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Answer 1 · 2018-02-10T08:21:47

#a/d = 3# or # a/d=-1#

Side of right angled triangle are #a,a+d,a+2d #

Applying Pythagoras rule,

#(a+2d)^2=a^2 +(a+d)^2#

#a^2 + 4ad +4d^2= a^2 + a^2 +2ad+d^2#

#2ad +3d^2-a^2=0#

#a^2 −3ad+ad−3d ^2 =0 #

#a(a+d)−3d(a+d)=0 #

#(3d−a)(d+a)=0 #

#=> a/d = 3# or # a/d=-1#

#a/d=-1# is inadmissible beacuse, #a>0;d>0#

therefore, #a:d=3:1#

-Sahar

Answer 2 · 2018-02-10T08:31:28

Using Pythagoras theorem and noting that largest side must represent hypotenuse we get

#a^2+(a+d)^2=(a+2d)^2#
#=>a^2+a^2+2ad+d^2=a^2+4ad+4d^2#
#=>a^2-2ad-3d^2=0#
#=>a^2+ad-3ad-3d^2=0#
#=>a(a+d)-3d(a+d)=0#
#=>(a+d)(a-3d)=0#

We get two equations as

#(a+d)=0# .....(1)
#(a-3d)=0# .....(2)

from (1) we get

#a=-d#

It is given that both # a and d# are positive. Therefore, this solution needs to be ignored.

From (2) we get

#a=3d#
#=>a:d=3:1#