Question

can anyone solve this please​

Answer 1

hope it helped you

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Answer 2

Step-by-step explanation:

according to question

the given triangle is right angle triangle

so by Pythagorous thereom we get

AB²+AC²=BC²

by putting values

AB²+(10√3)²=(5+x)²

AB² +300=(5+x)²

AB²= (5+x)²-300

now applying Pythagorous in ∆BAD

we get

BD² +AD²= AB²

by putting values

5²+AD²= (5+x)²-300

AD²=(5+x)²-300-5²

now applying Pythagorous in ∆ ADC

we get

AD²+DC²=AC²

by putting values

(5+x)²-300-5²+x²=300

25+x²+10x-300-25+x²=300

2x²+10x-600=0

x²+5x-300=0

x²+20x-15x-300=0

x(x+20)-15(x+20)=0

(x-15)(x+20)=0

x=15 and x=-20

length never negative so

x is 15

thank you

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