Question

Heres my question.....✅


❤CLASS 10☄

PLS EXPLAIN IT BY THE HELP OF BASIC PROPORTIONALITY THEOREM...✔


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Answer 1

Answer:

Length of BD is 3.6 cm and EC is 4.8 cm.

Step-by-step explanation:

Given,

Lengths of AD, AE, DE, BC are 2.4 cm, 3.2 cm, 2 cm, 5 cm respectively.

Using Basic Proportionality Theorem : In a ∆PQR, having XY intersecting lines PQ and PR and parallel to QR. Then

$\implies\boxed{\dfrac{PX}{XQ}=\dfrac{PY}{YR}}$

Here,

= > AD / DB = AE / EC

= > DB / AD = EC / AE

= > DB / AD + 1 = EC / AE + 1

= > ( DB + AD ) / AD = ( EC + AE ) / AE

= > AB / AD = AC / AE { DB + AD = AB ; EC + AE = AC, given }

We know, given triangles ∆ABC and ADE are similar, by AA.

So,

= > AB / AD = AC / AE = BC / DE

= > ( DB+ AD ) / AD = ( EC + AE ) / AE = 5 / 2

= > BD / AD + 1 = EC / AE + 1 = 5 / 2

= > BD / AD = ( 5 / 2 - 1 ) And EC / AE = ( 5 / 2 - 1 )

= > BD / 2.4 = ( 5 - 2 ) / 2 And EC / 3.2 cm = ( 5 - 2 ) / 2

= > BD = ( 3 / 2 ) ( 2.4 ) cm And EC = ( 3 / 2 ) ( 3.2 ) cm

= > BD = 3.6 cm and EC = 4.8 cm

Hence,

length of BD is 3.6 cm and EC is 4.8 cm.

Answer 2

Given :- In  ∆ ABC, AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm.

To Find :- DB = ? and EC = ?

Solution :-

In  ∆ ABC,

By Basic Proportionality theorem ( BPT) Theorem, We get

AD/ DB = AE/ EC

⇒ DB/ AB + 1 = EC/ AE +1

AD + DB / AD = AE + EC / AE

Therefore, AB / AD = AC / AE

In ∆ ABC and ∆ ADE

Angle ADE = Angle DBC

Angle AED = Angle ECB [ Corresponding angles ]

Therefore, ∆ ABC is similar to triangle ∆ADE by AA similarity.

AB/ AD = AC/ AE = BC/ DE

DB + AD/ AD = EC + AE / AE = 5/2

DB / AD + 1 = EC/AE + 1 = 5/2

BD / AD = 5/2-1 & EC/AE = 5/2-1

BD / 2.4 = 5-2/2 & EC/3.2 = 5-2/2

BD = 3/2(2.4) = 3.6 cm

EC = 3/2(3.2) = 4.8 cm

Therefore, the Lenght of Side BD = 3.6 cm and EC = 4.8 cm.