Question

What value(s) of x will make DE || AB, in the given figure ?

AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x.​

Answer 1

ॐ $\red{R}$$\orange{a}$$\green{d}$$\blue{h}$$\purple{e}$ $\red{R}$$\orange{a}$$\green{d}$$\blue{h}$$\purple{e}$ ॐ

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⎟⎟ ✪✪ ＱＵＥＳＴＩＯＮ ✪✪ ⎟⎟

What value(s) of x will make DE || AB, in the given figure ?

AD = 8x + 9, CD = x + 3,

BE = 3x + 4, CE = x.

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⎟⎟ ✰✰ ＡＮＳＷＥＲ ✰✰ ⎟⎟

Gɪᴠᴇɴ :- In ∆ABC, DE // AB

AD = 8x + 9, CD = x + 3,

BE = 3x + 4, CE = x

By Basic Proportionality theorem,

If DE // AB then we should have

CD/DA = CE/EB

∴ x + 3 / 8x + 9 = x / 3x + 4

☛ (x + 3) (3x + 4) = x (8x + 9)

☛ x (3x + 4) + 3 (3x + 4) = 8x² + 9x

☛ 3x² + 4x + 9x + 12 = 8x² + 9x

☛ 8x² + 9x - 3x² - 4x - 9x - 12 = 0

☛ 5x² - 4x - 12 = 0

☛ 5x² - 10x + 6x - 12 = 0

☛ 5x (x - 2) + 6 (x - 12) = 0

☛ (5x + 6) (x - 2) = 0

☛ 5x + 6 = 0 or x - 2 = 0

☛ x = -6/5 or x = 2

x cannot be negative.

∴ The value x = 2 will make DE//AB.

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Answer 2

Given, DE//AB

$\frac{AD}{DC} = \frac{BE}{EC} \\ \\ \frac{8x + 9}{x + 3} = \frac{3x + 4}{x} \\ \\ {8x}^{2} +9x= {3x}^{2} +13x+12 \\ \\ {5x}^{2} −4x−12=0 \\ \\ {5x}^{2} −10x+6x−12=0 \\ \\ 5x(x−2)+6(x−2)=0 \\ \\ (5x+6)(x−2)=0 \\ \\ x = \frac{ - 6}{5} \: and \: 2$

Therefore x is 2.