Welcome dear students! Today we are going to learn about Understanding Quadrilaterals from Class 8 Maths.
You know that the paper is a model for a plane surface. When you join a number of points without lifting a pencil from the paper, and without retracing any portion of the drawing other than single points, you get a plane curve.
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Let us look at convex and concave polygons. A simple closed curve made up of only line segments is called a polygon. Some curves are polygons, and some are not. We can see convex polygons and concave polygons. Can you find how these types of polygons differ from one another? Polygons that are convex have no portions of their diagonals in their exteriors, or any line segment joining any two different points in the interior of the polygon lies wholly in the interior of it. Is this true with concave polygons? Study the figures given. Then try to describe in your own words what we mean by a convex polygon and what we mean by a concave polygon. Give two rough sketches of each kind. In our work in this class, we will be dealing with convex polygons only.
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Now, regular and irregular polygons. A regular polygon is both equiangular and equilateral. For example, a square has sides of equal length and angles of equal measure. Hence it is a regular polygon. A rectangle is equiangular but not equilateral. Is a rectangle a regular polygon? No. Is an equilateral triangle a regular polygon? Yes, because it is both equiangular and equilateral. In the previous classes, have you come across any quadrilateral that is equilateral but not equiangular? Recall the quadrilateral shapes you saw in earlier classes like Rectangle, Square, Rhombus. Is there a triangle that is equilateral but not equiangular? No.
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Let us solve Exercise 3.1. Question 1 asks to classify eight given figures on five criteria. Classifying each on all five criteria: Figure 1 is a simple curve only. Figure 2 is a simple curve and simple closed curve. Figure 3 is a simple curve, simple closed curve, polygon, and convex polygon. Figure 4 is a simple curve, simple closed curve, polygon, and convex polygon. Figure 5 is a simple curve, simple closed curve, polygon, and concave polygon. Figure 6 is a simple curve and simple closed curve. Figure 7 is a simple curve only. Figure 8 is not a simple curve. Question 2 asks what a regular polygon is. A regular polygon is both equiangular and equilateral. The name of a regular polygon of 3 sides is an equilateral triangle. For 4 sides, it is a square. For 6 sides, it is a regular hexagon.
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Moving to section 3.2, Sum of the Measures of the Exterior Angles of a Polygon. On many occasions a knowledge of exterior angles may throw light on the nature of interior angles and sides. Let us do an activity. Draw a polygon on the floor, using a piece of chalk. In the figure, a pentagon ABCDE is shown. We want to know the total measure of angles, that is, m∠1 + m∠2 + m∠3 + m∠4 + m∠5. Start at A. Walk along AB. On reaching B, you need to turn through an angle of m∠1, to walk along BC. When you reach at C, you need to turn through an angle of m∠2 to walk along CD. You continue to move in this manner, until you return to side AB. You would have in fact made one complete turn. Therefore, m∠1 + m∠2 + m∠3 + m∠4 + m∠5 = 360°. This is true whatever be the number of sides of the polygon. Therefore, the sum of the measures of the external angles of any polygon is 360°.
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Example 1: Find measure x in the given quadrilateral figure. The exterior angles are given as 50°, 90°, 110°, and x. Solution: x + 90° + 50° + 110° = 360°. Why? Because the sum of exterior angles of any polygon is 360°. x + 250° = 360°. x = 110°. Let us do the Try These activity. Take a regular hexagon. Question 1: What is the sum of the measures of its exterior angles x, y, z, p, q, r? The sum is 360°. Question 2: Is x = y = z = p = q = r? Why? Yes, because it is a regular polygon, so all exterior angles are equal. Question 3: What is the measure of each? Exterior angle is 360°/6 = 60°. Interior angle is 180° - 60° = 120°. Question 4: Repeat for a regular octagon and a regular 20-gon. For octagon, exterior is 360°/8 = 45°. Interior is 135°. For 20-gon, exterior is 360°/20 = 18°. Interior is 162°.
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Example 2: Find the number of sides of a regular polygon whose each exterior angle has a measure of 45°. Solution: Total measure of all exterior angles = 360°. Measure of each exterior angle = 45°. Therefore, the number of exterior angles = 360°/45 = 8. The polygon has 8 sides. Now Exercise 3.2. Question 1: Find x. Figure a is a triangle with exterior angles 125° and 125°. Sum of exterior angles is 360°. So 125° + 125° + x = 360°. 250° + x = 360°. x = 110°. Figure b is a quadrilateral with interior angles 70°, a right angle, 60°, and x°. Sum of interior angles is 360°. 70° + 90° + 60° + x = 360°. 220° + x = 360°. x = 140°.
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Question 2: Measure of each exterior angle of regular polygon. (i) 9 sides: 360°/9 = 40°. (ii) 15 sides: 360°/15 = 24°. Question 3: Exterior angle is 24°. Number of sides = 360°/24 = 15. Question 4: Interior angle is 165°. Exterior angle = 180° - 165° = 15°. Number of sides = 360°/15 = 24. Question 5: (a) Is it possible to have exterior angle 22°? 360°/22 is not an integer, so no. (b) Can 22° be an interior angle? Exterior would be 158°. 360°/158 is not an integer, so no. Question 6: (a) Minimum interior angle possible for a regular polygon is 60°, which occurs in an equilateral triangle. (b) For a regular polygon, the maximum exterior angle possible is 120°. The exterior angle equals 360°/n. As n increases, the exterior angle decreases. The minimum number of sides is 3. Therefore, the maximum exterior angle occurs for a triangle, which is 360°/3 = 120°.
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Section 3.3: Kinds of Quadrilaterals. Based on the nature of the sides or angles of a quadrilateral, it gets special names. 3.3.1 Trapezium. A trapezium is a quadrilateral with a pair of parallel sides. The arrow marks in the figures indicate parallel lines. Some figures have exactly one pair of parallel sides, so they are trapeziums. Others have no parallel sides or two pairs, so they are not trapeziums. Let us do an activity. Take identical cut-outs of congruent triangles of sides 3 cm, 4 cm, 5 cm. Arrange them to form a trapezium ABCD. The parallel sides are AB and DC. The non-parallel sides are AD and BC, which are equal in this case. You can get two more trapeziums using the same triangles. If the non-parallel sides of a trapezium are of equal length, we call it an isosceles trapezium.
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3.3.2 Kite. A kite is a special type of a quadrilateral. The sides with the same markings are equal. For example, AB = AD and BC = CD. A kite has 4 sides. There are exactly two distinct consecutive pairs of sides of equal length. Check whether a square is a kite. A square has two pairs of equal consecutive sides, so it meets the basic criteria, but it is typically studied as a special rectangle and rhombus. You can explore this further by comparing their diagonal and angle properties. Let us do an activity. Take a thick white sheet. Fold the paper once. Draw two line segments of different lengths. Cut along them and open up. You get a kite. It has line symmetry along one diagonal. The diagonals are perpendicular to one another. One diagonal bisects the other. One pair of opposite angles are equal. In the standard kite figure, ∠B = ∠D. We can show that ΔABC ≅ ΔADC by SSS criterion. This implies corresponding angles are equal.
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3.3.3 Parallelogram. A parallelogram is a quadrilateral. As the name suggests, it has something to do with parallel lines. In the figures, AB || DC, and AD || BC. These are parallelograms. Others are not because they lack two pairs of parallel sides. A rectangle is also a parallelogram. Let us do an activity. Take two different rectangular cardboard strips. Place one horizontally, draw lines. Place the other slant, draw lines. The four lines enclose a quadrilateral made of two pairs of parallel lines. It is a parallelogram. A parallelogram is a quadrilateral whose opposite sides are parallel.
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3.3.4 Elements of a parallelogram. There are four sides and four angles. In parallelogram ABCD, AB and DC are opposite sides. AD and BC are opposite sides. ∠A and ∠C are opposite angles. ∠B and ∠D are opposite angles. AB and BC are adjacent sides. ∠A and ∠B are adjacent angles. Let us do an activity. Take two identical parallelograms. Place one over the other. You find that opposite sides coincide. Property: The opposite sides of a parallelogram are of equal length. We can prove this logically. Draw diagonal AC. In ΔABC and ΔADC, ∠1 = ∠2, and ∠3 = ∠4, because they are alternate interior angles. AC is common. By ASA congruency condition, ΔABC ≅ ΔCDA. This gives AB = DC and BC = AD.
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Example 3: Find the perimeter of parallelogram PQRS. In a parallelogram, the opposite sides have same length. Therefore, PQ = SR = 12 cm and QR = PS = 7 cm. So, Perimeter = PQ + QR + RS + SP = 12 cm + 7 cm + 12 cm + 7 cm = 38 cm. 3.3.5 Angles of a parallelogram. Let us do an activity. Copy parallelogram ABCD on tracing paper. Pin at diagonal intersection. Rotate 180°. The parallelograms coincide. ∠A lies on ∠C, and ∠B on ∠D. Property: The opposite angles of a parallelogram are of equal measure. We can prove this. Diagonals AC and BD intersect. ∠1 = ∠2, ∠3 = ∠4. By ASA, ΔABC ≅ ΔCDA. This shows ∠B = ∠D. Similarly, ∠A = ∠C.
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Example 4: In parallelogram BEST, find x, y, z. S is opposite to B. So, x = 100° (opposite angles property). y = 100°. Since side TS is parallel to side BE in parallelogram BEST, and line TE acts as a transversal, y and ∠x are corresponding angles. Therefore, y = 100°. z = 80° (since ∠y and ∠z form a linear pair, 180° - 100° = 80°). Now, adjacent angles. In parallelogram ABCD, ∠A and ∠D are supplementary since DC || AB and DA is a transversal. ∠A and ∠B are also supplementary. Property: The adjacent angles in a parallelogram are supplementary. Example 5: In parallelogram RING, if m∠R = 70°, find all other angles. Given m∠R = 70°. Then m∠N = 70° because ∠R and ∠N are opposite angles. Since ∠R and ∠I are supplementary, m∠I = 180° - 70° = 110°. Also, m∠G = 110° since ∠G is opposite to ∠I. Thus, m∠R = m∠N = 70° and m∠I = m∠G = 110°.
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3.3.6 Diagonals of a parallelogram. The diagonals are generally not equal. Let us do an activity. Take parallelogram ABCD. Diagonals AC and DB meet at O. Fold to find midpoint of AC. It coincides with O. So diagonal DB bisects AC. Repeat for DB. It bisects at O. Property: The diagonals of a parallelogram bisect each other. We can justify this. In ΔAOB and ΔCOD, ∠1 = ∠2, ∠3 = ∠4, and AB = CD. By ASA, ΔAOB ≅ ΔCOD. This gives AO = CO and BO = DO. Example 6: In parallelogram HELP, OE = 4 and HL is 5 more than PE. Find OH. If OE = 4 then OP also is 4 because diagonals bisect each other. So PE = 8. Therefore HL = 8 + 5 = 13. Hence OH = 13/2 = 6.5 cm.
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Exercise 3.3. Question 1: In parallelogram ABCD, (i) AD = BC (Opposite sides equal). (ii) ∠DCB = ∠DAB (Opposite angles equal). (iii) OC = OA (Diagonals bisect each other). (iv) m∠DAB + m∠CDA = 180° (Adjacent angles supplementary). Question 2: Find unknowns. (i) x = 100° (opposite to B), y = 80° (supplementary to B), z = 80° (opposite to y). (ii) x = 130° (supplementary to 50°), y = 50° (opposite to 50°), z = 130° (vertically opposite to x). (iii) In the rhombus diagram, diagonals intersect at 90°. Given an angle of 30°. Since opposite sides are parallel, y = 30° (alternate interior angles). In the right triangle formed by the diagonals, x = 180° - 90° - 30° = 60°. Since opposite sides are parallel, z = 60° (alternate interior to x). (iv) y = 100° (opposite to 80°), x = 80° (given), z = 100° (vertically opposite to y). (v) y = 40°. The angle adjacent to 112° is 180° - 112° = 68°. In the triangle, 40° + z + 68° = 180°. z = 180° - 108° = 72°. Since opposite sides are parallel, x = z = 72°.
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Question 3: Can ABCD be a parallelogram? (i) Yes, it can be a parallelogram. If it is a parallelogram, opposite angles are equal, so ∠D = ∠B. Given ∠D + ∠B = 180°, we have 2∠D = 180°, so ∠D = ∠B = 90°. This makes it a rectangle, which is a type of parallelogram. (ii) No, because AB = DC but AD ≠ BC. (iii) No, because in a parallelogram opposite angles must be equal, but here ∠A = 70° is not equal to ∠C = 65°. Question 4: A kite, specifically a non-rhombus kite, has one pair of opposite angles equal but is not a parallelogram. Question 5: Adjacent angles ratio 3:2. Let angles be 3x and 2x. 3x + 2x = 180°. 5x = 180°. x = 36°. Angles are 3x = 108° and 2x = 72°. The four angles are 108°, 72°, 108°, 72°. Question 6: Adjacent angles equal. x + x = 180°. 2x = 180°. x = 90°. All angles are 90°. Question 7: In HOPE, exterior angle at O is 70°, so interior ∠HOP = 110°. Opposite angle x = 110°. In ΔHOP, y + 40° + 110° = 180°. y = 30°. z = 40° (alternate interior angles). Question 8: (i) GS = NU, so 3x = 18, x = 6. GU = SN, so 3y - 1 = 26, 3y = 27, y = 9. (ii) Diagonals bisect. y + 7 = 20, so y = 13. x + y = 16, so x + 13 = 16, x = 3. Question 9: In parallelogram RISK, ∠K = 120°. Since adjacent angles are supplementary, ∠R = 180° - 120° = 60°. In parallelogram CLUE, ∠L = 70°. Since adjacent angles are supplementary, ∠E = 180° - 70° = 110°. The intersecting lines form a triangle with interior angles x, 60°, and 110°. By the angle sum property, x + 60° + 110° = 180°. x + 170° = 180°, giving x = 10°.
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Question 10: In figure KLMN, ∠NML + ∠KNM = 180°. Since interior angles on same side of transversal are supplementary, KL is parallel to NM. Hence it is a trapezium. Question 11: AB || DC. ∠B = 120°. ∠C = 180° - 120° = 60°. Question 12: SP || RQ. ∠R = 130°. ∠P = 130° (opposite angles). ∠S = 180° - 130° = 50° (adjacent supplementary). Section 3.4: Some Special Parallelograms. 3.4.1 Rhombus. A rhombus is obtained as a special case of a kite. If you draw AB = BC in the kite activity, you get a rhombus. Note that sides of rhombus are all of same length. A rhombus is a quadrilateral with sides of equal length. Since opposite sides are equal, it is also a parallelogram. It has all properties of a parallelogram and a kite. The most useful property is of its diagonals. Property: The diagonals of a rhombus are perpendicular bisectors of one another. We can justify this. In rhombus ABCD, OA = OC and OB = OD. By SSS, ΔAOD ≅ ΔCOD. So ∠AOD = ∠COD. They form a linear pair, so each is 90°.
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Example 7: RICE is a rhombus. Find x, y, z. x = OE = 5 (diagonals bisect). y = OR = 12 (diagonals bisect). z = side of rhombus = 13 (all sides of a rhombus are equal). 3.4.2 A rectangle. A rectangle is a parallelogram with equal angles. If each angle is x°, then 4x° = 360°, so x° = 90°. Each angle is a right angle. A rectangle is a parallelogram in which every angle is a right angle. It has opposite sides equal and diagonals bisect each other. Surprisingly, the diagonals are of equal length. Property: The diagonals of a rectangle are of equal length. We justify this. In rectangle ABCD, ΔABC ≅ ΔABD by SAS. AB is common, BC = AD, ∠A = ∠B = 90°. So AC = BD. Example 8: RENT is a rectangle. Diagonals meet at O. OR = 2x + 4, OT = 3x + 1. Find x. Diagonals are equal, so halves are equal. 3x + 1 = 2x + 4. x = 3.
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3.4.3 A square. A square is a rectangle with equal sides. It has all properties of a rectangle plus equal sides. Diagonals are equal. In a square, diagonals bisect one another, are equal, and are perpendicular. Property: The diagonals of a square are perpendicular bisectors of each other. We justify this. In square ABCD, OA = OC. By SSS, ΔAOD ≅ ΔCOD. So ∠AOD = ∠COD. They are a linear pair, so each is 90°. Exercise 3.4. Question 1: (a) False. (b) True. (c) True. (d) False. (e) False. (f) True. (g) True. (h) True. Question 2: (a) Rhombus and square. (b) Rectangle and square. Question 3: A square is (i) a quadrilateral because it has four sides. (ii) a parallelogram because opposite sides are parallel. (iii) a rhombus because all sides are equal. (iv) a rectangle because all angles are 90°. Question 4: (i) Parallelogram, rhombus, rectangle, square. (ii) Rhombus, square. (iii) Rectangle, square. Question 5: A rectangle is convex because all its diagonals lie completely inside it. Question 6: In right triangle ABC, O is midpoint of hypotenuse. Draw rectangle ABCD. Diagonals AC and BD are equal and bisect each other at O. So OA = OB = OC. O is equidistant from A, B, C.
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Think, Discuss and Write. Question 1: A mason can check right angles with a set square, check opposite sides are equal, and check diagonals are equal. Question 2: Yes, a rhombus with equal angles has all angles 90°, making it a square. Question 3: A trapezium with all angles equal to 90° would have both pairs of opposite sides parallel, making it a rectangle. The textbook defines a trapezium as having a pair of parallel sides and distinguishes it from figures with two pairs, so it would be classified as a rectangle, not a trapezium. What have we discussed? Let us summarize. A parallelogram is a quadrilateral with each pair of opposite sides parallel. Its properties are: opposite sides are equal, opposite angles are equal, and diagonals bisect one another. A rhombus is a parallelogram with sides of equal length. Its properties are: all properties of a parallelogram, and diagonals are perpendicular to each other. A rectangle is a parallelogram with a right angle. Its properties are: all properties of a parallelogram, each angle is a right angle, and diagonals are equal. A square is a rectangle with sides of equal length. It has all properties of a parallelogram, rhombus, and rectangle. A kite is a quadrilateral with exactly two pairs of equal consecutive sides. Its properties are: diagonals are perpendicular, one diagonal bisects the other, and one pair of opposite angles are equal.
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Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]