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## Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB)×ar (CPD) = ar (APD)×ar (BPC). [Hint : From A and C, draw perpendiculars to BD.] Q.6

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Please guide me the best way for solving the question of class 9th math of Areas of Parallelograms and Triangles ...Read more

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## In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(vi) ar (FED) = 1/8 ar (AFC) Q.5(6)

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What is the easiest way for solving the question of class 9th ncert math of exercise 9.4 of math of ...Read more

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## In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(v) ar (BFE) = 2 ar (FED) Q.5(5)

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what is the tricky way for solving the question of class 9th ncert math of Areas of Parallelograms and Triangles ...Read more

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## In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(iv) ar (BFE) = ar (AFD) Q.5(4)

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Yesterday i was doing the question from class 9th ncert book of math of Areas of Parallelograms and Triangles chapter ...Read more

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## In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(iii) ar (ABC) = 2 ar (BEC) Q.5(3)

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Please give me the best way for solving the problem of class 9th ncert math of Areas of Parallelograms and ...Read more

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## In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(ii) ar (BDE) = ½ ar (BAE) Q.5(2)

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Please guide me the best way for solving the question of class 9th math of Areas of Parallelograms and Triangles ...Read more

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## In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(i) ar (BDE) =1/4 ar (ABC) Q.5 (1)

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How i solve the question of class 9th ncert math of Areas of Parallelograms and Triangles chapter of exercise 9.4 ...Read more

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## In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [Hint : Join AC.] Q.4

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Hello sir i want to know the best solution of the question from exercise 9.4 of math of Areas of ...Read more

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## In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF). Q.3

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I want to know the best answer of the question from Areas of Parallelograms and Triangles chapter of class 9th ...Read more

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## In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area? Q.2

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What is the best way for solving the question from class 9th ncert math of Areas of Parallelograms and Triangles ...Read more