Math, asked by Neha5162, 11 months ago

in triangle abc D and e are points on AB and ac such that De parallel to BC ad equal to 2 cm, ab is equal to 6 CM AC equal to 9 cm find ae​

Answers

Answered by panchal444
2

Answer:

f a line is drawn parallel to one side of a triangle to intersect the

other two sides in distinct points, the other two sides are divided in the same ratio.

As DE\parallel BCDE∥BC 

\angle ADE=\angle ABC∠ADE=∠ABC

\angle AED=\angle ACB∠AED=∠ACB

So by AAAAAA \triangle AED\sim \triangle ACB△AED∼△ACB

Hence \dfrac{AE}{AC}=\dfrac{ED}{CB}ACAE=CBED

\therefore\dfrac{3.2}{AC}=\dfrac{2}{5}∴AC3.2=52

\Rightarrow AC=8cm⇒AC=8cm

AC=8cm=AE+EC=3.2+ECAC=8cm=AE+EC=3.2+EC

EC=8-3.2=4.8cmEC=8−3.2=4.8cm

Also \dfrac{AD}{AB}=\dfrac{ED}{CB}ABAD=CBED

\therefore \dfrac{2.4}{AB}=\dfrac{2}{5}∴AB2.4=52

AB=6cm=AD+DB=2.4+DBAB=6cm=AD+DB=2.4+DB

\Rightarrow DB=3.6cm⇒DB=3.6cm

So BD=3.6cmBD=3.6cm and CE=4.8cmCE=4.8cm

Step-by-step explanation:

hope u understand

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