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asked Jan 24, 2023 148k views

1 Answer

Qutron · Answer 1 · 2023-01-30T20:48:14+0000

Answer:

5

Explanation:

Refer to attachment for marking of sides.

In the given figure , ∆ABC , ∆ABD and ∆ADC are right angled triangles . Therefore here we can use the Pythagoras theorem , as ,

$\longrightarrow$ base² + perpendicular² = hypotenuse ² .

• In ∆ABD , we have ;

$\longrightarrow$ AB² + BD² = AD²

$\longrightarrow$ AB² + x² = 10²

$\longrightarrow$ AB² = 10² - x²

$\longrightarrow$ AB² = 100 - x²

$\rule{200}2$

• Again in ∆ADC , we have ;

$\longrightarrow$ AC² + AD² = CD²

$\longrightarrow$ AC² = 20² - 10²

$\longrightarrow$ AC² = 400 - 100

$\longrightarrow$ AC² = 300

$\rule{200}2$

Again in ∆ABC , we have ,

$\longrightarrow$ AC² = AB² + BC²

Substituting the values from above ,

$\longrightarrow$ 300 = 100-x² + (20-x)²

$\longrightarrow$ 300 = 100 - x² + 400 + x² - 40x

$\longrightarrow$ 40x = 500 - 300

$\longrightarrow$ 40x = 200

$\longrightarrow$ x = 200/40

$\longrightarrow$ x = 5