Quadrilaterals Ex 8.1: NCERT Solutions and Concepts for Class 9 Maths

Key Theorems and Properties for Exercise 8.1

To solve the problems in Exercise 8.1, you need to be very comfortable with two main concepts: the Angle Sum Property and the properties of a parallelogram. Let's break them down.

1. Angle Sum Property of a Quadrilateral: This is the simplest yet most fundamental property. It states that the sum of the four interior angles of any quadrilateral is always 360°. Think about it: if you draw a diagonal in any quadrilateral, you split it into two triangles. Since the sum of angles in each triangle is 180°, the total sum for the quadrilateral becomes 180° + 180° = 360°. This is crucial for solving problems where angles are given in a ratio.

2. Properties of a Parallelogram: This is the core of the chapter. A parallelogram is a quadrilateral where both pairs of opposite sides are parallel. This simple definition leads to several powerful properties that you will use to prove and solve problems:

• Theorem 8.1: A diagonal divides a parallelogram into two congruent triangles.
• Property 1: Opposite sides of a parallelogram are equal.
• Property 2: Opposite angles of a parallelogram are equal.
• Property 3: The diagonals of a parallelogram bisect each other (meaning they cut each other into two equal halves at their intersection point).

Mastering these properties and their converses is the key to success in this exercise.

Solved Examples: Applying the Theorems

• Problem 1 (Angle Sum Property): The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Solution: Step 1: Let the angles be 3x, 5x, 9x, and 13x. According to the Angle Sum Property of a quadrilateral, the sum of all angles is 360°. Step 2: Set up the equation: 3x + 5x + 9x + 13x = 360°. Step 3: Solve for x: 30x = 360°, which means x = 360 / 30 = 12°. Step 4: Calculate each angle: First angle = 3x = 3 12° = 36° Second angle = 5x = 5 12° = 60° Third angle = 9x = 9 12° = 108° Fourth angle = 13x = 13 12° = 156° * Final Answer: The angles are 36°, 60°, 108°, and 156°.
• Problem 2 (Parallelogram Property): Show that if the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Solution: Step 1: Diagram and Given Information. Consider a quadrilateral ABCD with diagonals AC and BD intersecting at point O. It is given that AO = CO and BO = DO. Step 2: Identify Triangles to Prove Congruent. We need to show that opposite sides are parallel. To do this, let's consider triangles ΔAOB and ΔCOD. Step 3: Prove Congruence using SAS. AO = CO (Given) ∠AOB = ∠COD (Vertically opposite angles are equal) BO = DO (Given) Therefore, by SAS (Side-Angle-Side) congruence rule, ΔAOB ≅ ΔCOD. Step 4: Use CPCTC to Find Parallel Lines. Since the triangles are congruent, their corresponding parts are equal. So, ∠OAB = ∠OCD (by CPCTC). These are alternate interior angles for lines AB and CD with transversal AC. When alternate interior angles are equal, the lines are parallel. Thus, AB || DC. Step 5: Repeat for the Other Pair of Sides. Similarly, we can prove ΔAOD ≅ ΔCOB (using SAS). This gives ∠OAD = ∠OCB (by CPCTC). These are alternate interior angles for lines AD and BC with transversal AC. Thus, AD || BC. Step 6: Conclude. Since both pairs of opposite sides are parallel (AB || DC and AD || BC), the quadrilateral ABCD is a parallelogram.

Exam Tips: Common Mistakes to Avoid in Quadrilaterals

When working on proofs in this chapter, students often make small errors that can cost them marks. Here’s what to watch out for:

1. Unjustified Assumptions: Never assume a figure is a parallelogram, rectangle, or square unless it's given or you have proven it. Always start with the properties of a general quadrilateral.
2. Forgetting Reasons: In a proof, every statement must have a valid reason. Don't forget to write "(Given)", "(Vertically opposite angles)", "(By SAS congruence)", or "(CPCTC)" next to your steps.
3. Confusing Properties and Converses: Knowing that "opposite sides of a parallelogram are equal" is a property. Knowing that "if opposite sides of a quadrilateral are equal, then it is a parallelogram" is its converse. Be clear about which one you are using to prove your point.
4. Incorrect Use of CPCTC: CPCTC (Corresponding Parts of Congruent Triangles are Congruent) can only be used after you have proven that two triangles are congruent. Do not use it as a reason for congruence itself.

Practice Questions with Solutions

• Q: In a parallelogram ABCD, if ∠A = (2x + 25)° and ∠B = (3x - 5)°, find the value of x and all the angles of the parallelogram. A: Step 1: Recall that consecutive angles of a parallelogram are supplementary (add up to 180°). So, ∠A + ∠B = 180°. Step 2: Set up the equation: (2x + 25) + (3x - 5) = 180. Step 3: Solve for x: 5x + 20 = 180 => 5x = 160 => x = 32. Step 4: Calculate the angles. ∠A = 2(32) + 25 = 64 + 25 = 89°. ∠B = 3(32) - 5 = 96 - 5 = 91°. Whoops, let's recheck the property! Consecutive angles of a parallelogram are supplementary. So ∠A + ∠B = 180°. Let's re-calculate: 2x+25+3x-5 = 180 => 5x+20=180 => 5x=160 => x=32. ∠A = 2(32)+25 = 64+25 = 89°. ∠B = 3(32)-5 = 96-5 = 91°. Let's use the property that opposite angles are equal, maybe the question has a typo and means ∠A and ∠C. Let's assume it's consecutive. Then ∠A = 89°, ∠B = 91°. Since opposite angles are equal, ∠C = ∠A = 89° and ∠D = ∠B = 91°. Final answer: x=32, Angles are 89°, 91°, 89°, 91°.
• Q: Prove that if one pair of opposite sides of a quadrilateral is equal and parallel, then it is a parallelogram. A: Step 1: Given a quadrilateral ABCD where AB = CD and AB || CD. Draw a diagonal AC. Step 2: In ΔABC and ΔCDA, we have: AB = CD (Given). Step 3: Since AB || CD and AC is the transversal, ∠BAC = ∠DCA (Alternate Interior Angles). Step 4: AC = CA (Common side). Step 5: By SAS congruence rule, ΔABC ≅ ΔCDA. Step 6: By CPCTC, we get ∠BCA = ∠DAC. These are alternate interior angles for lines BC and AD. Therefore, BC || AD. Step 7: Since we were given AB || CD and we have proved BC || AD, both pairs of opposite sides are parallel. Thus, ABCD is a parallelogram. Final answer: The quadrilateral ABCD is a parallelogram.
• Q: ABCD is a parallelogram and P and Q are mid-points of sides AB and CD respectively. Show that the line segments AQ and PC trisect the diagonal BD. A: Step 1: Given ABCD is a parallelogram. So AB || CD and AB = CD. P is the midpoint of AB, Q is the midpoint of CD. Step 2: Consider quadrilateral APCQ. Since P and Q are midpoints, AP = 1/2 AB and CQ = 1/2 CD. As AB=CD, we have AP = CQ. Also, since AB || CD, we have AP || CQ. A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. So, APCQ is a parallelogram. Step 3: This means AQ || PC. Let the diagonal BD intersect AQ at M and PC at N. Step 4: In ΔDQN, Q is the midpoint of CD and by construction, we need to show M and N are points of trisection. In ΔABM, P is the midpoint of AB, and PN || AM (since PC || AQ). By the converse of the midpoint theorem, N must be the midpoint of BM. So BN=NM. Step 5: In ΔCD N, Q is midpoint of CD and QM || DN. By converse of midpoint theorem M is the midpoint of DN. So DM=MN. From step 4, we have BN=NM. Combining these, we get DM = MN = NB. Thus AQ and PC trisect the diagonal BD. Final answer: The segments AQ and PC divide the diagonal BD into three equal parts.
• Q: The diagonals of a quadrilateral ABCD are perpendicular. Is it necessarily a parallelogram? Justify your answer. A: Step 1: Consider the property. The property for a parallelogram is that its diagonals bisect each other. The question states they are perpendicular. Step 2: Think of a counterexample. A kite is a quadrilateral where diagonals can be perpendicular, but it's not a parallelogram. A rhombus is a parallelogram with perpendicular diagonals, but not all quadrilaterals with perpendicular diagonals are rhombuses. Step 3: Formulate the justification. The condition that diagonals are perpendicular is not sufficient to prove a quadrilateral is a parallelogram. For example, in a kite, the diagonals are perpendicular, but opposite sides are not equal or parallel. Final answer: No, it is not necessarily a parallelogram. For a quadrilateral to be a parallelogram, its diagonals must bisect each other. Perpendicular diagonals are a property of a rhombus or a kite, not a general parallelogram.

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