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ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF (Figure), prove that ar (AER) = ar (AFR)

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Question

ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF (Figure), prove that ar (AER) = ar (AFR)

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Solution


Given: In square ABCD, E and F are the mid-points of BC and CD respectively. Also, R is the mid-point of EF.

To prove: ar (AER) = ar (AFR).

Construction: Draw AN ⊥ EF.

Proof: ar (ΔAER) = `1/2` × Base × Height

= `1/2 xx ER xx AN`   

= `1/2 xx FR xx AN`  ...[∵ R is the mid-point of EF, so ER = FR]

= ar (ΔAFR) 

Hence proved.

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Chapter 9: Areas of Parallelograms & Triangles - Exercise 9.3 [Page 90]

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NCERT Exemplar Mathematics Exemplar [English] Class 9
Chapter 9 Areas of Parallelograms & Triangles
Exercise 9.3 | Q 5. | Page 90

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