In AJKL, j = 75 cm, k = 23 cm and

just needed some points but try socratic

c.   x=2.4cm

step-by-step explanation:

similar triangles are triangles which have the same shape but not the same size. this means their angle measures are equal but their side lengths are not instead they are proportional. to solve, we will set up a proportion and solve.

a proportion is two equal ratios set equal to each other. we form the ratios by finding corresponding sides (sides which match to each other on both triangles by their position). out ratios will be one side of the small triangle over the corresponding side on the big triangle as shown below:

.

to solve the proportion, we'll cross multiply by multiplying numerator with denominator across the equal sign.

geometry: triangles

solvers

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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