Answer:
5
Step-by-step 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 ,
base² + perpendicular² = hypotenuse ² .
<u>•</u><u> </u><u>In </u><u>∆</u><u>A</u><u>B</u><u>D</u><u> </u><u>,</u><u> </u><u>we </u><u>have</u><u> </u><u>;</u>
AB² + BD² = AD²
AB² + x² = 10²
AB² = 10² - x²
AB² = 100 - x²
<u>•</u><u> </u><u>Again</u><u> </u><u>in </u><u>∆</u><u>A</u><u>D</u><u>C</u><u> </u><u>,</u><u> </u><u>we </u><u>have</u><u> </u><u>;</u>
AC² + AD² = CD²
AC² = 20² - 10²
AC² = 400 - 100
AC² = 300
<u>Again</u><u> </u><u>in </u><u>∆</u><u>A</u><u>B</u><u>C</u><u> </u><u>,</u><u> </u><u>we </u><u>have</u><u> </u><u>,</u>
AC² = AB² + BC²
Substituting the values from above ,
300 = 100-x² + (20-x)²
300 = 100 - x² + 400 + x² - 40x
40x = 500 - 300
40x = 200
x = 200/40
x = 5
