## In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that:(vii) ar(BCED) = ar(ABMN)+ar(ACFG) Q.8(7)

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Give me the best and simple way for solving the question of class 9th  of exercise 9.4 of Areas of ...Read more

## In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that:(vi) ar(CYXE) = ar(ACFG) Q.8(6)

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What is the simplest way for solving the question of class 9th of exercise 9.4 of math of question no.8(6) ...Read more

## In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that:(v) ar(CYXE) = 2ar(FCB) Q.8(5)

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How can i solve this tough question of class 9th ncert of Areas of Parallelograms and Triangles of math of ...Read more

## In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that: Ncert solutions class 9 chapter 9-38 (iv) ΔFCB ≅ ΔACE Q.8(4)

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Give me the best way for solving the question of class 9th of Areas of Parallelograms and Triangles of exercise ...Read more

## In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that:(iii) ar(BYXD) = ar(ABMN) Q.8(3)

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How i find the best tricky way for solving the question of class 9th of Areas of Parallelograms and Triangles ...Read more

## In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that: Ncert solutions class 9 chapter 9-38 (ii) ar(BYXD) = 2ar(MBC) Q.8(2)

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

## In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that:(i) ΔMBC ≅ ΔABD Q.8(1)

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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

## P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that:(iii) ar (PBQ) = ar (ARC) Q.7(3)

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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

## P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that:(ii) ar (RQC) = (3/8) ar (ABC) Q.7(2)

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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

## P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that: (i) ar (PRQ) = ½ ar (ARC) Q.7(1)

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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