Question

Find the value of x in the given figure.​

Answer 1

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Statement:-

• a line drawn parallel to one side of a triangle to intersect other two sides at distinct points, then they are divided in the same ratio

Solution:-

$\implies$ $\sf\dfrac{3x+19}{x+3} = \dfrac{3x+4}{x}$ ... (B.P.T)

$\implies$ $\sf (3x+19)(x) = (3x+4)(x+3)$

$\implies$ $\sf 3x²+19x = 3x²+9x+4x+12$

$\implies$ $\sf 19x = 13x+12$

$\implies$ $\sf 6x = 12$

$\implies$ $\sf x = \dfrac{12}{6}$

$\implies$ $\sf x = 2$

Answer 2

Answer:

BASIC PROPORTIONALITY THEOREM (BPT) : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio. That is also known as Thales theorem.

GIVEN:

In ∆ABC , DE || AB

AD= 3x+19, DC= x+3, BE= 3x+4, EC = x

AD /DC = BE/EC

[ By Thales theorem(BPT)]

3x+19/x+3 = 3x+4/x

(3x+19) × x = (3x+4) × (x+3)

3x² + 19x = 3x² + 9x + 4x +12

3x² + 19x = 3x² + 13x +12

3x² - 3x² +19x -13x = 12

6x = 12

x = 12/6= 2

x = 2

Hence, the value of x is 2 .

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Step-by-step explanation:

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