Math, asked by hamzu, 1 year ago

In the figure given below, ABCD is a parallelogram. E is a point on AB. CE intersects the diagonal BD at G and EF is parallel to BC. If AE : EB = 1 : 2 find (i) EF : AD
(ii) area of triangle BEF : area of triangle ABD [3] D C

F

G

A E B

Answers

Answered by rayner
51
(I)EF:AD=2:3
(II)AREA OF TRIANGLE BEF:AREA OF TRIANGLE ABC=4:9
Answered by Jiteshsuvarnak
37

Answer:

(i)EF:AD=2:3

(ii)area of triangle BEF:area of triangle ABD

=4:9

Step-by-step explanation:

(i) In triangle BEF & triangle BAD

Triangle BEF is similar to Triangle BAD

(Basic Proportionality Theorem)

Therefore,

BE/AB=EF/AD=BF/BD

BE/AE+AB=EF/AD

We know that,

AE/EB=1/2

Therefore,

EF/AD=2/2+1

EF:AD=2:3

(ii)ar(Triangle BEF)/ar(Triangle BAD)

=Square of (EF/AD)

=2^2/3^2

=4/9

Therefore,

ar(Triangle BEF):ar(Triangle BAD)

=4:9

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