Question

please yar help krooooooo spamming =10 answer report... ​

Answer 1

Answer:

• x = 3

Steps:

Consider ΔADE and ΔABC

It is given that DE || BC

Therefore,

• ∠ ADE = ∠ ABC
• ∠ AED = ∠ ACB

Hence by AA Similarity, we can say that: Δ ADE ~ Δ ABC

Hence by Basic Proportionality Theorem, we can say that:

⇒ AD/AB = AE/AC    ...(i)

According to the question,

• AD = x
• AB = AD + DB → x + x + 1 → ( 2x + 1 )
• AE = x + 3
• AC = AE + EC → x + 3 + x + 5 → ( 2x + 8 )

Substituting it in (i) we get:

$\implies \dfrac{x}{2x + 1} = \dfrac{x + 3}{2x+8}$

Cross multiplying we get:

⇒ x ( 2x + 8 ) = ( x + 3 ) ( 2x + 1 )

⇒ ( 2x² + 8x ) = ( 2x² + 6x + x + 3 )

2x² gets cancelled on both sides & we get,

⇒ 8x = 7x + 3

⇒ 8x - 7x = 3

⇒ x = 3

Hence the value of 'x' is 3.

Answer 2

सवाल –

${\mapsto}$ यदि ∆ ABC में DE || BC हो तो x ki कीमत ज्ञात कीजिए । चित्र लगाव में दिया गया है ( Given in attachment )

दिया गया है –

${\mapsto}$ ( लगाव देखना अनिवार्य है )

∆ ABC में DE || BC है ।

ज्ञात करना है –

${\mapsto}$ x की कीमत ।

हल –

${\mapsto}$ x की कीमत है = 3

समस्त हल –

$\rule{95}{2}$

~

∆ADE = ∆ABC

∆ ABC में DE || BC है

∠ ADE = ∠ ABC

∠ AED = ∠ ACB

$\rule{95}{2}$

~ आनुपातिक निष्क्रियता का उपयोग करके हमें मूल्यों को रखना होगा !

( Now we have to use Proportional theorm and have to put values ! )

${\bold{\sf{\mapsto \dfrac{AD}{AB} = \dfrac{AE}{AC}}}}$ Equation 1

$\rule{95}{2}$

~ सवाल के अनुसार –

${\bullet}$ AD = x

${\bullet}$ AE = x + 3

${\bullet}$ AB = AD + DB ${\longrightarrow}$ x + x + 1 = 2x + 1

${\bullet}$ AC = AE + EC ${\longrightarrow}$ x + 3 + x + 5 ${\longrightarrow}$ ( 2x + 8 )

$\rule{95}{2}$

${\bold{\sf{\mapsto \dfrac{x}{2x+1} = \dfrac{x+3}{2x+8}}}}$

${\mapsto}$ x(2x+8) = 3x+3(2x-1)

${\mapsto}$ 2x²+8x = 2x²+6x+x+3

${\mapsto}$ 8x = 6x + x + 3

${\mapsto}$ 8x = 7x + 3

${\mapsto}$ 8x - 7x = 3

${\mapsto}$ 1x = 3

${\mapsto}$ x = 3

$\rule{95}{2}$