Answer:
The perimeter of the larger square is
![150cm.](https://img.qammunity.org/2020/formulas/mathematics/middle-school/tzdcei7sqg2v139yvey8ssce6a8ssgpbhx.png)
Explanation:
First of all, let one part of it be "
"
and the other part of it "
"
Now to solve this :
The area of the square =
![L^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xk7jwjwwbmb5igu0xk89p70hniecz900a2.png)
The area of one part of the square =
![x^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lj2p7ilwuzg3rb119glsof4tows22y4e2d.png)
The area of the other part of the square =
![(200-x)^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/prnnipqwga2tt4kgxkmvhnrqggwmjiykpl.png)
![9x^2= (200-x)^2\\9x^2=40000-400x+x^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ov2i64qh1eaf6zvhwdr0bh4kixe383uxtb.png)
Now, add,
to both the sides :
![9x^2-9x^2=40000-400x+x^2-9x^2\\-1(8x^2+400x-40000)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8hdmf74w4b9wrzojrael4njyzf0xuthg29.png)
Now, take out the "8" which is common :
![-8(x^2+50x-5000)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/c002ll2ksxz2y1cjkor4m23u446ofghjgq.png)
Now, divide it by
:
![x^2+100x-50x-5000=0\\x(x+100)-50(x+100)=0\\(x+100)(x-50)=0\\](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2af72o90l657ed24we5e1g3gk51r6r6mba.png)
or
![50](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pn3hqg7e3dzo61faoa5ac4xrbkiwxi7agh.png)
So, now we know that :
One of the part is =
![50cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/w5i6hlamuv71ts8euzwaqbi3bmh4nr7xt4.png)
And the other part is =
![150cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k6ds16465q6skjecx8al8b9du7tzh73k26.png)
Thus the perimeter of the larger square is =
![150cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k6ds16465q6skjecx8al8b9du7tzh73k26.png)