1 A 1 B 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A1 15 A2 15 A3 15 A4 15 A5 15 B 15 C 16 A 16 B 16 C 16 D 16 E

West Bengal Board - Mathematics

22. Pythagoras Theorem

1. Quadratic Equations in One Variable
2. Simple Interest
3. Theorems Related to Circle
4. Rectangular Parallelopiped or Cuboid
5. Ratio and Proportion
6. Compound Interest and Uniform Rate of Increase or Decrease
7. Theorems Related to Angles in a Circle
8. Right Circular Cylinder
9. Quadratic Surd
10. Theorems Related to Cyclic Quadrilateral
11. Construction: Construction of Circumcircle and Incircle of a Triangle
12. Sphere
13. Variation
14. Partnership Business
15. Theorems Related to Tangent to a Circle
16. Right Circular Cone
17. Construction: Construction of Tangent to a Circle
18. Similarity
19. Real Life Problems Related to Different Solid Objects
20. Trigonometry: Concept of Measurement of Angle
21. Construction: Determination of Mean Proportional
22. Pythagoras Theorem
23. Trigonometric Ratios and Trigonometric Identities
24. Trigonometric Ratios of Complementary Angle
25. Application of Trigonometric Ratios: Heights and Distances
26. Statistics: Mean, Median, Ogive, Mode

Let us Work Out 22

1 A 1 B 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A1 15 A2 15 A3 15 A4 15 A5 15 B 15 C 16 A 16 B 16 C 16 D 16 E

Q. 11

ABC is and isosceles triangle whose ∠C is right angle. If D is mid point of AB, then let us prove that, AD² + DB² = 2CD²

Class 10th West Bengal Board - Mathematics 22. Pythagoras Theorem

Answer :

Given: ABC is an isosceles triangle. So, AC = BC

∠C = right angle

To prove: AD² + DB² = 2CD²

Proof:
By Pythagoras Theroem,

In ΔABC, We have

AB² = AC² + BC² [1]

In ΔADC and ΔADB, we have

AC = BC [Given]

CD = CD [Common]

AD = DB [D is the mid point]

⇒ ΔADC ≅ ΔADB [By SSS Congruency Criterion]

⇒ ∠ADC = ∠ADB [Corresponding parts of congruent triangles are equal]

Also, ∠ADC + ∠ADB = 180° [Linear Pair]

⇒ ∠ADC + ∠ADC = 180°

⇒ ∠ADC = ∠ADB = 90°

Hence, ADC and ADB are right triangles, ∴ By Pythagoras theorem

AC² = CD² + AD² [2]

BC² = CD² + BD² [3]

Adding [2] and [3], we have

AC² + BC² = AD² + BD² + 2CD²

⇒ AB² = AD² + BD² + 2CD²

⇒ (2AD)² = AD² + AD² + 2CD²

[∵ D is mid-point of AB, AD = BD and AB = 2AD = 2BD]

⇒ 4AD² = 2AD² + 2CD²

⇒ 2AD² = 2CD²

⇒ AD² + BD² = 2CD² [∵ AD = BD]

Hence, Proved!

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