Question

find the value of x in each of the following figure( using Pythagoras theorem)​
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Answer 1

Answer:

In ∆ADB,

By Pythagoras theorem,

(AB)^2 = (AD)^2+(DB)^2

(13)^2 = (12)^2 +(DB)^2

(DB)^2 = 169-144

(DB) = √25

DB = 5 cm

In ∆ADC,

By Pythagoras theorem,

(AC)^2 = (AD)^2+(DC)^2

(13)^2 = (12)^2 +(DC)^2

(DC)^2 = 169-144

(DC) = √25

DC = 5 cm

BC = DC + DB

BC = 5+5 = 10 CM

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Answer 2

Answer:

Triangle ABD

by pythagoras Therom

AB²=DB²+AB²

13²=DB²+12²

169=DB²+144

DB²=169-144

DB²=25 square root

DB=5

AC²=AD²+DC²

13²=12²+DC²

DC²=169-144

DC²=25 Square Root

DC=5²

X=DB+DC

X=5+5

X=10