ML Aggarwal Class 8 Maths Solutions Chapter 13 Understanding Quadrilaterals

Exercise 13(A)

Question 1. Some figures are given below. Classify each of them on the basis of the following: (a) Simple curve (b) Simple closed curve (c) Polygon (d) Convex polygon (e) Concave polygon
Answer: Based on the given figures, the classification is as follows:

(a) Simple curves: Figures (i), (ii), (iii), (v) and (vi). A simple curve is a line that does not intersect itself.

(b) Simple closed curves: Figures (iii), (v) and (vi). These are shapes that are closed by straight line segments or a curved line without crossing themselves.

(c) Polygons: Figures (iii) and (vi). A polygon is a two-dimensional figure made up of straight line segments.

(d) Convex polygon: Figure (iii). In a convex polygon, all diagonals lie completely inside the polygon.

(e) Concave polygon: Figure (vi). In a concave polygon, at least one diagonal passes outside the polygon's boundary.
In simple words: A simple curve doesn't cross itself. A closed curve creates a complete shape. A polygon uses only straight lines. Convex means all diagonals stay inside. Concave means at least one diagonal goes outside.

Exam Tip: Know the key differences - simple curves are just paths, closed curves form shapes, and polygons must have only straight sides. Remember that a convex polygon is bulging outward everywhere, while a concave polygon has at least one inward indent.

Question 2. How many diagonals does each of the following have? (a) A convex quadrilateral (b) A regular hexagon
Answer:
(a) A convex quadrilateral has two diagonals. [Diagram shows quadrilateral ABCD with both diagonals AC and BD drawn.]

(b) A regular hexagon has 9 diagonals as shown. [Diagram shows hexagon TSRQPRU with all diagonals connecting the vertices.]
In simple words: A four-sided shape has 2 lines joining opposite corners. A six-sided shape has 9 such lines.

Exam Tip: For a quadrilateral, remember there are only two ways to join opposite corners. For a hexagon, count carefully - you can use the formula n(n-3)/2 where n is the number of sides: for 6 sides, 6(3)/2 = 9.

Question 3. Find the sum of measures of all interior angles of a polygon with the number of sides: (i) 8 (ii) 12
Answer:
(i) For an 8-sided polygon:
Using the formula: Sum of interior angles = (2n - 4) × 90°, where n = 8
\( = ((2 \times 8) - 4) \times 90° \)
\( = (16 - 4) \times 90° \)
\( = 12 \times 90° \)
\( = 1080° \)

(ii) For a 12-sided polygon:
Using the formula: Sum of interior angles = (2n - 4) × 90°, where n = 12
\( = ((2 \times 12) - 4) \times 90° \)
\( = (24 - 4) \times 90° \)
\( = 20 \times 90° \)
\( = 1800° \)
In simple words: For any polygon, add up all the angles inside using this rule: multiply the number of sides by 2, subtract 4, then multiply by 90 degrees.

Exam Tip: Always use the formula (2n - 4) × 90° for the sum of interior angles. This works for any polygon, whether it has 3 sides or 100 sides.

Question 4. Find the number of sides of a regular polygon whose each exterior angles has a measure of (i) 24° (ii) 60° (iii) 72°
Answer:
(i) For an exterior angle of 24°:
Since the sum of all exterior angles of any polygon is 360°, the number of sides is calculated as:
\( n = \frac{360°}{24°} = 15 \)
Therefore, the polygon has 15 sides.

(ii) For an exterior angle of 60°:
\( n = \frac{360°}{60°} = 6 \)
Therefore, the polygon has 6 sides (hexagon).

(iii) For an exterior angle of 72°:
\( n = \frac{360°}{72°} = 5 \)
Therefore, the polygon has 5 sides (pentagon).
In simple words: To find how many sides a polygon has, divide 360 by the exterior angle. This works because all exterior angles together always add up to 360 degrees.

Exam Tip: Remember that the sum of exterior angles is always 360° for any polygon. Simply divide 360° by each exterior angle to get the number of sides instantly.

Question 5. Find the number of sides of a regular polygon if each of its interior angles is (i) 90° (ii) 108° (iii) 165°
Answer:
(i) For an interior angle of 90°:
Using the interior angle formula: interior angle = \( \frac{(2n - 4)}{n} \times 90° \)
\( 90° = \frac{(2n - 4)}{n} \times 90° \)
\( \frac{90°}{90°} = \frac{(2n - 4)}{n} \)
\( 1 = \frac{(2n - 4)}{n} \)
\( n = 2n - 4 \)
By rearranging: \( 2n - n = 4 \)
\( n = 4 \)
Therefore, the polygon has 4 sides (square).

(ii) For an interior angle of 108°:
\( 108° = \frac{(2n - 4)}{n} \times 90° \)
\( \frac{108°}{90°} = \frac{(2n - 4)}{n} \)
\( \frac{6}{5} = \frac{(2n - 4)}{n} \)
By cross multiplication: \( 5(2n - 4) = 6n \)
\( 10n - 20 = 6n \)
By rearranging: \( 10n - 6n = 20 \)
\( 4n = 20 \)
\( n = 5 \)
Therefore, the polygon has 5 sides (pentagon).

(iii) For an interior angle of 165°:
\( 165° = \frac{(2n - 4)}{n} \times 90° \)
\( \frac{165°}{90°} = \frac{(2n - 4)}{n} \)
\( \frac{11}{6} = \frac{(2n - 4)}{n} \)
By cross multiplication: \( 6(2n - 4) = 11n \)
\( 12n - 24 = 11n \)
By rearranging: \( 12n - 11n = 24 \)
\( n = 24 \)
Therefore, the polygon has 24 sides.
In simple words: Set each interior angle equal to the formula, then solve for n. Larger interior angles mean more sides.

Exam Tip: Use the formula interior angle = \( \frac{(2n - 4)}{n} \times 90° \) and solve step by step. A right angle (90°) gives 4 sides, and angles close to 180° mean many sides.

Question 6. Find the number of sides in a polygon if the sum of its interior angles is: (i) 1260° (ii) 1980° (iii) 3420°
Answer:
(i) For a sum of 1260°:
Using the formula: Sum = (2n - 4) × 90°
\( 1260 = (2n - 4) \times 90° \)
\( \frac{1260}{90} = 2n - 4 \)
\( 14 = 2n - 4 \)
By rearranging: \( 2n = 14 + 4 \)
\( 2n = 18 \)
\( n = 9 \)
Therefore, the polygon has 9 sides.

(ii) For a sum of 1980°:
\( 1980 = (2n - 4) \times 90° \)
\( \frac{1980}{90} = 2n - 4 \)
\( 22 = 2n - 4 \)
By rearranging: \( 2n = 22 + 4 \)
\( 2n = 26 \)
\( n = 13 \)
Therefore, the polygon has 13 sides.

(iii) For a sum of 3420°:
\( 3420 = (2n - 4) \times 90° \)
\( \frac{3420}{90} = 2n - 4 \)
\( 38 = 2n - 4 \)
By rearranging: \( 2n = 38 + 4 \)
\( 2n = 42 \)
\( n = 21 \)
Therefore, the polygon has 21 sides.
In simple words: Divide the sum by 90, add 4, then divide by 2 to get the number of sides.

Exam Tip: Always divide the angle sum by 90° first, then rearrange the formula to isolate n. This approach avoids calculation errors.

Question 7. If the angles of a pentagon are in the ratio 7 : 8 : 11 : 13 : 15, find the angles.
Answer: First, find the sum of all interior angles of the pentagon using the formula (2n - 4) × 90°, where n = 5:
\( = ((2 \times 5) - 4) \times 90° \)
\( = (10 - 4) \times 90° \)
\( = 6 \times 90° \)
\( = 540° \)

Let the angles be 7a, 8a, 11a, 13a, and 15a.
Then: \( 7a + 8a + 11a + 13a + 15a = 540° \)
\( 54a = 540° \)
\( a = 10° \)

Therefore, the five angles are:
\( 7a = 7 \times 10° = 70° \)
\( 8a = 8 \times 10° = 80° \)
\( 11a = 11 \times 10° = 110° \)
\( 13a = 13 \times 10° = 130° \)
\( 15a = 15 \times 10° = 150° \)
In simple words: Add all ratio parts (7 + 8 + 11 + 13 + 15 = 54). Divide the total angle sum (540°) by 54 to get a. Then multiply each ratio part by a.

Exam Tip: When angles are in a ratio, always find the constant multiplier a first by adding all ratio parts and dividing the angle sum by this total. This method works for any polygon with angles in ratio form.

Question 8. The angles of a pentagon are x°, (x - 10)°, (x + 20)°, (2x - 44)° and (2x - 70)°. Calculate x.
Answer: The sum of all interior angles in a pentagon is (2n - 4) × 90°. For n = 5:
\( = ((2 \times 5) - 4) \times 90° \)
\( = 540° \)

Adding all five angles:
\( x + (x - 10) + (x + 20) + (2x - 44) + (2x - 70) = 540° \)
\( x + x - 10 + x + 20 + 2x - 44 + 2x - 70 = 540° \)
\( 7x - 104 = 540° \)
By rearranging: \( 7x = 540° + 104° \)
\( 7x = 644° \)
\( x = 92° \)
In simple words: Combine all the x terms together, combine all the numbers, then solve for x by dividing.

Exam Tip: Always collect like terms (all x terms on one side, all constants on the other) before solving. Double-check by substituting your answer back into all five angle expressions to verify they sum to 540°.

Question 9. The exterior angles of a pentagon are in ratio 1 : 2 : 3 : 4 : 5. Find all the interior angles of the pentagon.
Answer: Since the sum of all exterior angles of any polygon is 360°, let the exterior angles be 1a, 2a, 3a, 4a, and 5a.
\( 1a + 2a + 3a + 4a + 5a = 360° \)
\( 15a = 360° \)
\( a = 24° \)

The exterior angles are:
\( 1a = 1 \times 24° = 24° \)
\( 2a = 2 \times 24° = 48° \)
\( 3a = 3 \times 24° = 72° \)
\( 4a = 4 \times 24° = 96° \)
\( 5a = 5 \times 24° = 120° \)

To find each interior angle, use the fact that interior angle + exterior angle = 180°:
\( 180° - 24° = 156° \)
\( 180° - 48° = 132° \)
\( 180° - 72° = 108° \)
\( 180° - 96° = 84° \)
\( 180° - 120° = 60° \)

Therefore, the five interior angles are: 156°, 132°, 108°, 84°, and 60°.
In simple words: Find each exterior angle by dividing 360° by the ratio sum. Then subtract each exterior angle from 180° to get the interior angle.

Exam Tip: Remember that an interior and exterior angle at the same vertex always add up to 180°. This relationship is the fastest way to convert between exterior and interior angles.

Question 10. In a quadrilateral ABCD, AB || DC. If ∠A : ∠D = 2:3 and ∠B : ∠C = 7 : 8, find the measure of each angle.
Answer: Since AB || DC, angles A and D are co-interior angles (also called consecutive interior angles), so they add up to 180°:
\( \angle A + \angle D = 180° \)

Let ∠A = 2a and ∠D = 3a.
\( 2a + 3a = 180° \)
\( 5a = 180° \)
\( a = 36° \)

Therefore:
\( \angle A = 2a = 2 \times 36° = 72° \)
\( \angle D = 3a = 3 \times 36° = 108° \)

Similarly, angles B and C are co-interior angles, so:
\( \angle B + \angle C = 180° \)

Let ∠B = 7b and ∠C = 8b.
\( 7b + 8b = 180° \)
\( 15b = 180° \)
\( b = 12° \)

Therefore:
\( \angle B = 7b = 7 \times 12° = 84° \)
\( \angle C = 8b = 8 \times 12° = 96° \)
In simple words: When two sides are parallel, angles on the same side of a line cutting through them add to 180°. Use this property along with the given ratios to find each angle.

Exam Tip: Recognize when parallel lines create co-interior angle pairs. These angles always sum to 180°, which gives you a key equation to solve ratio problems in quadrilaterals.

Question 11. From the adjoining figure, find (i) x (ii) ∠DAB (iii) ∠ADB
Answer:
(i) In quadrilateral ABCD, the sum of all interior angles is 360°:
\( \angle A + \angle B + \angle C + \angle D = 360° \)
\( (3x + 4) + (50 + x) + (5x + 8) + (3x + 10) = 360° \)
\( 3x + 4 + 50 + x + 5x + 8 + 3x + 10 = 360° \)
\( 12x + 72 = 360° \)
By rearranging: \( 12x = 360° - 72 \)
\( 12x = 288 \)
\( x = 24 \)

(ii) ∠DAB = (3x + 4)°
\( = (3 \times 24) + 4 \)
\( = 72 + 4 \)
\( = 76° \)

(iii) In triangle ABD, the sum of interior angles equals 180°:
\( \angle DAB + \angle ABD + \angle ADB = 180° \)
\( 76° + 50° + \angle ADB = 180° \)
\( 126° + \angle ADB = 180° \)
\( \angle ADB = 180° - 126° \)
\( \angle ADB = 54° \)
In simple words: First find x using the quadrilateral angle sum. Then find ∠DAB by substituting x. Finally, use triangle angles summing to 180° to find ∠ADB.

Exam Tip: Break complex figures into simpler shapes - use quadrilateral properties first, then triangle properties. Always substitute the found value of x back to get numerical angle measures.

Question 12. Find the angle measure x in the following figures: (i) [Quadrilateral with angles 40°, 100°, 140°, x] (ii) [Pentagon MNOPQ with angles 40°, x, x, 60°, 80°] (iii) [Quadrilateral with angles 60°, 100°, 90°, x]
Answer:
(i) In the quadrilateral, the sum of all angles is 360°:
\( 40° + 100° + 140° + x = 360° \)
\( 280° + x = 360° \)
\( x = 80° \)

(ii) In the pentagon MNOPQ, first find the angles at the base. Since the angles on a straight line sum to 180°:
\( \angle 1 + 60° = 180° \)
\( \angle 1 = 120° \)
\( \angle 2 + 80° = 180° \)
\( \angle 2 = 100° \)

The sum of interior angles in a pentagon is (2 × 5 - 4) × 90° = 540°:
\( 120° + 100° + x + x + 40° = 540° \)
\( 260° + 2x = 540° \)
\( 2x = 280° \)
\( x = 140° \)

(iii) The quadrilateral has angles 60°, 100°, and another angle. Since there's a linear pair with 90°, the fourth interior angle is 90°. Therefore:
\( 60° + 100° + 90° + x = 360° \)
\( 250° + x = 360° \)
\( x = 110° \)
In simple words: For any quadrilateral, all four angles add to 360°. For a pentagon, they add to 540°. When you see a line with angles on both sides, they add to 180°.

Exam Tip: Always identify which polygon you're working with, then use the correct angle sum formula. Watch for angles on a straight line - these always sum to 180° and are your key to finding interior angles of the polygon.

Question 12 (continued). Find the angle measure x in the following figures: (iv) [Quadrilateral with angles 90°, 110°, 83°, and exterior angle x]
Answer: In quadrilateral MNOP, the sum of interior angles is 360°:
\( 90° + y + 83° + 110° = 360° \)
\( 283° + y = 360° \)
\( y = 77° \)

Since x and y are supplementary angles (they form a linear pair on a straight line):
\( y + x = 180° \)
\( 77° + x = 180° \)
\( x = 103° \)
In simple words: First use quadrilateral angle sum to find y. Then use the linear pair property (two angles on a line sum to 180°) to find x.

Exam Tip: When an angle x is outside the polygon on a line, it forms a linear pair with an interior angle. Find the interior angle first using polygon angle sums, then subtract from 180°.

Question 13. (i) In the given figure, find x + y + z.
Answer: In triangle MNO, the sum of interior angles equals 180°:
\( \angle M + \angle N + \angle O = 180° \)
\( \angle M + 70° + 90° = 180° \)
\( \angle M = 20° \)

Since x and 90° form a linear pair:
\( x + 90° = 180° \)
\( x = 90° \)

Since y and 70° form a linear pair:
\( y + 70° = 180° \)
\( y = 110° \)

Since z and 20° form a linear pair:
\( z + 20° = 180° \)
\( z = 160° \)

Therefore:
\( x + y + z = 90° + 110° + 160° = 360° \)
In simple words: Find each interior angle of the triangle. Then use linear pairs (angles on a line) to find x, y, and z. Finally, add all three.

Exam Tip: When exterior angles are marked around a triangle, use linear pairs to relate them to interior angles. The sum of the three exterior angles always equals 360°.

Question 13. (ii) In the given figure, find x + y + z + w
Answer: In quadrilateral MNOP, the sum of interior angles is 360°:
\( \angle M + \angle N + \angle O + \angle P = 360° \)
\( 130° + 80° + 70° + \angle P = 360° \)
\( 280° + \angle P = 360° \)
\( \angle P = 80° \)

Since each of x, y, z, and w forms a linear pair with its corresponding interior angle:
\( x + 130° = 180° \implies x = 50° \)
\( y + 80° = 180° \implies y = 100° \)
\( z + 70° = 180° \implies z = 110° \)
\( w + 80° = 180° \implies w = 100° \)

Therefore:
\( x + y + z + w = 50° + 100° + 110° + 100° = 360° \)
In simple words: Find the missing interior angle of the quadrilateral. Then find each exterior angle by subtracting from 180°. Add all four exterior angles.

Exam Tip: The sum of exterior angles of any polygon is always 360°. This is a quick check - if your exterior angles don't sum to 360°, you've made an error somewhere.

Question 14. A heptagon has three equal angles each of 120° and four equal angles. Find the size of equal angles.
Answer: The sum of interior angles in a heptagon (7-sided polygon) is calculated using the formula (2n - 4) × 90° = (2 × 7 - 4) × 90° = 900°. Since three angles measure 120° each, their total is 360°. The remaining four equal angles must sum to 900° - 360° = 540°. When you divide 540° by 4, you get 135° for each of the remaining angles.
In simple words: Add up all three known angles to get 360°. Take that away from 900° to find what the other four angles must add up to. Then divide by 4 to find each angle's size.

Exam Tip: Always start by finding the total sum of interior angles using (2n - 4) × 90°. Then account for the given angles before solving for unknowns.

Question 15. The ratio between an exterior angle and the interior angle of a regular polygon is 1 : 5. Find (i) the measure of each exterior angle (ii) the measure of each interior angle (iii) the number of sides in the polygon.
Answer:
(i) Let the exterior angle be y and the interior angle be 5y. Since any exterior angle and its adjacent interior angle sum to 180°, we have y + 5y = 180°, which gives 6y = 180°, so y = 30°. Therefore, each exterior angle measures 30°.

(ii) Each interior angle = 5y = 5 × 30° = 150°.

(iii) For a regular polygon, the interior angle formula is \( \frac{(2n-4) \times 90°}{n} \). Setting this equal to 150°: \( \frac{(2n-4) \times 90°}{n} = 150° \). Simplifying: \( \frac{5}{3} = \frac{2n-4}{n} \). Cross-multiplying gives 3(2n - 4) = 5n, which leads to 6n - 12 = 5n. Solving: n = 12. The polygon has 12 sides.
In simple words: Find the exterior angle from the ratio. The interior angle is five times bigger. Use these to find how many sides the shape has.

Exam Tip: Always use the linear pair property (exterior + interior = 180°) as your starting point. The ratio relationship gives you a system you can solve directly.

Question 16. Each interior angle of a regular polygon is double of its exterior angle. Find the number of sides in the polygon.
Answer: Let the exterior angle be y, making the interior angle 2y. Since an interior and exterior angle form a linear pair, y + 2y = 180°, giving 3y = 180° and y = 60°. So the interior angle is 2 × 60° = 120°. Using the interior angle formula for a regular polygon: \( 120° = \frac{(2n-4) \times 90°}{n} \). This simplifies to \( \frac{4}{3} = \frac{2n-4}{n} \). Cross-multiplying: 3(2n - 4) = 4n, which gives 6n - 12 = 4n. Solving yields 2n = 12, so n = 6. The polygon has 6 sides (a regular hexagon).
In simple words: The exterior angle and interior angle add up to 180°. If the interior angle is twice the exterior angle, you can find both. Then count the sides using the angle formula.

Exam Tip: When a relationship like "double" or "half" is given between interior and exterior angles, always express both in terms of a single variable and use the 180° sum rule.

Exercise 13.2

Question 1. In the given figure, ABCD is a parallelogram. Complete each statement along with the definition or property used.
(i) AD = .........
(ii) DC = .........
(iii) ∠DCB = .........
(iv) ∠ADC = .........
(v) ∠DAB = .........
(vi) OC = .........
(vii) OB = .........
(viii) m∠DAB + m∠CDA = .........
Answer:
(i) AD = 6 cm - because in a parallelogram, opposite sides are equal in length
(ii) DC = 9 cm - because opposite sides of a parallelogram are equal
(iii) ∠DCB = 60°
(iv) ∠ADC = ∠ABC = 120°
(v) ∠DAB = ∠DCB = 60°
(vi) OC = AO = 7 cm - the diagonals of a parallelogram bisect each other
(vii) OB = OD = 5 cm - the diagonals bisect each other
(viii) m∠DAB + m∠CDA = 180° - consecutive angles in a parallelogram are supplementary
In simple words: In a parallelogram, opposite sides are equal, opposite angles are equal, and consecutive angles add to 180°. The diagonals cut each other in half.

Exam Tip: Always state the property you use for each answer. Examiners value the reasoning as much as the final value.

Question 2. Consider the following parallelograms. Find the values of x, y, z in each.
Answer:

(i) In parallelogram MNOP, the exterior angle at O is 120°. The angle ∠PON and the exterior angle form a linear pair, so ∠PON + 120° = 180°, giving ∠PON = 60°. Since opposite angles in a parallelogram are equal, ∠M = ∠O = 60°. The angle corresponding to the 120° exterior angle is ∠MNO = 120° (by the linear pair relationship). Since opposite angles are equal, y = 120°. Also, corresponding angles show z = 120°. Therefore, x = 60°, y = 120°, z = 120°.

(ii) From the figure, ∠PQO = 100°, ∠OMN = 30°, ∠PMO = 40°. Using the alternate angle property (since MNOP is a parallelogram, alternate angles where a transversal crosses parallel lines are equal), z = 40°. The angle ∠NMO = 30° by alternate angles. In triangle PQO, the angles sum to 180°: x + 100° + 30° = 180°, so x = 50°. Using the exterior angle theorem, the exterior angle ∠OQP equals y + z, so 100° = y + 40°, giving y = 60°. Therefore, x = 50°, y = 60°, z = 40°.

(iii) In this case, ∠PQO = 100°, ∠OMN = 30°, ∠PMO = 40°. By the alternate angle property, z = 40° and ∠NMO = 30°. In triangle PQO: x + 100° + 30° = 180°, so x = 50°. The exterior angle relationship gives y = 60°. Therefore, x = 50°, y = 60°, z = 40°.

(iv) Given ∠SPR = 35°, ∠PQR = 120°. By alternate angles, ∠PRQ = 35°. In triangle PQR: z + 120° + 35° = 180°, giving z = 25°. Since alternate angles are equal, x = 25°. By the opposite angles property of parallelograms, y = 120°. Therefore, x = 25°, y = 120°, z = 25°.

(v) Given ∠SPR = 67°, ∠PQR = 70°. By alternate angles, ∠PRQ = 67°. In triangle PQR: x + 70° + 67° = 180°, giving x = 43°. Since opposite angles are equal, z = 70°. The exterior angle ∠SRT = ∠PSR + ∠SPR = 70° + 67° = 137°, so y = 137°. Therefore, x = 43°, y = 137°, z = 70°.
In simple words: Use alternate angle rules for parallel lines, properties of triangles, and parallelogram angle rules to find each unknown.

Exam Tip: When a diagonal divides a parallelogram, look for alternate angles and triangle angle sums. Clearly label which property you apply at each step.

Question 3. Two adjacent sides of a parallelogram are in the ratio 5 : 7. If the perimeter of a parallelogram is 72 cm, find the length of its sides.
Answer: In a parallelogram, opposite sides are equal, so the perimeter is 2(SP + RQ) = 72 cm, which means SP + RQ = 36 cm. Let the adjacent sides be 5y and 7y. Then 5y + 7y = 36, so 12y = 36, and y = 3. Therefore, one side is 5 × 3 = 15 cm, and the adjacent side is 7 × 3 = 21 cm. Since opposite sides are equal, the parallelogram has two sides of 15 cm and two sides of 21 cm.
In simple words: In a parallelogram, add up the two different side lengths using the ratio. Use the perimeter to find the scale factor, then multiply to get each side.

Exam Tip: Always express sides in terms of a variable when given a ratio. The perimeter formula for a parallelogram is 2(a + b), not 4s like a square.

Question 4. The measure of two adjacent angles of a parallelogram is in the ratio 4 : 5. Find the measure of each angle of the parallelogram.
Answer: Let the two adjacent angles be 4y and 5y. Since consecutive angles in a parallelogram are supplementary (they add to 180°), we have 4y + 5y = 180°, which gives 9y = 180° and y = 20°. Therefore, the first angle is 4 × 20° = 80° and the second is 5 × 20° = 100°. In a parallelogram, opposite angles are equal, so all four angles are: 80°, 100°, 80°, 100°.
In simple words: Two angles next to each other add up to 180°. Use the ratio to write them as multiples, then solve for the multiplier.

Exam Tip: Remember that consecutive (adjacent) angles are supplementary, while opposite angles are equal. These two facts are key to solving parallelogram angle problems.

Question 5. Can a quadrilateral ABCD be a parallelogram, give reasons in support of your answer.
(i) ∠A + ∠C = 180°?
(ii) AD = BC = 6 cm, AB = 5 cm, DC = 4.5 cm?
(iii) ∠B = 80°, ∠D = 70°?
(iv) ∠B + ∠C = 180°?
Answer:
(i) If ∠A + ∠C = 180°, the quadrilateral may or may not be a parallelogram. This condition alone is not sufficient because in a parallelogram, opposite angles must be equal, not supplementary.

(ii) If AD = BC = 6 cm, AB = 5 cm, DC = 4.5 cm, this cannot form a parallelogram. In a parallelogram, opposite sides must be equal, but here AB = 5 cm ≠ 4.5 cm = DC.

(iii) If ∠B = 80° and ∠D = 70°, this cannot form a parallelogram. Opposite angles in a parallelogram must be equal, but 80° ≠ 70°.

(iv) If ∠B + ∠C = 180°, the quadrilateral may or may not be a parallelogram. This condition means that B and C are consecutive angles summing to 180°, which is a property of parallelograms, but it is not enough by itself to guarantee the shape is a parallelogram.
In simple words: A parallelogram needs opposite sides equal and opposite angles equal. Check each condition carefully against these rules.

Exam Tip: Students often confuse sufficient and necessary conditions. Always test whether a single property alone guarantees a parallelogram or just suggests it might be one.

Question 6. In the following figures, HOPE and ROPE are parallelograms. Find the measures of angles x, y and z. State the properties you use to find them.
Answer:

(i) For parallelogram HOPE:
The exterior angle at O is 70°. Since an exterior angle and its adjacent interior angle form a linear pair, ∠HOP + 70° = 180°, so ∠HOP = 110°. By the opposite angles property of parallelograms, x = ∠HEP = ∠HOP = 110°. By the alternate angles property (where a diagonal acts as a transversal across parallel sides), y = ∠OPH = 40°. In triangle HOP, the sum of interior angles equals 180°: z + 110° + 40° = 180°, so z = 30°. Therefore, x = 110°, y = 40°, z = 30°.

(ii) For parallelogram ROPE:
From the figure, ∠POM = 80° and ∠POE = 60°. By the alternate angles property, y = ∠OPE = ∠POM = 80°. Since angles on a straight line sum to 180°: x + 60° + 80° = 180°, so x = 40°. By the alternate angles property again, z = ∠ROE = x = 40°. Therefore, x = 40°, y = 80°, z = 40°.
In simple words: Use the linear pair rule for exterior angles. Use alternate angle rules where the diagonal crosses parallel sides. Use triangle angle sums to find remaining angles.

Exam Tip: Always name the property (opposite angles, alternate angles, linear pair, triangle angle sum) next to each step. This shows understanding and earns full marks.

Question 7. In the given figure TURN and BURN are parallelograms. Find the measures of x and y (lengths are in cm).
Answer:

(i) For parallelogram TURN:
In a parallelogram, opposite sides are equal. So TU = RN and NT = RU. From TU = RN: 4x + 2 = 28, which gives 4x = 26, so x = 6.5 cm. From NT = RU: 5y - 1 = 24, which gives 5y = 25, so y = 5 cm. Therefore, x = 6.5 cm and y = 5 cm.

(ii) For parallelogram BURN:
The diagonals of a parallelogram bisect each other. So BO = OR and UO = ON. From BO = OR: x + y = 20 (equation i). From UO = ON: x + 3 = 18, which gives x = 15. Substituting into equation (i): 15 + y = 20, so y = 5. Therefore, x = 15 and y = 5.
In simple words: In a parallelogram, opposite sides are equal. The diagonals cut each other exactly in half at the center point.

Exam Tip: Remember two key facts: (1) opposite sides are equal, (2) diagonals bisect each other (cut each other exactly in half). These directly give you equations to solve for unknowns.

Question 8. In the following figure, both ABCD and PQRS are parallelograms. Find the value of x.
Answer: In parallelogram ABCD, since consecutive angles are supplementary: ∠A + ∠B = 180°, so 120° + ∠B = 180°, giving ∠B = 60°. In triangle MPB, the sum of interior angles equals 180°: x + 50° + 60° = 180°, so x + 110° = 180°. Therefore, x = 70°.
In simple words: Find the angle at B using the consecutive angle rule. Then use the triangle angle sum to find x.

Exam Tip: When two parallelograms overlap or share a triangle, always find all angles of one parallelogram first, then apply triangle rules to the shared region.

Question 9. In the given figure, ABCD is a parallelogram and diagonals intersect at O. Find: (i) ∠CAD (ii) ∠ACD (iii) ∠ADC
Answer:
(i) ∠CBD = 46° and ∠BDA = 46° (by the alternate angles property, since AB and DC are parallel). Therefore, ∠CAD = ∠BDA = 46°.

(ii) We need to find ∠ACD. In triangle ACD, we use the fact that ∠AOD = 68°. Since opposite angles at the intersection point are equal, ∠COB = 68° as well. In triangle OCD, ∠BDC = 30° (given), and using angle relationships in the configuration, ∠ACD = 104°. (Alternatively, from the given angle measures and triangle properties, ∠ACD can be determined to be 104°.)

(iii) ∠ADC can be found by summing the parts: ∠ADC = ∠ADB + ∠BDC = 46° + 30° = 76°.
In simple words: Use alternate angle rules for parallel lines cut by diagonals. Break angles into smaller parts using the diagonal, then add them back together.

Exam Tip: When diagonals are drawn in a parallelogram, always use the alternate angles property for the parallel sides and carefully identify which angles can be added or which are equal by symmetry.

Question 10. In the given figure, ABCD is a parallelogram. Perpendiculars DN and BP are drawn on diagonal AC. Prove that:
(i) ∆DCN ≅ ∆BAP
(ii) AN = CP
Answer: Looking at triangles DCN and BAP, we observe that AB and DC are opposite sides of the parallelogram, so they are equal. Both angle N and angle P measure 90 degrees since DN and BP are perpendiculars. Since AB is parallel to DC, angles BAP and DCN are alternate angles and hence equal. By the AAS (Angle-Angle-Side) criterion, triangle DCN is congruent to triangle BAP.

From this congruence, corresponding sides NC and AP are equal. If we subtract the length NP from both sides of this equation, we get NC - NP = AP - NP, which simplifies to AN = CP. Thus both statements are proved.
In simple words: Two triangles are congruent because they share matching angles and equal sides from the parallelogram. When triangles match, their remaining parts are also equal.

Exam Tip: Always identify which sides/angles of the parallelogram you can use as given facts - opposite sides equal and parallel sides create alternate angles. State the congruence criterion clearly (AAS, SAS, SSS) for full marks.

Question 11. In the given figure, ABC is a triangle. Through A, B and C lines are drawn parallel to BC, CA and AB respectively, which forms a ∆PQR. Show that 2(AB + BC + CA) = PQ + QR + RP.
Answer: Since line through A is parallel to BC and line through C is parallel to AB, the quadrilateral ABCR becomes a parallelogram. Thus, AB = CR and CB = AR.

Similarly, since line through A is parallel to BC and line through B is parallel to CA, the quadrilateral ABPC forms a parallelogram. This gives us AB = PC and AC = PB.

Again, with line through B parallel to CA and line through C parallel to AB, the quadrilateral ACBQ is a parallelogram. Therefore, AC = BQ and AQ = BC.

Adding all these six equations together:
AB + AB + BC + BC + AC + AC = PB + PC + CR + AR + BQ + BC

The left side simplifies to 2AB + 2BC + 2AC, and the right side becomes PQ + QR + RP (where PQ = PB + BQ, QR = CR + AR + remaining parts, and RP accounts for remaining segments). Factoring out 2 from the left side, we get 2(AB + BC + AC) = PQ + QR + RP.
In simple words: Three parallelograms are formed by drawing parallel lines. Each original side appears twice when we add all the equations, while the outer triangle's sides contain all the pieces exactly once.

Exam Tip: Identify all three parallelograms first and write down the equal opposite sides for each. The key is to add all equations and then group the right side into the perimeter segments PQ, QR, and RP.

Exercise 13.3

Question 1. Identify all the quadrilaterals that have
(i) four sides of equal length
(ii) four right angles.
Answer: A square and a rhombus both have all four sides of equal length. A square and a rectangle are the quadrilaterals with all four right angles.
In simple words: Equal sides appear in a square and rhombus. Right angles (90 degrees at every corner) are found only in a square and rectangle.

Exam Tip: Remember that a square satisfies both conditions, making it special. A rhombus has equal sides but not right angles, while a rectangle has right angles but not all equal sides.

Question 2. Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle.
Answer: A square qualifies as a quadrilateral because it possesses four sides and four angles whose measures add up to 360 degrees. It meets the definition of a parallelogram since both pairs of opposite sides are equal in length and parallel to each other. A square fulfils the properties of a rhombus because all four of its sides have the same length. Finally, it satisfies the conditions of a rectangle because its opposite sides are both equal and parallel, and each of its interior angles measures exactly 90 degrees.
In simple words: A square is a four-sided shape (quadrilateral) where opposite sides match and run parallel (parallelogram), all four sides are the same length (rhombus), and all four corners are 90 degrees (rectangle).

Exam Tip: Connect each classification to one key property - four sides for quadrilateral, parallel opposite sides for parallelogram, equal sides for rhombus, and 90-degree angles for rectangle.

Question 3. Name the quadrilaterals whose diagonals
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal.
Answer: The diagonals bisect each other in a rectangle, square, rhombus, and parallelogram. The diagonals that serve as perpendicular bisectors of each other occur only in a square and a rhombus. The quadrilaterals having equal diagonals are the square and the rectangle.
In simple words: Four shapes have diagonals that cut each other in half (rectangle, square, rhombus, parallelogram). Two shapes have diagonals that cross at right angles and divide each other equally (square and rhombus). Two shapes have diagonals of the same length (square and rectangle).

Exam Tip: A square has all three diagonal properties together. Memorize which property belongs to which shape - bisecting, perpendicular, and equal are three distinct features.

Question 4. One of the diagonals of a rhombus and its sides are equal. Find the angles of the rhombus.
Answer: When one diagonal of a rhombus equals the length of its sides, the diagonal divides the rhombus into two congruent equilateral triangles. In an equilateral triangle, all three angles measure 60 degrees. Therefore, the angles of the rhombus are 60° and 120°. The two acute angles each measure 60 degrees, and the two obtuse angles each measure 120 degrees.
In simple words: If a diagonal equals the side length, it creates two equilateral triangles. Equilateral triangles have all 60-degree angles, so the rhombus angles are 60° and 120°.

Exam Tip: Sketch the rhombus with the diagonal drawn - you will see the two equilateral triangles immediately, making the 60° and 120° angles obvious.

Question 5. In the given figure, ABCD is a rhombus, find the values of x, y and z
Answer: In a rhombus, the diagonals cut each other at right angles and also bisect each other. Since the diagonals bisect each other, the segment AO equals the segment OC. Therefore, x = 8 cm. Similarly, BO equals OD, so y = 6 cm.

To find z (which represents the side length AB), we apply the Pythagorean theorem to the right triangle AOB. We have AB² = AO² + BO² = 8² + 6² = 64 + 36 = 100. Taking the square root, AB = √100 = 10 cm. Therefore, z = 10 cm.
In simple words: The diagonals split each other in half, so x and y are half-lengths of the diagonals. The side length comes from using the Pythagorean theorem on the right triangle formed by half-diagonals.

Exam Tip: Always apply the bisecting property first to find x and y, then use the right-angle property and Pythagorean theorem for the side length z.

Question 6. In the given figure, ABCD is a trapezium. If ∠A : ∠D = 5 : 7, ∠B = (3x + 11)° and ∠C = (5x - 31)°, then find all the angles of the trapezium.
Answer: Since ABCD is a trapezium with AB parallel to CD, angles B and C are co-interior angles and thus sum to 180 degrees. Setting up the equation (3x + 11) + (5x - 31) = 180, we get 8x - 20 = 180, which gives 8x = 200 and x = 25. Substituting back, ∠B = 3(25) + 11 = 86° and ∠C = 5(25) - 31 = 94°.

For angles A and D, let ∠A = 5y and ∠D = 7y. Since AD is parallel to BC, angles A and D are also co-interior and sum to 180 degrees: 5y + 7y = 180, giving 12y = 180 and y = 15. Therefore, ∠A = 5(15) = 75° and ∠D = 7(15) = 105°.

The four angles are ∠A = 75°, ∠B = 86°, ∠C = 94°, and ∠D = 105°.
In simple words: Co-interior angles in a trapezium add to 180 degrees. Use this fact twice - once for angles B and C with the given expressions, and once for angles A and D with the given ratio.

Exam Tip: Identify which sides are parallel first to know which angle pairs are co-interior. The two conditions give you two separate equations to solve for x and y independently.

Question 7. In the given figure, ABCD is a rectangle. If ∠CEB : ∠ECB = 3 : 2 find
(i) ∠CEB,
(ii) ∠DCF
Answer: In rectangle ABCD, angle B is a right angle (90 degrees). Within triangle BCE, the two acute angles CEB and ECB must sum to 90 degrees. Let ∠CEB = 3y and ∠ECB = 2y. Then 3y + 2y = 90, giving 5y = 90 and y = 18. Therefore, ∠CEB = 3(18) = 54°.

Since AB is parallel to DC, angle CEB and angle ECD are alternate angles and thus equal. So ∠ECD = 54°. Points E, C, and F lie on a straight line, making ∠ECD and ∠DCF supplementary angles that sum to 180 degrees. Therefore, ∠DCF = 180 - 54 = 126°.
In simple words: The ratio 3:2 and the 90-degree angle in the rectangle let you find ∠CEB = 54°. The alternate angles property then helps find ∠ECD, and the straight line gives ∠DCF = 126°.

Exam Tip: Use the right angle at B to set up the ratio equation. Then use parallel sides to bring in alternate angles, and linear pairs for supplementary angles.

Question 8. In the given figure, ABCD is a rectangle and diagonals intersect at O. If ∠AOB = 118°, find
(i) ∠ABO
(ii) ∠ADO
(iii) ∠OCB
Answer: In triangle AOB, sides OA and OB are equal (since diagonals of a rectangle bisect each other, creating equal segments). When two sides of a triangle are equal, the angles opposite them are also equal, so ∠OAB = ∠OBA. Let this angle be y. The sum of angles in triangle AOB is 180 degrees: y + y + 118 = 180, giving 2y = 62 and y = 31. Therefore, ∠ABO = 31°.

Since angles AOB and AOD form a linear pair on the diagonal, ∠AOD = 180 - 118 = 62°. In triangle AOD, since OA = OD, we have ∠ADO = ∠DAO. Let this angle be x. Then 62 + x + x = 180, giving 2x = 118 and x = 59. Therefore, ∠ADO = 59°.

Angles OCB and OAD are alternate angles formed by the parallel sides AD and BC cut by diagonal AC. Therefore, ∠OCB = ∠OAD = 59°.
In simple words: Equal segments from the bisecting diagonals create isosceles triangles. Use the isosceles triangle property to find base angles, then apply linear pairs and alternate angles.

Exam Tip: Remember that in an isosceles triangle, base angles are equal. Use this to set up simple equations rather than treating each angle independently.

Question 9. In the given figure, ABCD is a rhombus and ∠ABD = 50°. Find :
(i) ∠CAB
(ii) ∠BCD
(iii) ∠ADC
Answer: In a rhombus, the diagonals intersect at right angles. Consider triangle AOB, where O is the intersection point and ∠AOB = 90°. Given that ∠ABO = 50°, we can find ∠OAB using the angle sum property: ∠OAB + 90 + 50 = 180, which gives ∠OAB = 40°. Since the diagonal AC passes through O and bisects angle A, ∠CAB = 40°.

In a rhombus, the diagonal bisects the vertex angle. Therefore, ∠BCD = 2 × ∠ACD = 2 × 40 = 80°. Similarly, since the diagonal BD bisects angle D and ∠ABD = ∠BDC (alternate angles in the rhombus), we have ∠ADC = 2 × ∠BDC = 2 × 50 = 100°.
In simple words: The right-angle property of rhombus diagonals helps find one angle in a triangle. Each diagonal bisects the opposite angles, so doubling the small angle gives the full vertex angle.

Exam Tip: Draw and mark the diagonals clearly. The right angle at the center is your starting point. Remember that "bisect" means divide into two equal parts, so multiply by 2 to recover the full angle.

Question 10. In the given isosceles trapezium ABCD, ∠C = 102°. Find all the remaining angles of the trapezium.
Answer: In an isosceles trapezium, the non-parallel sides are equal in length. Since AB is parallel to CD, angles B and C are co-interior angles on the same side of the trapezium and thus sum to 180 degrees. With ∠C = 102°, we get ∠B = 180 - 102 = 78°.

In an isosceles trapezium, the base angles are equal. Since AD = BC (equal non-parallel sides), the angles at the bases must match: ∠A = ∠B = 78°. Using the angle sum property for all angles in a quadrilateral (sum = 360°), we have 78 + 78 + 102 + ∠D = 360, giving ∠D = 360 - 258 = 102°.

The four angles are ∠A = 78°, ∠B = 78°, ∠C = 102°, and ∠D = 102°.
In simple words: Co-interior angles with parallel sides sum to 180 degrees. Equal non-parallel sides mean matching angles at the two bases. The angles appear in two pairs.

Exam Tip: Identify which sides are parallel (the bases) and which are equal (the legs). Use co-interior angles for adjacent angles and the isosceles property for base angles.

Question 11. In the given figure, PQRS is a kite. Find the values of x and y.
Answer: In a kite, one pair of opposite angles are equal. From the given figure, ∠Q = 120° and ∠R = 50°. The kite property states that ∠Q = ∠S, so x = 120°.

The sum of all interior angles in any quadrilateral is 360 degrees. Therefore, ∠P + ∠Q + ∠R + ∠S = 360. Substituting the known values: y + 120 + 50 + 120 = 360, which gives y + 290 = 360 and y = 70°.
In simple words: A kite has two pairs of adjacent equal sides, which means one pair of opposite angles must be equal. Once you know three angles, the fourth follows from the 360-degree rule.

Exam Tip: Identify which angles are opposite in a kite - these should be equal. Use the angle sum rule as the final step once you've applied all kite properties.