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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

$a=3d$
$=>a:d=3:1$