The rectangle shown has a perimeter of 58 cm and the given area. its length is 8 more than twice its width. write and solve a system of equations to find the dimensions of the rectangle. the area is 154cm squared

The length of the rectangle is 22 cm and width is 7 cm.

Step-by-step explanation:

Let the width of the rectangle be 'w'.

Given:

Perimeter (P) = 58 cm

Area of the rectangle (A) = 154 cm²

As per question:

Length = 8 more than twice its width

Now, perimeter is given as:

Now, we have to solve the given system of equations.

For that, we plug in 'L' from equation 1 to equation 2. This gives,

Therefore, the length of the rectangle is 22 cm and width is 7 cm.

Area = Length × Width = 22 × 7 = 154 cm².

if you explain the the question more correctly i might be able to answer this better

step-by-step explanation:

geometry: triangles

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lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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