Question

If ∆ABC ∼ ∆DEF such that AB = 5 cm, area (∆ABC) = 20 cm² and area (∆DEF) = 45 cm², determine DE.

Answer 1

SOLUTION :  

Given : ΔABC∼ΔDEF , AB = 5cm , ar (ΔABC) = 20cm²  and ar(ΔDEF)  = 45cm².

We know that the ratio of the areas of the two similar triangles is equal to the ratio of the squares of their corresponding sides.

arΔ(ABC) / ar(ΔDEF) = (AB/DE)²

20/45 = 5²/DE²

20/45 = 25/DE

DE² = (25×45) / 20

DE² = 225/4

DE =√225/4

DE = 15/2

DE = 7.5 cm

Hence, the length of  DE is 7.5 cm.

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

$<b><u>$ Solution

area of f two similar ΔABC = 20cm², ΔDEF = 45cm² respectively and AB = 5cm.

To find: measure of DE 

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{(area \: abc)}{(area \: def)} = \frac{ ({ab}^{2} }{de}) \\ \\ \frac{20}{45} = \frac{5}{de ^{2} } \\ \\ \frac{20}{45} = \frac{5}{de {}^{2} } \\ \\ \: {de}^{2} = \frac{25 \times 45}{20} \\ {de}^{2} = \frac{225}{4} \\ \\de = 7.5cm$

Hence the answer is 7.5cm