Question

2- des triangles semblables les quatre triangles abc suivants sont semblables entre eux. trouve les valeurs manquantes dans chacun des triangles 2, 3 et 4. triangle #1: a to b: 35 dm, b to c: 20 dm, c to a: 43 dm, angle at b: 100° triangle #2: angle at b: 100°, b to c: 38 dm a) ab? b) ac? triangle #3: a to b: 40 dm, angle at b: 100° a) ac? b) bc? triangle #4: b to c: 50 dm, angle at b: 100° a) ab? b) bc?

Explanation:

Step1: Find similarity ratio

First, identify corresponding sides: In Triangle #1, $AC=43$ dm, which corresponds to $AC=38$ dm in Triangle #2, $AB=40$ dm in Triangle #3 corresponds to $AB=35$ dm in Triangle #1, and $AB=50$ dm in Triangle #4 corresponds to $AB=35$ dm in Triangle #1.

For Triangle #2:

Ratio $r_2 = \frac{AC_2}{AC_1} = \frac{38}{43}$

Step2: Calculate AB for Triangle #2

$AB_2 = AB_1 \times r_2$
$AB_2 = 35 \times \frac{38}{43} = \frac{1330}{43} \approx 30.93$ dm

Step3: Calculate BC for Triangle #2

$BC_2 = BC_1 \times r_2$
$BC_2 = 20 \times \frac{38}{43} = \frac{760}{43} \approx 17.67$ dm

For Triangle #3:

Ratio $r_3 = \frac{AB_3}{AB_1} = \frac{40}{35} = \frac{8}{7}$

Step4: Calculate AC for Triangle #3

$AC_3 = AC_1 \times r_3$
$AC_3 = 43 \times \frac{8}{7} = \frac{344}{7} \approx 49.14$ dm

Step5: Calculate BC for Triangle #3

$BC_3 = BC_1 \times r_3$
$BC_3 = 20 \times \frac{8}{7} = \frac{160}{7} \approx 22.86$ dm

For Triangle #4:

Ratio $r_4 = \frac{AB_4}{AB_1} = \frac{50}{35} = \frac{10}{7}$

Step6: Calculate AC for Triangle #4

$AC_4 = AC_1 \times r_4$
$AC_4 = 43 \times \frac{10}{7} = \frac{430}{7} \approx 61.43$ dm

Step7: Calculate BC for Triangle #4

$BC_4 = BC_1 \times r_4$
$BC_4 = 20 \times \frac{10}{7} = \frac{200}{7} \approx 28.57$ dm

Answer:

Triangle #2:

a) $AB = \frac{1330}{43} \approx 30.93$ dm
b) $BC = \frac{760}{43} \approx 17.67$ dm

Triangle #3:

a) $AC = \frac{344}{7} \approx 49.14$ dm
b) $BC = \frac{160}{7} \approx 22.86$ dm

Triangle #4:

a) $AC = \frac{430}{7} \approx 61.43$ dm
b) $BC = \frac{200}{7} \approx 28.57$ dm