The diagonals must bisect each other at their midpoints. 3. The diagonals must intersect each other at right angles ($90^\circ$) to ensure all sides are equal. **Answer:** The diagonals must be of equal length, bisect each other, and intersect at an angle of $90^\circ$. > Common mistake: Stating that diagonals only need to be equal without mentioning that they must intersect at right angles. ## To find the angle formed by the diagonals, what are the two triangles we should consider for congruence? ### Question 1 *3 marks · Short answer* Can this be used to find the angles $\angle \text{BOA}$ and $\angle \text{BOC}$ formed by the diagonals? **Solution** 1. Consider triangles $\Delta \text{BOA}$ and $\Delta \text{BOC}$. 2. By the SSS condition for congruence, $\Delta \text{BOA} \cong \Delta \text{BOC}$. 3. Since these angles are corresponding parts of congruent triangles, $\angle \text{BOA} = \angle \text{BOC}$. 4. Further, these angles form a straight angle, so $\angle \text{BOA} + \angle \text{BOC} = 180^\circ$. 5. Thus, each angle is $\frac{180^\circ}{2} = 90^\circ$. **Answer:** Yes, $\Delta \text{BOA}$ and $\Delta \text{BOC}$ are considered, and the angles are $90^\circ$ each. > Common mistake: Forgetting that the angles form a linear pair on a straight line. ## Using this fact, construct a square with a diagonal of length 8 cm. ### Question 1 *3 marks · Short answer* Using this fact, construct a square with a diagonal of length 8 cm. **Solution** 1. Draw a line segment $AB = 8\text{\,cm}$. 2. Draw the perpendicular bisector of $AB$ intersecting it at the midpoint $O$. 3. Mark points $C$ and $D$ on the perpendicular bisector on either side of $O$ such that $OC = OD = 4\text{\,cm}$. 4. Join $AC$, $BC$, $BD$, and $AD$ to obtain the required square $ACBD$. **Answer:** ACBD is the required square with diagonal of length 8 cm. > Common mistake: Drawing the perpendicular lines at an angle other than 90 degrees or not bisecting the diagonal equally. ## Verify if this is true by going through geometric reasoning in Deduction 1 and Deduction 2, and see if they apply to a square as well. ### Question 1 *3 marks · Short answer* Verify if this is true by going through geometric reasoning in Deduction 1 and Deduction 2, and see if they apply to a square as well. **Solution** 1. Since a square is a special type of rectangle, all properties of a rectangle apply to a square. 2. Deduction 1 shows that diagonals of a rectangle are equal in length, which holds for a square as $\text{AC} = \text{BD}$. 3. Deduction 2 shows that diagonals bisect each other, which also holds for a square since $\text{OA} = \text{OC}$ and $\text{OB} = \text{OD}$. **Answer:** The geometric reasoning in Deduction 1 and Deduction 2 applies to a square because every square is a rectangle. > Common mistake: Thinking that properties of rectangles do not apply to squares. ## What are the measures of $\angle 1, \angle 2, \angle 3,$ and $\angle 4$? See if you can reason and/or experiment to figure this out! ### Question 1 *3 marks · Short answer* What are the measures of $\angle 1, \angle 2, \angle 3,$ and $\angle 4$? See if you can reason and/or experiment to figure this out! **Solution** 1. In triangle ADC, we have $\angle 1 + \angle 3 + 90^\circ = 180^\circ$. 2. Since $AD = DC$ in a square, the base angles are equal, so $\angle 1 = \angle 3$. 3. Thus, $\angle 1 = \angle 3 = 45^\circ$ and similarly $\angle 2 = \angle 4 = 45^\circ$. **Answer:** $\angle 1 = 45^\circ, \angle 2 = 45^\circ, \angle 3 = 45^\circ, \angle 4 = 45^\circ$ > Common mistake: Assuming the diagonal bisects the corner angle without verifying the isosceles triangle property. ## Figure it Out ### Question 1 *3 marks · Short answer* Find all the other angles inside the following rectangles. **Part (i)** 1. In rectangle (i), the diagonals are equal and bisect each other, forming isosceles triangles. 2. Using the given angle of $30^\circ$, we find alternate interior angles and base angles of isosceles triangles. 3. The remaining angles are $\angle \text{ABD} = 30^\circ$, $\angle \text{BDC} = 30^\circ$, $\angle \text{CAD} = 60^\circ$, $\angle \text{ACD} = 30^\circ$, $\angle \text{ADB} = 60^\circ$, and $\angle \text{ACB} = 60^\circ$. Answer (i): $\angle \text{ABD} = 30^\circ$, $\angle \text{BDC} = 30^\circ$, $\angle \text{CAD} = 60^\circ$, $\angle \text{ACD} = 30^\circ$, $\angle \text{ADB} = 60^\circ$, $\angle \text{ACB} = 60^\circ$ **Part (ii)** 1. In rectangle (ii), using vertically opposite angles at the intersection O and base angles of isosceles triangles formed by the diagonals. 2. The angles between the diagonals are $110^\circ$, $70^\circ$, $70^\circ$, and $110^\circ$. 3. The remaining interior vertex angles are $\angle \text{OQR} = 35^\circ$, $\angle \text{ORQ} = 35^\circ$, $\angle \text{OQP} = 55^\circ$, $\angle \text{OPQ} = 55^\circ$, $\angle \text{ORS} = 55^\circ$, and $\angle \text{OSR} = 55^\circ$. Answer (ii): $\angle \text{POS} = 110^\circ$, $\angle \text{QOP} = 70^\circ$, $\angle \text{ROS} = 70^\circ$, $\angle \text{OQR} = 35^\circ$, $\angle \text{ORQ} = 35^\circ$, $\angle \text{OQP} = 55^\circ$, $\angle \text{OPQ} = 55^\circ$, $\angle \text{ORS} = 55^\circ$, $\angle \text{OSR} = 55^\circ$ **Answer:** Refer to the parts for angles in each rectangle. > Common mistake: Confusing base angles of the isosceles triangles formed by the diagonals of a rectangle. ### Question 2 *3 marks · Short answer* 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° **Part (i)** 1. Draw a line segment $AB = 8~\text{cm}$ representing one diagonal. 2. Mark the midpoint $O$ of $AB$, so that $OA = OB = 4~\text{cm}$. 3. At $O$, draw an angle of $30^\circ$ and draw the second diagonal line such that $OC = OD = 4~\text{cm}$. 4. Join $AD$, $DB$, $BC$, and $AC$ to get the required rectangle. Answer (i): Draw diagonal $AB = 8~\text{cm}$, bisect at $O$, draw $CD = 8~\text{cm}$ at angle $30^\circ$ intersecting at midpoints, and join endpoints. **Part (ii)** 1. Draw a line segment of length $8~\text{cm}$ and take its midpoint. 2. Draw the second diagonal of length $8~\text{cm}$ intersecting at their midpoints at an angle of $40^\circ$. 3. Join the four endpoints to form the quadrilateral. Answer (ii): Follow the same steps with an angle of $40^\circ$ between diagonals. **Part (iii)** 1. Draw a line segment of length $8~\text{cm}$ and take its midpoint. 2. Draw the second diagonal of length $8~\text{cm}$ intersecting at their midpoints at a right angle ($90^\circ$). 3. Join the four endpoints to form a square. Answer (iii): Follow the same steps with an angle of $90^\circ$ between diagonals. **Answer:** Refer to the parts for the construction steps. > Common mistake: Not bisecting the diagonals correctly at their intersection point. ### Question 3 *3 marks · Short answer* 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** 1. Given that PL and AM are two diameters of the circle with centre O, their lengths are equal (both are diameters). 2. Since they are diameters, they pass through the centre O and bisect each other at O ($OA = OM$ and $OP = OL$). 3. The angle between them is given as $90^\circ$ since they are perpendicular diameters. 4. A quadrilateral whose diagonals are equal and bisect each other at right angles is a square. **Answer:** APML is a square. > Common mistake: Conclude it is only a rectangle instead of a square, forgetting that perpendicular diagonals make it a square. ### Question 4 *3 marks · Short answer* 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** 1. Let the two equal sticks be modeled as line segments representing the diagonals of a rectangle. 2. Place them together such that their midpoints coincide at a point O. 3. Pass a thread through their endpoints to form a quadrilateral whose diagonals are equal and bisect each other. 4. Since the diagonals are equal and bisect each other, the resulting quadrilateral is a rectangle, making the angle at the vertices $90^\circ$. **Answer:** Place the two equal sticks such that their midpoints coincide, and stretch a thread through their endpoints to form a rectangle with $90^\circ$ angles. > Common mistake: Not ensuring that the sticks bisect each other at their exact midpoints. ### Question 5 *3 marks · Short answer* 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** 1. A rectangle is defined as a quadrilateral in which all angles are right angles ($90^\circ$). 2. A quadrilateral with opposite sides parallel and equal is a parallelogram, not necessarily a rectangle, because its interior angles do not have to be $90^\circ$. 3. Therefore, having opposite sides parallel and equal cannot be chosen as the definition of a rectangle. **Answer:** No, this cannot be taken as a definition of a rectangle because such a quadrilateral is a parallelogram and its angles need not be $90^\circ$. > Common mistake: Confusing a rectangle with a general parallelogram. ## Is it possible to construct a quadrilateral with three angles equal to 90° and the fourth angle not equal to 90°? ### Question 1 *2 marks · Very short answer* Is it possible to construct a quadrilateral with three angles equal to 90° and the fourth angle not equal to 90°? **Solution** 1. The sum of all four interior angles in any quadrilateral is always equal to $360^\circ$. 2. If three angles are $90^\circ$ each, their sum is $90^\circ + 90^\circ + 90^\circ = 270^\circ$, leaving the fourth angle to be $360^\circ - 270^\circ = 90^\circ$. 3. Therefore, it is not possible to have a quadrilateral with three right angles and a fourth angle different from $90^\circ$. **Answer:** No, it is not possible because the sum of all angles in a quadrilateral is always $360^\circ$. > Common mistake: Thinking that three angles can be $90^\circ$ independently without affecting the fourth angle. ## Are there quadrilaterals that have parallel opposite sides that are not rectangles? ### Question 1 *3 marks · Short answer* Construct such a figure by recalling how parallel lines can be constructed using a ruler and a set-square, or a compass and a ruler. **Solution** 1. Draw a line and mark two points A and B on it at a distance of a chosen length, say $4\text{ cm}$. 2. Using a ruler and a set-square, draw a line parallel to AB at a desired distance, and use a compass to mark the side lengths (such as $5\text{ cm}$) at an angle to form the vertices C and D. 3. Join the points to obtain the required quadrilateral ABCD with parallel opposite sides that is not a rectangle. **Answer:** A parallelogram ABCD constructed with parallel opposite sides. > Common mistake: Drawing the sides at right angles, which results in a rectangle instead of a general parallelogram. ## Draw a parallelogram with adjacent sides of lengths 4 cm and 5 cm, and an angle of 30° between them. ### Question 1 *3 marks · Short answer* What are the remaining angles of the parallelogram? What are the lengths of the remaining sides? See if you can reason out and/or experiment to figure these out. **Solution** 1. In a parallelogram, opposite sides are equal in length, so the remaining sides are of lengths $4\text{ cm}$ and $5\text{ cm}$. 2. The opposite angles of a parallelogram are equal, so the angle opposite to the $30^\circ$ angle is also $30^\circ$. 3. The adjacent angles of a parallelogram add up to $180^\circ$, so the remaining two angles are each $180^\circ - 30^\circ = 150^\circ$. **Answer:** The remaining sides are $4\text{ cm}$ and $5\text{ cm}$, and the remaining angles are $30^\circ$, $150^\circ$, and $150^\circ$. > Common mistake: Adding adjacent angles to $90^\circ$ instead of $180^\circ$. ## What about the opposite angles? Will they be equal in all parallelograms? If yes, how can we be sure? ### Question 1 *3 marks · Short answer* What about the opposite angles? Will they be equal in all parallelograms? If yes, how can we be sure? Let us take one of the angles to be x. What are the other angles? **Solution** 1. Let one of the angles of the parallelogram be $x$. 2. Since the adjacent angles of a parallelogram add up to $180^\circ$, the adjacent angle is $180^\circ - x$. 3. The opposite angle to the first angle is also supplementary to this adjacent angle, giving $180^\circ - (180^\circ - x) = x$. 4. Thus, the other angles of the parallelogram are $180^\circ - x$, $x$, and $180^\circ - x$, showing that opposite angles are always equal. **Answer:** The other angles are $180^\circ - x$, $x$, and $180^\circ - x$, showing that opposite angles in a parallelogram are always equal. > Common mistake: Assuming opposite angles are supplementary instead of equal. ## Are the diagonals of a parallelogram always equal? Check with the parallelogram that you have constructed. ### Question 1 *2 marks · Very short answer* Are the diagonals of a parallelogram always equal? Check with the parallelogram that you have constructed. **Solution** 1. No, the diagonals of a parallelogram are not always equal. 2. They are equal only in special types of parallelograms such as rectangles and squares. **Answer:** No, the diagonals of a parallelogram are not always equal. > Common mistake: Assuming all parallelograms have equal diagonals like rectangles. ## Do they bisect each other (do they intersect at their midpoints)? Reason and/or experiment to figure this out. ### Question 1 *3 marks · Short answer* Do they bisect each other (do they intersect at their midpoints)? Reason and/or experiment to figure this out. **Solution** 1. Consider the parallelogram EASY with diagonals intersecting at O, by considering $\triangle AOE$ and $\triangle YOS$. 2. AE = YS as they are opposite sides of the parallelogram, and alternate interior angles are equal for the parallel lines. 3. By the ASA congruence condition, $\triangle AOE \cong \triangle YOS$, which gives OA = OY and OE = OS as corresponding parts. **Answer:** Yes, the diagonals of a parallelogram bisect each other at their point of intersection. > Common mistake: Confusing the order of vertices when writing the congruent triangles. ## Do the diagonals of a parallelogram intersect at a particular angle? ### Question 1 *2 marks · Very short answer* Do the diagonals of a parallelogram intersect at a particular angle? **Solution** 1. No, the diagonals of a general parallelogram do not intersect at any particular angle. 2. They can intersect at any angle depending on the adjacent sides and the angles of the parallelogram. **Answer:** No, the diagonals of a parallelogram do not intersect at any particular angle. > Common mistake: Confusing a general parallelogram with a rhombus or a square, whose diagonals do intersect at right angles. ## Are squares the only quadrilaterals that have equal sidelengths? Let us explore this question through construction. ### Question 1 *3 marks · Short answer* Can we complete this quadrilateral so that all its sides are of the same length? **Solution** 1. Yes, we can complete the quadrilateral such that all its sides are of the same length. 2. Measure the length of the given side $AB$ using a compass. 3. Keeping this length as the radius, draw arcs from vertices $B$ and $D$ to intersect at a point $C$, forming a rhombus. **Answer:** Yes, by using a compass with radius equal to the side length and drawing intersecting arcs from the other vertices. > Common mistake: Taking different radii for the arcs instead of keeping the same length as the given side. ## What are the other angles of the rhombus ABCD that we have constructed? Reason and/or experiment to figure this out. ### Question 1 *3 marks · Short answer* What are the other angles of the rhombus ABCD that we have constructed? Reason and/or experiment to figure this out. **Solution** 1. In the constructed rhombus ABCD, let the given angle at vertex A be $50^\circ$. 2. The adjacent angles of a rhombus add up to $180^\circ$, so angle B = $180^\circ - 50^\circ = 130^\circ$. 3. The opposite angles of a rhombus are equal, so angle C = $50^\circ$ and angle D = $130^\circ$. **Answer:** The angles of the rhombus ABCD are $50^\circ$, $130^\circ$, $50^\circ$, and $130^\circ$. > Common mistake: Confusing adjacent angles with opposite angles in a rhombus. ## It can be seen that $\Delta \text{GAE} \cong \Delta \text{MAE}$ (How?) ### Question 1 *3 marks · Short answer* It can be seen that $\Delta \text{GAE} \cong \Delta \text{MAE}$ (How?) **Solution** 1. In $\Delta \text{GAE}$ and $\Delta \text{MAE}$, we have $\text{GE} = \text{GM}$ (sides of a rhombus are equal). 2. Also, $\text{AE} = \text{AE}$ (common side). 3. Since a rhombus is a parallelogram, opposite sides are equal, so $\text{GA} = \text{MA}$. 4. Therefore, by the SSS congruence condition, $\Delta \text{GAE} \cong \Delta \text{MAE}$. **Answer:** $\Delta \text{GAE} \cong \Delta \text{MAE}$ by SSS congruence condition since $\text{GE} = \text{GM}$, $\text{AE} = \text{AE}$, and $\text{GA} = \text{MA}$. > Common mistake: Forgetting to state that the common side and all sides of the rhombus are equal. ## So a rhombus is a parallelogram, and a rectangle is also a parallelogram. How can this be represented using a Venn diagram? ### Question 1 *3 marks · Short answer* Where will the set of squares occur in this diagram? **Solution** 1. A square is a special type of rectangle because all its angles are $90^\circ$. 2. A square is also a rhombus because all its sides are of equal length. 3. Therefore, the set of squares will occur in the overlapping region (intersection) of the set of rectangles and the set of rhombuses. **Answer:** The set of squares occurs in the intersection (overlapping region) of the rectangle and rhombus sets in the Venn diagram. > Common mistake: Placing squares inside only rectangles or only rhombuses, forgetting that a square possesses properties of both. ## Are the diagonals of a rhombus equal? ### Question 1 *2 marks · Very short answer* Are the diagonals of a rhombus equal? **Solution** 1. No, the diagonals of a rhombus are not always equal in length. 2. They are equal only in the case of a square, which is a special type of rhombus. **Answer:** No, the diagonals of a rhombus need not be equal. > Common mistake: Assuming all sides being equal in a rhombus implies its diagonals are equal, confusing it with a square or rectangle. ## Do the diagonals of a rhombus intersect at any particular angle? Reason out and/or experiment to figure this out! ### Question 1 *3 marks · Short answer* Do the diagonals of a rhombus intersect at any particular angle? Reason out and/or experiment to figure this out! **Solution** 1. Consider a rhombus GAME where diagonals intersect at O. 2. In triangles $\triangle GEO$ and $\triangle MEO$, $GE = ME$ (sides of rhombus), $EO$ is common, and $\triangle GEO \cong \triangle MEO$ by SSS condition. 3. Therefore, $\angle GOE = \angle MOE$ as corresponding parts of congruent triangles. 4. Since $\angle GOE + \angle MOE = 180^\circ$ (linear pair), we have $\angle GOE = \angle MOE = 90^\circ$. 5. Thus, diagonals of a rhombus intersect each other at an angle of $90^\circ$. **Answer:** Yes, the diagonals of a rhombus intersect each other at $90^\circ$ (at right angles). > Common mistake: Assuming diagonals are equal without proving congruence first. ## In the rhombus GAME, we have $\Delta \text{GEO} \cong \Delta \text{MEO}$ (why?). ### Question 1 *3 marks · Short answer* In the rhombus GAME, we have $\Delta \text{GEO} \cong \Delta \text{MEO}$ (why?). **Solution** 1. In $\Delta \text{GEO}$ and $\Delta \text{MEO}$, we have $GE = ME$ as all sides of a rhombus are equal. 2. The diagonal $EO$ is common to both triangles, so $EO = EO$. 3. The diagonals of a rhombus bisect each other, so $GO = MO$. 4. Therefore, by the SSS congruence condition, $\Delta \text{GEO} \cong \text{MEO}$. **Answer:** $\Delta \text{GEO} \cong \Delta \text{MEO}$ by the SSS condition since $GE = ME$, $EO = EO$, and $GO = MO$. > Common mistake: Forgetting to mention that all sides of a rhombus are equal or that diagonals bisect each other. ## Figure it Out ### Question 1 *3 marks · Short answer* Find the remaining angles in the following quadrilaterals. **Part (i)** 1. Opposite angles of a parallelogram are equal, so $\angle E = 40^\circ$. 2. Adjacent angles add up to $180^\circ$, so $\angle R = 180^\circ - 40^\circ = 140^\circ$. 3. The remaining angles are $\angle R = 140^\circ$, $\angle E = 40^\circ$, and $\angle A = 40^\circ$. Answer (i): $\angle E = 140^\circ$, $\angle R = 140^\circ$, and $\angle A = 40^\circ$ **Part (ii)** 1. Adjacent angles of a parallelogram add up to $180^\circ$, so $\angle Q = 180^\circ - 110^\circ = 70^\circ$. 2. Opposite angles are equal, so $\angle S = 70^\circ$ and $\angle R = 110^\circ$. Answer (ii): $\angle Q = 70^\circ$, $\angle S = 70^\circ$, and $\angle R = 110^\circ$ **Part (iii)** 1. Opposite sides are parallel, so alternate interior angles are equal giving $\angle XVU = \angle XVW = 30^\circ$. 2. Adjacent angles add up to $180^\circ$, so $\angle U = 180^\circ - 60^\circ = 120^\circ$. 3. Opposite angles are equal, so $\angle W = 120^\circ$. Answer (iii): $\angle U = 120^\circ$, $\angle W = 120^\circ$, and remaining interior angles are $30^\circ$ and $60^\circ$ **Part (iv)** 1. Diagonals of a parallelogram bisect each other, and opposite angles are equal. 2. Using vertically opposite angles and alternate interior angles, we get $\angle OEI = 20^\circ$, $\angle AOE = 20^\circ$, $\angle EOI = 20^\circ$. 3. The angles of the parallelogram are $\angle A = 140^\circ$ and $\angle I = 140^\circ$. Answer (iv): $\angle OEI = 20^\circ$, $\angle AOE = 20^\circ$, $\angle EOI = 20^\circ$, $\angle A = 140^\circ$, and $\angle I = 140^\circ$ **Answer:** Refer to each part for the remaining angles. > Common mistake: Confusing adjacent angles with opposite angles. ### Question 2 *3 marks · Short answer* 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** 1. Draw a line segment AB = 7 cm. 2. Take the midpoint O of AB and draw an angle of $140^\circ$ at O on AB. 3. Draw arcs of radius $2.5\text{ cm}$ (half of 5 cm) from O on both sides of the line to mark points C and D. 4. Join AC, AD, BD, and BC to form the parallelogram ADBC. **Answer:** ADBC is the required parallelogram with diagonals 7 cm and 5 cm intersecting at $140^\circ$. > Common mistake: Taking the full length of 5 cm for the radius instead of half ($2.5\text{ cm}$). ### Question 3 *3 marks · Short answer* Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm. **Solution** 1. Draw a line segment AB = 5 cm. 2. Take the midpoint O of AB and draw a perpendicular line at O. 3. Mark points C and D on the perpendicular line such that OC = OD = 2 cm (half of 4 cm). 4. Join AD, BD, CB, and AC to form the rhombus ADBC. **Answer:** ADBC is the required rhombus with diagonals 5 cm and 4 cm. > Common mistake: Taking 4 cm as the radius instead of half of it. ## What is the quadrilateral that you get? Justify your answer. Extend one of the diagonals on both sides by 2 cm. ### Question 1 *3 marks · Short answer* What quadrilateral will you get now? Justify your answer. **Solution** 1. Place two perpendicular rubber bands of unequal lengths acting as diagonals that bisect each other. 2. Join their endpoints to form a quadrilateral. 3. The diagonals are unequal in length and bisect each other at right angles, which are the properties of a rhombus. **Answer:** A rhombus, because the diagonals are unequal and bisect each other at right angles. > Common mistake: Confusing a rhombus with a square or rectangle when the diagonals are unequal. ## Can you join them to get a quadrilateral? ### Question 1 *3 marks · Short answer* What type of a quadrilateral is this? Justify your answer. **Solution** 1. Take two identical equilateral triangles of side length $8\text{ cm}$. 2. Join them along one of their common sides to form a four-sided figure. 3. Since all four sides of the resulting quadrilateral are equal to $8\text{ cm}$, the quadrilateral is a rhombus. **Answer:** The quadrilateral is a rhombus because all its four sides are equal. > Common mistake: Assuming the figure is a square without checking its interior angles. ## What are the different ways they can be joined to get a quadrilateral? ### Question 1 *3 marks · Short answer* What quadrilaterals are these? Justify your answers. **Solution** 1. When two isosceles triangles with sides $8\text{ cm}$, $8\text{ cm}$, and $6\text{ cm}$ are joined along their equal sides or unequal bases, they form different quadrilaterals. 2. Joining them along their base of length $6\text{ cm}$ creates a rhombus because all four sides are equal to $8\text{ cm}$. 3. Joining them edge-to-edge along one of the equal sides of length $8\text{ cm}$ creates a kite with two pairs of adjacent equal sides of lengths $8\text{ cm}$ and $6\text{ cm}$. **Answer:** A rhombus and a kite. > Common mistake: Confusing the types of quadrilaterals formed by not checking the side lengths properly. ## What are the different ways they can be joined to get a quadrilateral? ### Question 1 *5 marks · Long answer* Are you able to identify the different quadrilaterals that are obtained by joining the triangles? Justify your answer whenever you identify a quadrilateral. **Solution** 1. Take two congruent scalene triangles with sides 6 cm, 9 cm, and 12 cm. 2. Join the two triangles along their 6 cm sides so that the 9 cm and 12 cm sides form the outer boundary, obtaining a kite. 3. Join the two triangles along their 9 cm sides so that the 6 cm and 12 cm sides form the outer boundary, obtaining another kite. 4. Join the two triangles along their 12 cm sides so that the 6 cm and 9 cm sides form the outer boundary, obtaining a third kite. 5. In all these cases, the resulting quadrilaterals are kites because each figure has two pairs of adjacent sides of equal length ($6\text{ cm}$ and $6\text{ cm}$, or $9\text{ cm}$ and $9\text{ cm}$, or $12\text{ cm}$ and $12\text{ cm}$). 6. It is also possible to form trapeziums by arranging the triangles such that a pair of sides become parallel. **Answer:** Three different types of kites and trapeziums can be obtained depending on which matching sides of the scalene triangles are joined together. > Common mistake: Assuming only one quadrilateral can be formed regardless of which side is joined. ## Construct a trapezium. Measure the base angles (marked in the figure). ### Question 1 *3 marks · Short answer* Can you find the remaining angles without measuring them? **Solution** 1. Since $PQ \parallel SR$, the adjacent angles on the same side of the transversal are supplementary. 2. Therefore, $\angle S + \angle P = 180^\circ$ and $\angle R + \angle Q = 180^\circ$. 3. Using these relations, the remaining angles can be found by subtracting the given base angles from $180^\circ$. **Answer:** The remaining angles are found using the property that adjacent interior angles on the same side of a transversal sum to $180^\circ$. > Common mistake: Confusing adjacent angles with opposite angles in a trapezium. ## How do we construct an isosceles trapezium? ### Question 1 *3 marks · Short answer* Construct an isosceles trapezium UVWX, with UV||XW. Measure $\angle \text{U}$. **Solution** 1. Draw a line segment UV of suitable length. 2. Draw a line parallel to UV at a certain distance and mark points X and W on it such that UX = VW. 3. Join UW and XV to complete the isosceles trapezium UVWX where UV || XW and non-parallel sides UX and VW are equal. **Answer:** UVWX is the required isosceles trapezium constructed with UV || XW and equal non-parallel sides UX and VW. > Common mistake: Drawing the non-parallel sides of unequal lengths, which fails to make it an isosceles trapezium. ## Can you find the remaining angles without measuring them? ### Question 1 *3 marks · Short answer* Can you find the remaining angles without measuring them? **Solution** 1. In the isosceles trapezium UVWX with $UV \parallel XW$, the adjacent angles on the same side of a transversal add up to $180^\circ$. 2. Therefore, $\angle U + \angle X = 180^\circ$ and $\angle V + \angle W = 180^\circ$. 3. Since the angles opposite to the equal sides of an isosceles trapezium are equal, we have $\angle U = \angle V$ and $\angle X = \angle W$. Using these properties, the remaining angles can be found without measuring. **Answer:** The remaining angles can be found using the properties that adjacent interior angles sum to $180^\circ$ and base angles of an isosceles trapezium are equal. > Common mistake: Confusing the base angles or assuming non-parallel opposite sides are equal. ## Now, it can be shown that $\Delta \text{UXY} \cong \Delta \text{VWZ}$. (How?) ### Question 1 *3 marks · Short answer* Now, it can be shown that $\Delta \text{UXY} \cong \Delta \text{VWZ}$. (How?) **Solution** 1. In right-angled triangles $\Delta \text{UXY}$ and $\Delta \text{VWZ}$, side $\text{UX} = \text{VW}$ because $\text{UVWX}$ is an isosceles trapezium. 2. Side $\text{XY} = \text{WZ}$ because opposite sides of rectangle $\text{XWZY}$ are equal. 3. Angle $\text{XYU} = \text{WZV} = 90^\circ$ since $\text{XY}$ and $\text{WZ}$ are perpendicular to $\text{UV}$. 4. Therefore, $\Delta \text{UXY} \cong \Delta \text{VWZ}$ by the RHS congruence condition. **Answer:** $\Delta \text{UXY} \cong \Delta \text{VWZ}$ by the RHS congruence condition. > Common mistake: Forgetting to state that the right angles and the hypotenuse are equal before applying RHS congruence. ## Figure it Out ### Question 1 *3 marks · Short answer* Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm. **Solution** 1. Given: Two equilateral triangles with sides 4 cm joined together. 2. All four sides of the resulting quadrilateral are equal to 4 cm. 3. All the interior angles are formed by combining two angles of 60° each, so the angles are 60°, 120°, 60°, and 120°. **Answer:** Sides: 4 cm, 4 cm, 4 cm, 4 cm; Angles: 60°, 120°, 60°, 120° > Common mistake: Writing all angles as 60 degrees without summing them up when triangles are joined. ### Question 2 *3 marks · Short answer* Construct a kite whose diagonals are of lengths 6 cm and 8 cm. **Solution** 1. Draw a line segment PQ = 6 cm. 2. Draw a perpendicular bisector of PQ, cutting PQ at point T. 3. Cut arcs of length such that the total length of the second diagonal RS is 8 cm, with T as the midpoint. Join PR, RQ, QS, and PS. **Answer:** PRQS is the required kite with diagonals 6 cm and 8 cm. > Common mistake: Not making the diagonals perpendicular bisectors of each other properly. ### Question 3 *3 marks · Short answer* Find the remaining angles in the following trapeziums— **Solution** 1. Given: Trapeziums with parallel opposite sides. 2. Using the property that adjacent interior angles on the same side of a transversal sum to 180°, we find the missing angles. 3. For figure (i), the angles are 75° and 45°. For figure (ii), the angles are 80° each. **Answer:** (i) 75° and 45°; (ii) 80° and 80° > Common mistake: Confusing adjacent angles with opposite angles. ### Question 4 *3 marks · Short answer* 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? **Part (i)** 1. A quadrilateral that has opposite sides parallel (parallelogram) and adjacent sides equal (kite) is a rhombus and a square. Answer (i): Rhombus and Square **Part (ii)** 1. A rectangle has all angles equal to 90°, whereas a kite has perpendicular diagonals and equal adjacent sides, so no general rectangle is a kite unless it is a square. Answer (ii): No **Part (iii)** 1. A rhombus has all four sides equal, but a kite only has two pairs of adjacent sides equal. Thus, every rhombus is a kite, but every kite is not a rhombus. Answer (iii): No. A rhombus is a kite whereas a kite need not be a rhombus. **Answer:** Refer to the Venn diagram and sub-part answers. > Common mistake: Assuming all kites are rhombuses because they have equal sides. ### Question 5 *3 marks · Short answer* If PAIR and RODS are two rectangles, find $\angle \text{IOD}$. **Solution** 1. Given: PAIR and RODS are two rectangles with an angle of 30°. 2. From the properties of rectangles and alternate interior angles, the angle ∯IOD is equal to the given angle. 3. Therefore, ∯IOD = 30°. **Answer:** 30° > Common mistake: Misidentifying the alternate angles in the overlapping region. ### Question 6 *3 marks · Short answer* Construct a square with diagonal 6 cm without using a protractor. **Solution** 1. Draw a line segment AB = 6 cm. 2. Construct a perpendicular bisector to AB through its midpoint O, and mark points C and D on it such that OC = OD = 3 cm. 3. Join AC, BC, BD, and AD to obtain the square ACBD. **Answer:** ACBD is the required square with diagonal 6 cm. > Common mistake: Using a protractor instead of compass constructions for the perpendicular bisector. ### Question 7 *3 marks · Short answer* 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** 1. Let the side of the square CASE be $x$. 2. The segments joining the midpoints $U, V, W, X$ form four right-angled triangles with legs of lengths $\frac{x}{2}$. 3. By the Pythagoras theorem, each side of the quadrilateral UVWX is equal to $\frac{x}{\sqrt{2}}$, and all its angles are $90^\circ$. 4. Therefore, quadrilateral UVWX is a square. **Answer:** The quadrilateral UVWX is a square. > Common mistake: Assuming the inner quadrilateral is only a rhombus without checking the angles. ### Question 8 *3 marks · Short answer* 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** 1. A quadrilateral with four equal sides is a rhombus. 2. In a rhombus, opposite angles are equal and adjacent angles are supplementary. 3. Since one angle is given as $90^\circ$, its adjacent angle and opposite angle must also be $90^\circ$, making all four angles $90^\circ$. 4. Therefore, a quadrilateral with four equal sides and one right angle is a square. **Answer:** Yes, it will be a square. > Common mistake: Confusing a rhombus with a square when only one angle is $90^\circ$. ### Question 9 *3 marks · Short answer* What type of a quadrilateral is one in which the opposite sides are equal? Justify your answer. Hint: Draw a diagonal and check for congruent triangles. **Solution** 1. Let the quadrilateral be ABCD with opposite sides equal, so $AB = CD$ and $BC = AD$. 2. Draw the diagonal AC, dividing the quadrilateral into $\triangle ABC$ and $\triangle CDA$. 3. By the SSS congruence condition, $\triangle ABC \cong \triangle CDA$, which gives $\angle 1 = \angle 4$ and $\angle 2 = \angle 3$. 4. Since alternate interior angles are equal, the opposite sides are parallel, making the quadrilateral a parallelogram. **Answer:** The quadrilateral is a parallelogram. > Common mistake: Forgetting to state that alternate angles being equal implies parallel lines. ### Question 10 *3 marks · Short answer* 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** 1. Join diagonal BD to divide the quadrilateral ADCB into two triangles, $\triangle ADB$ and $\triangle BDC$. 2. The sum of the angles in $\triangle ADB$ is $180^\circ$ and the sum of the angles in $\triangle BDC$ is $180^\circ$. 3. Adding the angle sums of both triangles gives $180^\circ + 180^\circ = 360^\circ$, showing that the sum of the angles in any quadrilateral is $360^\circ$. **Answer:** Yes, the sum of the angles in this quadrilateral is also $360^\circ$. > Common mistake: Assuming that concave quadrilaterals have a different angle sum than convex quadrilaterals. ### Question 11 *5 marks · Case-based* 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. (ii) A quadrilateral having three right angles must be a rectangle. (iii) A quadrilateral whose diagonals bisect each other must be a parallelogram. (iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus. (v) A quadrilateral in which the opposite angles are equal must be a parallelogram. (vi) A quadrilateral in which all the angles are equal is a rectangle. (vii) Isosceles trapeziums are parallelograms. **Part (i)** 1. A quadrilateral whose diagonals are equal and bisect each other forms a rectangle, not necessarily a square. 2. For it to be a square, the diagonals must also be perpendicular to each other. Answer (i): False. In a square, the diagonals are also perpendicular to each other. **Part (ii)** 1. If a quadrilateral has three right angles, the sum of all four interior angles being $360^\circ$ forces the fourth angle to be $360^\circ - (90^\circ + 90^\circ + 90^\circ) = 90^\circ$. 2. Thus, all four angles are right angles, making it a rectangle. Answer (ii): True. The fourth angle will also be equal to $90^\circ$ because the sum of all angles is equal to $360^\circ$. **Part (iii)** 1. If the diagonals of a quadrilateral bisect each other, opposite sides are equal and parallel. 2. Therefore, the quadrilateral must be a parallelogram. Answer (iii): True. **Part (iv)** 1. A quadrilateral whose diagonals are perpendicular to each other can be a kite or a rhombus. 2. For it to be a rhombus, the diagonals must also bisect each other. Answer (iv): False. It may be a kite also. **Part (v)** 1. If the opposite angles of a quadrilateral are equal, each pair of opposite sides is parallel. 2. Therefore, the quadrilateral must be a parallelogram. Answer (v): True. **Part (vi)** 1. A quadrilateral in which all angles are equal must have each angle equal to $360^\circ / 4 = 90^\circ$. 2. A quadrilateral with all right angles is defined as a rectangle. Answer (vi): True. **Part (vii)** 1. An isosceles trapezium has one pair of opposite sides parallel, but the other pair of non-parallel sides are equal in length but not parallel. 2. Thus, it is not a parallelogram. Answer (vii): False. The other pair of opposite sides may not be parallel. **Answer:** State whether statements (i) to (vii) are true or false with justifications. > Common mistake: Confusing properties of rectangles and squares, or assuming perpendicular diagonals always imply a rhombus without bisecting. ## Frequently asked questions ### How many questions and topics are covered in NCERT Solutions for Class 8 Maths Chapter 4 Quadrilaterals? The chapter covers various topics including the definition of a quadrilateral, properties of rectangles, squares, parallelograms, rhombuses, trapeziums, and their diagonal properties. The solutions provide detailed answers for all the textbook questions based on the new NCERT book for the 2026-27 session. You can access SwaVid's free PDF and step-by-step solutions on this page only. ### Are SwaVid's NCERT Solutions for Class 8 Maths Chapter 4 based on the latest syllabus? Yes, all solutions are strictly prepared according to the new NCERT book based on the NCF 2023 for the 2026-27 session. They cover concepts like triangle congruence in rectangles, properties of diagonals in a square, and angle sum properties. SwaVid's free PDF and step-by-step solutions are available on this page only to help you prepare effectively. ### What are the hardest question types in this chapter and how should we approach them? The most challenging questions involve geometric reasoning and proofs, such as establishing triangle congruence like $\Delta \text{AOD} \cong \Delta \text{COB}$ to prove opposite sides of a rectangle are equal. To score full marks, students should clearly state the given properties, apply theorems step by step, and justify every deduction logically. SwaVid's free PDF and step-by-step solutions on this page only explain these difficult proofs clearly. ### How should I write answers to get full marks in Class 8 Maths Chapter 4 exams? To secure full marks, always write down the given information first, state the geometric theorem or property you are using, and show all intermediate steps clearly. For construction questions, write the steps of construction methodically alongside the rough sketch. SwaVid's free PDF and step-by-step solutions on this page only provide ideal answer formats for board and school exams. ### Is the free PDF for Class 8 Maths Chapter 4 Quadrilaterals available for download? Yes, comprehensive solutions covering all exercises and activities from the new NCERT textbook are provided for student reference. SwaVid's free PDF and step-by-step solutions are available on this page only, making it easy to revise concepts like parallelogram properties and rhombus diagonals offline. ## Related pages - [Class 8 Maths chapters](https://www.swavid.com/maths/class/8) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. 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