ANSWER KEY Class 8 – Ganita Prakash – CHAPTER 4 QUADRILATERALS

 

Class 8 Maths (Ganita Prakash) - Chapter 4: QUADRILATERALS

Study Material & Competency-Based Worksheet

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Answer Key & Explanations

 Multiple Choice Questions

1. (c) 360° - This is a fundamental property of quadrilaterals.

2. (c) Trapezium - Definition of a trapezium.

3. (b) Perpendicular - The bisectors of adjacent angles in a parallelogram are perpendicular because adjacent angles are supplementary.

4. (d) Are equal and perpendicular - A square has the properties of both a rectangle (equal diagonals) and a rhombus (perpendicular diagonals).

5. (b) 32 - Adjacent angles are supplementary: (2x+25)+(3x-5)=180 → 5x+20=180 → 5x=160 → x=32.

6. (d) Square - Only a square satisfies all these conditions.

7. (b) 144° - Let angles be x, 2x, 3x, 4x. Sum = 10x = 360 → x=36. Largest angle = 4×36=144°.

8. (a) 60° and 120° - When a diagonal equals the side, it forms an equilateral triangle, giving 60° angles.

9. (d) Parallelogram - This is a standard result from the Midpoint Theorem.

10. (a) 13 cm - Use Pythagoras theorem: √(12² + 5²) = √(144+25) = √169 = 13 cm.

11. (b) 2 - A kite has two pairs of adjacent equal sides.

12. (c) Rectangle - A parallelogram with one right angle has all right angles.

13. (b) Are equal - This is a defining property of rectangles.

14. (c) Parallelogram - This is a test for a parallelogram.

15. (c) 8 cm - By Midpoint Theorem, DE = ½ BC → BC = 2×DE = 2×4 = 8 cm.

16. (c) Rectangle and Rhombus - A square satisfies all properties of both.

17. (b) 2 - A quadrilateral has 2 diagonals.

18. (a) 70° - In a parallelogram, opposite angles are equal.

19. (c) Diagonals are equal - This is true only for rectangles and squares, not all parallelograms.

20. (b) Square - Only a square has all sides equal and diagonals equal.

 Assertion & Reasoning Questions

1. (a) Both are true and R correctly explains A.

2. (c) A is true but R is false (in a rectangle, both pairs of opposite sides are parallel).

3. (c) A is true but R is false (diagonals of a kite are perpendicular but do not bisect each other).

4. (d) A is false but R is true (diagonals are equal only in rectangles/squares).

5. (a) Both are true and R correctly explains A.

6. (a) Both are true and R correctly explains A.

7. (b) Both are true but R is not the correct explanation (the correct reason is that all angles become 90°).

8. (a) Both are true and R correctly explains A.

9. (d) A is false but R is true (a kite doesn't need all sides equal).

10. (a) Both are true and R correctly explains A.

11. (b) Both are true but R is not the correct explanation for A.

12. (b) Both are true but R is not the correct explanation for A (they are related but not directly explanatory).

13. (d) A is false but R is true.

14. (d) A is false but R is true (a rhombus also has perpendicular diagonals).

15. (a) Both are true and R correctly explains A.

16. (a) Both are true and R correctly explains A.

17. (a) Both are true and R correctly explains A.

18. (d) A is false but R is true.

19. (a) Both are true and R correctly explains A.

20. (a) Both are true and R correctly explains A.

 True/False Questions

1. True - All squares have all properties of rectangles.

2. False - A rhombus may not have all angles equal to 90°.

3. False - Diagonals are equal only in rectangles and squares.

4. False - Rhombus also has perpendicular diagonals.

5. True - The sum of exterior angles of any polygon is 360°.

6. False - They are supplementary (sum to 180°), not complementary (sum to 90°).

7. True - This forms a right-angled trapezium.

8. False - This is true only for a square, not a general rectangle.

9. True - This is a standard result.

10. True - All rectangles have both pairs of opposite sides parallel.

 Short Answer Type I

1. Let angles be 4x and 5x. Since adjacent angles are supplementary: 4x+5x=180 → 9x=180 → x=20. Angles are 80°, 100°, 80°, 100°.

2. Adjacent angles: (2x+15)+(3x-25)=180 → 5x-10=180 → 5x=190 → x=38.

3. Diagonals of a rhombus are perpendicular bisectors of each other, and they bisect the interior angles.

4. In kite ABCD, ΔABC is isosceles with AB=BC. So ∠BCA=∠BAC=(180-50)/2=65°. Similarly, in ΔADC, AD=CD, so ∠DCA=∠DAC=65°. Thus ∠ADC=180-2×65=50°.

5. Proof: In ΔABC, with D and E midpoints of AB and AC, by Basic Proportionality Theorem or using coordinate geometry, DE ∥ BC and DE=½BC.

6. 2x+3x+4x+6x=360 → 15x=360 → x=24. Smallest angle=2×24=48°.

7. No. If all angles were acute (<90°), their sum would be less than 360°, but the angle sum must be exactly 360°.

8. 10 cm. In a rectangle, diagonals are equal.

9. 65°. In an isosceles trapezium, base angles are equal.

10. A square is a quadrilateral with all sides equal and all angles 90°. One unique property: Diagonals are equal, perpendicular, and bisect each other.

11. Parallelogram. This is a standard test.

12. Proof: Let ABCD be a quadrilateral with AC=BD and diagonals bisecting each other. Then it's a parallelogram. In ΔABC and ΔDCB: AB=DC, BC=BC, AC=DB → ΔABC≅ΔDCB (SSS) → ∠ABC=∠DCB. But these are consecutive angles, so each must be 90°. Thus, ABCD is a rectangle.

13. Perimeter=2(a+b)=60. Given a=12, then 2(12+b)=60 → 24+2b=60 → 2b=36 → b=18 cm.

14. By Midpoint Theorem, ST=½QR=½×14=7 cm.

15. Side of rhombus=400/4=100 m. Half of given diagonal=160/2=80 m. Other half-diagonal=√(100²-80²)=√(10000-6400)=√3600=60 m. Other diagonal=120 m. Area=½×d1×d2=½×160×120=9600 m².

 Short Answer Type II

1. Proof: In rectangles ABCD, consider ΔABC and ΔDCB. AB=DC, BC=CB, ∠ABC=∠DCB=90° → ΔABC≅ΔDCB (SAS) → AC=DB.

2. Proof: In rhombus ABCD, AB=BC=CD=DA. Diagonal AC is common to ΔABC and ΔADC. So ΔABC≅ΔADC (SSS) → ∠BAC=∠DAC and ∠BCA=∠DCA. Thus AC bisects ∠A and ∠C.

3. Proof: Join AC. In ΔABC, E and F are midpoints, so EF∥AC and EF=½AC. In ΔADC, G and H are midpoints, so GH∥AC and GH=½AC. Thus EF∥GH and EF=GH, making EFGH a parallelogram.

4. Let ABCD be a parallelogram. Bisectors of ∠A and ∠B meet at O. ∠A+∠B=180° → ½∠A+½∠B=90°. In ΔAOB, ∠OAB+∠OBA=90° → ∠AOB=90°.

5. Let sides be 5x and 4x. Perimeter=2(5x+4x)=18x=90 → x=5. Sides are 25 cm, 20 cm, 25 cm, 20 cm.

6. The bisectors form a quadrilateral with each angle being 90° (since they are bisectors of supplementary angles), hence it is a rectangle.

7. Let ∠A=3x, ∠D=2x. Since AB∥CD, ∠A+∠D=180° → 5x=180 → x=36. So ∠A=108°, ∠D=72°. Similarly, ∠B=4y, ∠C=5y, and ∠B+∠C=180° → 9y=180 → y=20. So ∠B=80°, ∠C=100°.

8. Proof: In parallelogram ABCD, diagonal AC. AB=CD, AD=BC, AC=AC → ΔABC≅ΔCDA (SSS).

9. In kite ABCD, AB=AD=5 cm. BD=8 cm, so OB=OD=4 cm. In ΔAOB, AO=√(AB²-OB²)=√(25-16)=√9=3 cm. So AD=5 cm. Area=½×AC×BD=½×(3+3)×8=½×6×8=24 cm².

10. Proof: Using Midpoint Theorem, DE∥AB, EF∥BC, FD∥AC, and DE=½AB, EF=½BC, FD=½AC. Also, the four triangles are formed by three segments joining midpoints, creating four congruent triangles by SSS congruence.

 Long Answer Questions

1. Construction steps: Draw AC=7 cm. With A as center, radius 4.5 cm, draw arc. With C as center, radius 4 cm, draw arc to intersect first arc at D. Similarly, with A as center, radius 6 cm, and with C as center, radius 5.5 cm, draw arcs to get B. Join all points. BD should measure approximately 6.5-7 cm.

2. Proof: Diagonals bisect each other → parallelogram. Diagonals equal → rectangle. Diagonals perpendicular → rhombus. A figure that is both a rectangle and a rhombus is a square.

3. Proof: (i) AC bisects ∠A and ∠C. In parallelogram, ∠A=∠C. So ∠DAC=∠DCA → AD=CD. Thus adjacent sides equal → rhombus. But also a rectangle → square. (ii) In a square, diagonals bisect angles.

4. Midpoint Theorem. Proof: In quadrilateral ABCD, join A to midpoint M of CD. The line AM divides the quadrilateral into two parts of equal area because triangles AMD and AMC have equal bases (DM=MC) and same height.

5. Proof similar to Short Answer II Q4, but generalized for any two consecutive angles.

6. Diagonals: d1=16 cm, d2=30 cm. Side=√((d1/2)²+(d2/2)²)=√(8²+15²)=√(64+225)=√289=17 cm. Perimeter=4×17=68 cm. Area=½×16×30=240 cm².

7. Explanation: The sum of angles around any point in the pattern is 360°. Parallelograms have opposite angles equal, rhombuses have adjustable angles. By arranging them properly, we can ensure no gaps.

8. Proof: Draw diagonal SQ. In ΔPSQ, M is midpoint of PS and MN∥PQ (given). By converse of Midpoint Theorem, N is midpoint of SQ. Similarly, in ΔSQR, N is midpoint of SQ and NO∥SR → O is midpoint of QR.

9. Proof: (i) In ΔABC, M is midpoint of AB, MD∥BC → by converse of Midpoint Theorem, D is midpoint of AC. (ii) MD∥BC and BC⊥AC → MD⊥AC. (iii) In right ΔABC, midpoint of hypotenuse is equidistant from vertices → CM=MA=½AB.

10. Construction: Draw AC=6 cm. Find midpoint O. Draw line through O making 60° with AC. Mark OB=OD=4 cm. Join A,B,C,D. Sides will be approximately 5 cm each.

 Case-Based Questions

Case 1: The Playground

1. (c) Parallelogram

2. (b) Rhombus

3. (a) Rectangle

4. (c) Midpoint Theorem

Case 2: The Kite Festival

1. (b) Perpendicular

2. (a) 120 cm² (Area=½×24×10=120 cm²)

3. (b) 55° (In ΔAOB, ∠BAO=180-90-35=55°)

4. (b) BD

Case 3: The Picture Frame

1. (b) Diagonals are equal

2. (a) 126 cm (Width=√(45²-36²)=√(2025-1296)=√729=27 cm; Perimeter=2(36+27)=126 cm)

3. (c) They become unequal

4. (d) Both (b) and (c)

Case 4: The Garden Plot

1. (b) Two congruent triangles

2. (a) Right-angled triangle (15²+20²=225+400=625=25²)

3. (a) The diagonals

4. (b) SAS

Case 5: The Tile Design

1. (c) Measure of angles

2. (a) 96 cm² (Area=½×16×12=96 cm²)

3. (a) The sum of angles around a point is 360°

4. (b) Have gaps (Rectangles cannot tessellate with squares in the same pattern without gaps)