Question

In the given figure, AD = 2.5 cm, DB = 9.5 cm, DE = 3.5 cm and BC = x cm. Prove that ∆ADE is similar to ∆ACB. find the value of x. ​

Answer 1

Sσℓꪊէเòɳ

$\longrightarrow$$\bf\dfrac{AD}{AC}$ = $\bf\dfrac{2.5 cm}{(4+3.5)cm}$ = $\bf\dfrac{2.5 cm}{7.5 cm}$ = $\bf\dfrac{1}{3}$

$\longrightarrow$$\bf\dfrac{AE}{AB}$ = $\bf\dfrac{4 cm}{(2.5+9.5)cm}$ = $\bf\dfrac{4 cm}{12 cm}$ = $\bf\dfrac{1}{3}$

∴⠀⠀⠀⠀⠀$\bf\dfrac{AD}{AC}$ = $\bf\dfrac{AE}{AB}$ = $\bf\dfrac{1}{3}$⠀⠀⠀⠀ ⠀⠀..(1)

${\bold{\underline{\underline{In\:\: ∆ADE\:\:and\:\:∆ACB,}}}}$

$\longrightarrow$⠀⠀⠀ $\bf\dfrac{AD}{AC}$ = $\bf\dfrac{AE}{AB}$⠀⠀⠀⠀⠀⠀[∠A = ∠A]

∴ ∆ADE ~ ∆ACB⠀⠀⠀ ⠀⠀⠀[By SAS]

∴⠀⠀⠀⠀⠀$\bf\dfrac{AD}{AC}$ = $\bf\dfrac{AE}{AB}$ = $\bf\dfrac{DE}{CB}$

$\longrightarrow$⠀⠀⠀ $\bf\dfrac{1}{3}$ = $\bf\dfrac{3.5 cm}{x cm}$⠀⠀ ⠀⠀⠀⠀[From (1)]

⠀⠀⠀ $\huge\underline{\overline{\mid{\bold{\pink{x = 10.5\:cm}}\mid}}}$

Answer 2

Answer:

The triangle ∆ADE is similar to ∆ACB. i.e Δ ADE ≈ ΔABC and the value of x is 10.5 cm

Step-by-step explanation:

Given: AD = 2.5 cm, DB = 9.5 cm, DE = 3.5 cm

To find:  ∆ADE is similar to ∆ACB and the value of X i.e BC

Solution:

In the given triangle

From the given figure,

Δ ADE and ΔACB both are right angled triangle.

According to the theorem,

Two right angled triangles are similar to each other.

Δ ADE ≈ ΔABC

And  also ratio of corresponding sides are equal.

$\frac{DE}{BC} = \frac{BA}{AC} = \frac{AE}{AB}$

Given that, AD = 2.5 cm, DB = 9.5 cm, DE = 3.5 cm

⇒ $\frac{3.5}{X} = \frac{2.5}{7.5} = \frac{4}{12}$

⇒ $\frac{3.5}{X} = \frac{1}{3} = \frac{1}{3}$

⇒ $\frac{3.5}{x} = \frac{1}{3}$

⇒3.5(3) = x

x = 10.5 cm

Final answer:

The triangle ∆ADE is similar to ∆ACB. i.e Δ ADE ≈ ΔABC and the value of x is 10.5 cm

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