# Question dd576

Feb 10, 2018

$\frac{a}{d} = 3$ or $\frac{a}{d} = - 1$

#### Explanation:

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

Applying Pythagoras rule,

${\left(a + 2 d\right)}^{2} = {a}^{2} + {\left(a + d\right)}^{2}$

${a}^{2} + 4 a d + 4 {d}^{2} = {a}^{2} + {a}^{2} + 2 a d + {d}^{2}$

$2 a d + 3 {d}^{2} - {a}^{2} = 0$

a^2 −3ad+ad−3d ^2 =0 

a(a+d)−3d(a+d)=0 

(3d−a)(d+a)=0 

$\implies \frac{a}{d} = 3$ or $\frac{a}{d} = - 1$

$\frac{a}{d} = - 1$ is inadmissible beacuse, a>0;d>0#

therefore, $a : d = 3 : 1$

-Sahar

Feb 10, 2018

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

${a}^{2} + {\left(a + d\right)}^{2} = {\left(a + 2 d\right)}^{2}$
$\implies {a}^{2} + {a}^{2} + 2 a d + {d}^{2} = {a}^{2} + 4 a d + 4 {d}^{2}$
$\implies {a}^{2} - 2 a d - 3 {d}^{2} = 0$
$\implies {a}^{2} + a d - 3 a d - 3 {d}^{2} = 0$
$\implies a \left(a + d\right) - 3 d \left(a + d\right) = 0$
$\implies \left(a + d\right) \left(a - 3 d\right) = 0$

We get two equations as

$\left(a + d\right) = 0$ .....(1)
$\left(a - 3 d\right) = 0$ .....(2)

from (1) we get

$a = - d$

It is given that both $a \mathmr{and} d$ are positive. Therefore, this solution needs to be ignored.

From (2) we get

$a = 3 d$
$\implies a : d = 3 : 1$