By midpoint theorem,
DE=$\dfrac{1}{2}$ BC
DE = $\dfrac{1}{2}$ of 6
DE = 3 cm
Here, half of the diagonals of a rhombus are the sides of the triangle and side of the rhombus is the hypotenuse.
By Pythagoras theorem,
$(\dfrac{16}{2})^2+(\dfrac{12}{2})^2=side^2$
$8^2+6^2=side^2$
64+36=$side^2$
$side^2$ = 100
side=10 cm
Let A1 and A2 are areas of the small and large triangle.
Then,
$\dfrac{A2}{A1}$=($\dfrac{side \:\: of \:\: large \:\: triangle}{side \:\: of \:\: small \:\: triangle}$)
$\dfrac{A2}{48}=(\dfrac{3}{2})^2$
A2=108 sq.cm.
Perimeter of triangle = sum of all its sides
P = 30+40+x
100=70+x
x=30 cm
If ABC and DEF are two triangles and $\dfrac{AB}{DE}=\dfrac{BC}{FD}$, then the two triangles are similar if $\angle B=\angle D$.
All circles, squares, and equilateral triangles are similar figures.
Given that, in triangles ABC and PQR, $\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}$
If sides of one triangle are proportional to the side of the other triangle, and their corresponding angles are also equal, then both the triangles are similar by SSS similarity. Therefore, ΔPQR ~ ΔCAB
The main criteria for similarity of two triangles are AAA, AA, SAS and SSS.