Question

find value of x in the following figure given

Answer 1

by Pythagoras theorem
(ab)^2 =(ac)^2+(BC)^2
13^2=12^2+(BC)^2
169=144+(BC)^2
√25=BC
5=BC
Now (ad)^2=(ac)^2+(cd)^2
13^2=12^2+(CD)^2
169-144=(CD)^2
√25=CD
5=CD
now BC+cd=5+5=10


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Answer 2

In the ∆ABD

There are 2 triangles

∆ACB & ∆ACD

(The two triangles are right angled triangles)

In ∆ACB

AB=13cm;AC=12cm;BC=BC

By Pythagoras Theorem

AB^2=AC^2+BC^2

13^2=12^2+BC^2

169-144=BC^2

BC=√25

BC=5cm


In ∆ACD

AD=13cm;AC=12cm;CD=CD

AD^2=AC^2+CD^2

13^2=12^2+CD^2

169-144=CD^2

CD=√25

CD=5cm

BC+CD=BD

BC+CD=x. {BD=x}

x=5+5

x=10cm

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