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Get Answers to all your Questions
- #Mathematics Textbook for Class IX
- #Areas Of Parallelograms And Triangles
- #NCERT
- #Maths
- #CBSE 9 Class
Q : 8 In Fig. , 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
meets BC at Y. Show that:
(v)
Answers (1)
Since ||gm CYXE and ACE lie on the same base CE and between the same parallels CE and AX.
Therefore, 2. ar(ACE) = ar (||gm CYXE)
But, FCB
ACE (in iv part)
Since the congruent triangle has equal areas. So,
Hence proved
View full answer
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