POMOCY ZAD 6 7 8 ????????.... Question from @agrudzinska7maq

agrudzinska7maq @agrudzinska7maq

June 2022 1 9 Report

POMOCY ZAD 6 7 8 ????????

Roma

Zad. 6

ΔA₁B₁C₁ ∼ ΔABC

k = 3 > 1 (ΔA₁B₁C₁ ma większy obwód od ΔABC) i r = 14

$\frac{L_{\Delta A_1B_1C_1}}{L_{\Delta ABC}} = 3 \\ L_{\Delta A_1B_1C_1} = 3 \cdot L_{\Delta ABC} \\\\ L_{\Delta A_1B_1C_1} - L_{\Delta ABC} = 14 \\ 3 \cdot L_{\Delta ABC} - L_{\Delta ABC} = 14 \\ 2 \cdot L_{\Delta ABC} = 14 \ \ \ |: 2 \\ L_{\Delta ABC} =7 \\\\ L_{\Delta A_1B_1C_1} = 3 \cdot L_{\Delta ABC} \\ L_{\Delta A_1B_1C_1} = 3 \cdot 7 \\ L_{\Delta A_1B_1C_1} = 21$

k = ¹/₄ < 1 (ΔABC ma większy obwód od A₁B₁C₁ i r = 9

$\frac{L_{\Delta A_1B_1C_1}}{L_{\Delta ABC}} =\frac{1}{4} \\ L_{\Delta ABC} = 4 \cdot L_{\Delta A_1B_1C_1} \\\\ L_{\Delta ABC}- L_{\Delta A_1B_1C_1} = 9 \\ 4 \cdot L_{\Delta A_1B_1C_1} - L_{\Delta A_1B_1C_1} = 9 \\ 3 \cdot L_{\Delta A_1B_1C_1} = 9 \ \ \ |:3 \\ L_{\Delta A_1B_1C_1} = 3 \\\\ L_{\Delta ABC} = 4 \cdot L_{\Delta A_1B_1C_1} \\ L_{\Delta ABC} = 4 \cdot 3 \\ L_{\Delta ABC} = 12$

k = ⁵/₆ < 1 (ΔABC ma większy obwód od A₁B₁C₁ i r = 2,5

$\frac{L_{\Delta A_1B_1C_1}}{L_{\Delta ABC}} =\frac{5}{6} \\ 5 \cdot L_{\Delta ABC} = 6 \cdot L_{\Delta A_1B_1C_1} \ \ \ |:5 \\ L_{\Delta ABC} = 1,2 \cdot L_{\Delta A_1B_1C_1} \\\\ L_{\Delta ABC}- L_{\Delta A_1B_1C_1} =2,5 \\ 1,2 \cdot L_{\Delta A_1B_1C_1} - L_{\Delta A_1B_1C_1} = 2,5 \\ 0,2 \cdot L_{\Delta A_1B_1C_1} = 2,5 \ \ \ | \cdot 10 \\ 2 \cdot L_{\Delta A_1B_1C_1} = 25 \ \ \ |:2 \\ L_{\Delta A_1B_1C_1} = 12,5$

$L_{\Delta ABC} = 1,2 \cdot L_{\Delta A_1B_1C_1} \\ L_{\Delta ABC} = 1,2 \cdot 12,5 \\ L_{\Delta ABC} = 15$

k = 2¹/₃ = ⁷/₃ > 1 (ΔA₁B₁C₁ ma większy obwód od ΔABC) i r = 2,8 = ²⁸/₁₀ = ¹⁴/₅

$\frac{L_{\Delta A_1B_1C_1}}{L_{\Delta ABC}} =\frac{7}{3} \\ L_{\Delta A_1B_1C_1} = \frac{7}{3} \cdot L_{\Delta ABC} \\\\ L_{\Delta A_1B_1C_1} - L_{\Delta ABC} = \frac{14}{5} \\ \frac{7}{3} \cdot L_{\Delta ABC} - L_{\Delta ABC} =\frac{14}{5} \\ \frac{4}{3} \cdot L_{\Delta ABC} =\frac{14}{5} \ \ \ | \cdot \frac{3}{4} \\ L_{\Delta ABC} =\frac{\not{14}^7}{5} \cdot \frac{3}{\not{4}_2} \\ L_{\Delta ABC} =\frac{21}{10} \\ L_{\Delta ABC} =2,1$

$L_{\Delta A_1B_1C_1} = \frac{7}{3} \cdot L_{\Delta ABC} \\ L_{\Delta A_1B_1C_1} = \frac{7}{\not{3}_1} \cdot \frac{\not{21}^7}{10} \\ L_{\Delta A_1B_1C_1} = \frac{49}{10} \\ L_{\Delta A_1B_1C_1} =4,9$

Zad. 7

ΔABC ∼ ΔA₁B₁C₁

$L_{\Delta A_1B_1C_1} = L_{\Delta ABC} - 20\% \cdot L_{\Delta ABC} = L_{\Delta ABC} - 0,2 \cdot L_{\Delta ABC} = 0,8 \cdot L_{\Delta ABC} = \\ = \frac{8}{10} \cdot L_{\Delta ABC} = \frac{4}{5} \cdot L_{\Delta ABC} \\\\ k =\dfrac{L_{\Delta ABC}}{L_{\Delta A_1B_1C_1}} =\dfrac{L_{\Delta ABC}}{\frac{4}{5} \cdot L_{\Delta ABC}} =\dfrac{1}{\frac{4}{5}} =1 : \frac{4}{5} = 1 \cdot \frac{5}{4}= \frac{5}{4} = 1,25$

Odp. Skala tego podobieństwa jest równa k = 1,25.

Zad. 8

ΔABC ∼ ΔA₁B₁C₁

k = ⁵/₇

$\frac{L_{\Delta ABC}}{L_{\Delta A_1B_1C_1}} =\frac{5}{7} \\ 5 \cdot L_{\Delta A_1B_1C_1} = 7 \cdot L_{\Delta ABC} \ \ \ |:5 \\ L_{\Delta A_1B_1C_1} = 1,4 \cdot L_{\Delta ABC} \\\\ \frac{L_{\Delta A_1B_1C_1} - L_{\Delta ABC}}{L_{\Delta ABC}} =\frac{1,4 \cdot L_{\Delta ABC} - L_{\Delta ABC}}{L_{\Delta ABC}} =\frac{0,4 \cdot L_{\Delta ABC}}{L_{\Delta ABC}} = 0,4 = 0,4 \cdot 100\% = 40 \%$