![\bold{\huge{\underline{ Solution }}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/jdi2w7914cic76zpb2xuxp7e51pz44d9g8.png)
Here, We have given
- 2 squares , In which 1 square is enclosed within the another square and it arranged in a form that it forms 4 right angled triangle
- The height and base of the given right angled triangles are 6 and 3 each.
We know that,
Area of triangle
![{\sf{=}}{\sf{(1)/(2)}}{\sf{ {*} base {*} height }}](https://img.qammunity.org/2023/formulas/mathematics/high-school/5jwxkpdno6uswn6xbdvau7rzscya5fiix8.png)
Subsitute the required values,
![{\sf{=}}{\sf{(1)/(2)}}{\sf{ {*} 3 {*} 6}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/v3lstcdpos3p6uusvehd0r4vzsz2c99t1d.png)
![\sf{ = 3 {*} 3 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/7de6rzzpeie25vmszvyvqrxfmzkkqw4w3v.png)
![\bold{ = 9 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/8i0r557d1l34uy3eqi2jsw8pp74d92isfv.png)
Therefore,
Area covered by 4 right angled triangles
![\sf{ = 4 {*} 9 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/lt7gvdh9hcjo6tfwroasxatbsmtjjv5tzs.png)
![\bold{ = 36}](https://img.qammunity.org/2023/formulas/mathematics/high-school/1x18x841x4b9zgzlyi8nir9kger4izyecm.png)
Now,
We have to find the area of the big square
- The length of the side of the big square
![\sf{ = 6 + 3 = 9 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/cbxdgxlo7y0j8o2bk0wvpf33emp2g9hfow.png)
We know that,
Area of square
![\sf{ = Side {*} Side }](https://img.qammunity.org/2023/formulas/mathematics/high-school/l53d995i8pcak9oeknc2t7etxuutn33rkg.png)
Subsitute the required values,
![\sf{ = 9 {*} 9 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/b3zu4mriond6d2nmak6sa8c9tor5lhzzg9.png)
![\bold{ = 81 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/3yp6vmu2ztlttwfoq09l76hbixkbhumh6d.png)
Therefore,
The total area of shaded region
= Area of big square - Area covered by 4 right angled triangle
![\sf{ = 81 - 36 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/7ajrynsqg387ohjb2xpme4dpe79rvx28hb.png)
![\bold{ = 45 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/qremua7wjqplsx8f4rffminsx364uu3rgi.png)
Hence, The total area of shaded region is 45 .
Part 2 :-
Here,
We have to find the area of non shaded region
According to the question
- Hypotenuse = The length of square
Let the hypotenuse of the given right angled triangle be x
Therefore,
By using Pythagoras theorem,
- This theorem states that the sum of the squares of the base and perpendicular height is equal to the square of hypotenuse.
That is,
![\sf{ (Perpendicular)^(2) + (Base)^(2) = (Hypotenuse)^(2) }](https://img.qammunity.org/2023/formulas/mathematics/high-school/321p62ye811fsforum78zhsm01x36cqy9z.png)
Subsitute the required values
![\sf{ (6)^(2) + (3)^(2) = (x)^(2) }](https://img.qammunity.org/2023/formulas/mathematics/high-school/y1vy8mn7xjkiovdyvgacjxx4yyi2l8gf31.png)
![\sf{ 36 + 9 = (x)^(2) }](https://img.qammunity.org/2023/formulas/mathematics/high-school/1uifg3k9wgff3n8tj6d1rauylx3crchgek.png)
![\sf{ x = √(45)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ldy39vta8uexgmtc3i0yqrjd0lln5txprm.png)
![\bold{ x = 6.7 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/czi2z7dm7vq2e5wmsu3h2x08p8qgpiqxn5.png)
That means,
- The length of the small square = 6.7
We know that ,
Area of square
![\sf{ = Side {*} Side }](https://img.qammunity.org/2023/formulas/mathematics/high-school/l53d995i8pcak9oeknc2t7etxuutn33rkg.png)
Subsitute the required values,
![\sf{ = 6.7 {*} 6.7 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/vu6ok8ak9njc219kviy7z9c11ttjxgal72.png)
![\bold{ = 44.89 \:\: or \:\: 44.9 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/hbtlk8pag0y50pvnyho9ksee28iir9r4o3.png)
Therefore ,
Area of non shaded region
= Area of big square - Area of small square
![\sf{ = 81 - 44.9 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/urpfy9qdda6bs373sg6grtcp0hkni73uil.png)
![\bold{ = 36.1 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/nlx97s6lp5jbvo2jmjvv3mrwvhyq4lxy2t.png)
Hence, The total area of non shaded region is 36.1 or 36 (approx) .
Part 3 :-
Here, we have to
- find the total area of the figure
Therefore,
The total area of the figure
= Non shaded region + Shaded region
![\sf{= 36 + 45 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/pm771kw5q86zl3aco247et3o8c8cro04kj.png)
![\bold{= 81}](https://img.qammunity.org/2023/formulas/mathematics/high-school/lrk73qjo8vyk7y8kr7inwj4ut0xhj9vste.png)
Hence, The total area of the given figure is 81 .