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Relevant wiki: Length and Area Problem Solving

Notice the two triangles form a square with area $24-15=9$

Hence, the height of rectangle $=$ length of the square $=\sqrt{9}=3$

The length of the rectangle $=24\div3=8$

$\therefore$ Perimeter $=2(3+8)=\boxed{22}$

Area of the rectangle is $ab=24$

From this we take off two right triangles with legs equal to $a$ , so the area will become $ab-a^2=15$

Combining he equations will lead to $a^2=24-15=9$

So that the sides of the rectangle $a=3, b=8$

And the perimeter is $2(a+b)=2\times11=\boxed{22}$

Let the longer and shorter side of the original rectangle be of length $a$ and $b$ respectively.

Then, in the original rectangle, Area $=a\times b = 24$ . $\color{#3D99F6}{\cdots (1)}$

In the second polygon (i.e. parallelogram), the height will be same that is $b$ . As they are folded at $45^o$ ,

then the base will be equal to $a-b$ , this gives us the area of parallelogram to be $(a-b)(b) = 15$ .

From $\color{#3D99F6}{(1)}$ , we have $24-b^2=15$

$\implies b=3$ Thus this gives $a=\dfrac{24}{3} = 8$ . Therefore, the perimeter becomes $\boxed{22}$

$A_{Rectangle} = xy = 24$

$A_{parallelogram} = 15 = (x - y)y = xy - y^2 \implies$

$15 = 24 - y^2 \implies y = 3 \implies x = 8 \implies$ Perimeter $P_{Rectangle} = 2(11) = \boxed{22}$ .

Let V be the length of the vertical side of the rectangle and H the length of the horizontal side of the rectangle.

Given the 45 degree angles shown, two congruent right isosceles triangle areas are removed from each corner making in total a square with side length V with area 24 - 15 = 9. Thus V = 3, H = 8 (to get original area 24) and the perimeter is 22.

Let the side of the original rectangle be b and h

So, b×h=24 Now the rectangle is folded into a parallelogram at 45° So the triangle which is folded has one angle 45°,and the other 90° as angle of the rectangle was 90° so that is an isosceles triangle

Since I have taken the breadth of the rectangle as b so base of parallelogram is (h-b)

So the area of the parallelogram is

(h-b)×b=15 So now again bh-b^2= 15

24-b^2=15

b=3 So h=24÷3=8

Therefore perimeter of these rectangle is

( 8+3)×2=22

Let the dimensions of the rectangle be a < b. By inspection the missing area is a^2 = 24-15 = 9, making a=3 and b=8. Perimeter is 2(a+b) = 22.

The only way I found 15 as area was 3*5

The height is 3 and logically 3*8 = 24

The perimeter is 2(3+8) = 22.

Might not be how it's supposed to be solved but it worked.

Since 2 right-angled triangles were taken off from the sides, we were left with a parallelogram. Both the parallelogram and the rectangle, therefore, should have the same height. So, to find this proposed 'height', just find the Highest Common Factor (HCF) of 24 and 15. Your answer should be 3. Since we are told to find the perimeter of the rectangle, we need two numbers: The width = height = 3 The length = area divided by height = 24 divided by 3 = 8 Now all we need to do is to find the perimeter: P = 2 x (length + width) = 2 (8 + 3) = 2 x 11 = 22