What is the length of the ad to the nearest tenth of a centimeter?

Firstly we will get the third side AB by PGT

H^2=P^2+B^2

H^2=6^2+8^2

H^2=36+64

H^2=100

H= 10 CM

Now by herons formula

s=a+b+c/2

=10+8+6/2

=24/2

=12cm

now

Area of ABC=root(s(s-a)(s-b)(s-c)

=root(12*2*4*6

=root(2*6*2*2*2*6

=2*2*6

=24cm^2

Now altitude is DC

and Area of triangle=1/2 * B * H

24=1/2 * 10 * DC

24*2 * 1/10=DC

DC=4.8cm

ADC is right angled triangle with angle D=90

BY PGT

H^2 = P^2 + B^2

6^2 = 4.8^2 + x^2

36-23.04 = x^2

x^2 = 12.96

x= root(12.96

x= 3.6 (approximately)

Hope it helped

Its equal to d=7.07in