Question

In the ∆ABC,
angle B = 90degree ,
AB: BD: DC = 3:1:3,
If AC = 20cm,

then what is length of AD

as soon as possible please tomorrow is my exam!!!!!!!!!!​

Answer 1

Step-by-step explanation:

By the question we get to know that triangle ABC

is a right angled triangle.

given,

AB:BD:DC=3:1:3

---->AB:BD=3:1

AB/BD=3/1

AB=3BD----(1)

---->BD:DC=1:3

BD/DC=1/3

DC=3BD

FROM THIS WE GET AB=DC

AND,

BD=AB/3 (or)BD=DC/3

FROM QUESTION,

BC=BD+DC

BC=DC/3+DC

BC=4DC/3

WE KNOW THAT FOR A RIGHT ANGLED TRIANGLE ABC,

AC^2=AB^2+BC^2

20^2=AB^2+(4DC/3)^2

400=AB^2+16DC^2/9

AB^2=400-16DC^2/9

AB^2=(3600-16DC^2)/9

AB=√(3600-16DC^2)/9

AB=(60-4DC)/3----(3)

FROM RIGHT ANGLED TRIANGLE ABD,

AD^2=AB^2+BD^2

AD^2=[(60-4DC)/3]^2+(DC/3)^2

AD^2=3600-16DC^2/9+DC^2/9

AD^2=3600-16DC^2+DC^2

9

AD^2=3600-15DC^2

9

AD^2=(3600/9)-(15DC^2/9)

AD^2=400-5DC^2/3

AD=√400-5DC^2/3

AD=20-(√5)DC/√3

RATIONALIZING (√5)DC/√3

WE GET,

AD=20-DC(√15)/3

NOW FROM (1)

AD=20-3BD(√15)/3

AD=20-BD(√15)