A B C D is a parallelogram such that ∠ B A D = 120 ∘ and A C bisects ∠ B A D show that A B C D is a rhombus.

Given ∠ B A D = 120 ∘ and A C bisects ∠ B A D ∠ B A C = 1 2 × 120 ∘ = 60 ∘ ∠ 1 = ∠ 2 = 60 ∘ A D ‖ B C and A C is the traversal ∠ 2 = ∠ 4 = 60 ∘ ∆ A B C is isosceles triangle ∵ ∠ 1 = ∠ 4 = 60 ∘ ⇒ A B = B C Parallelogram A B C D is a rhombus.

E M B I B E

Mathematics Standard 9>Chapter 4 - Geometry>Exercise

Tamil Nadu Board Solutions for Chapter: Geometry, Exercise 9: Exercise

Author:Tamil Nadu Board

Tamil Nadu Board Mathematics Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Geometry, Exercise 9: Exercise

Attempt the free practice questions on Chapter 4: Geometry, Exercise 9: Exercise with hints and solutions to strengthen your understanding. Mathematics Standard 9 solutions are prepared by Experienced Embibe Experts.

Questions from Tamil Nadu Board Solutions for Chapter: Geometry, Exercise 9: Exercise with Hints & Solutions

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9th Tamil Nadu Board

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Mathematics Standard 9>Chapter 4 - Geometry>Exercise>Q 6

Two circles of radii $5 cm$ and $3 cm$ intersect at two points and the distance between their centres is $4 cm$ . If the length of the common chord is $k c m$ then find the value of $k$ .