Question

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$\longrightarrow$ find the value of x in the attached figure.

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Answer 1

What to do in this?

1)Take∆ABC and ∆BDC,and prove them similar

2)Now if they will be similar then ratio of their corresponding sides will be equal

3)Make the ratios equal and take out the value of x by further calculation

In ∆ABC and ∆BDC,

angle(ABC)=angle(BDC) (each equal to 90)..i)

angle(BCD)=angleACB) (common)..ii)

From AA property,we can say ∆ABC~∆BDC

now ,we know that if two triangles are similar then the ratios of their corresponding sides are equal.

BC/DC=AC/BC

(x+13)/x=52/(x+13)

(x+13)^2=52x

x^2+169+26x=52x

x^2-26x+169=0

x^2-13x-13x+169=0

x(x-13)-13(x-13)=0

(x-13)(x-13)=0

x=13

hence,the value of x is 13

Answer 2

EXPLANATION.

In right angled triangle,

⇒ BC = x + 13.

⇒ Dc = x.

⇒ Ac = 52.

By using, Similarity of triangle, we get.

⇒ AC/BC = BC/CD.

⇒ 52/(x + 13) = (x + 13)/x.

⇒ 52x = (x + 13)².

⇒ 52x = x² + 169 + 26x.

⇒ x² + 169 + 26x - 52x.

⇒ x² - 26x + 169 = 0.

As we know that,

Factorizes the equation into middle term split, we get.

⇒ x² - 13x - 13x + 169 = 0.

⇒ x(x - 13) - 13(x - 13) = 0.

⇒ (x - 13)(x - 13) = 0.

⇒ (x - 13)² = 0.

⇒ x = 13.

Value of x = 13.