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Get Answers to all your Questions
- #Mathematics Textbook for Class IX
- #Maths
- #Areas Of Parallelograms And Triangles
- #CBSE 9 Class
- #NCERT
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:
(vi)
Answers (1)
Since ar( ||gm CYXE) = 2. ar(ACE) {in part (v)}...................(i)
Also, FCB and quadrilateral ACFG lie on the same base FC and between the same parallels FC and BG.
Therefore, ar (quad. ACFG) = 2.ar(FCB) ................(ii)
From eq (i) and eq (ii), we get
Hence proved
View full answer
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