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complete the proportion 1) $\frac{rs}{square}=\frac{jr}{kj}$ 2) $\frac{…

Question

complete the proportion

  1. $\frac{rs}{square}=\frac{jr}{kj}$
  2. $\frac{ql}{pm}=\frac{sq}{square}$
  3. $\frac{kl}{lw}=\frac{square}{mw}$
  4. $\frac{kq}{qp}=\frac{ks}{square}$
  5. $\frac{pt}{square}=\frac{tq}{kq}$
  6. $\frac{square}{kp}=\frac{lq}{kq}$
  7. $\frac{c}{a}=\frac{?}{d}$
  8. $\frac{d}{a}=\frac{e}{?}$
  9. $\frac{f}{d + e}=\frac{c}{?}$
  10. $\frac{a + b + c}{a}=\frac{?}{?}$

solve for the variables.
11)
12)
13)
14)
15)
16)

Explanation:

Step1: Use the property of similar - triangles or parallel - lines in a triangle for proportion problems.

For example, in similar triangles, the ratios of corresponding sides are equal.

Step2: Solve problem 11.

Since $\triangle{GKH}\sim\triangle{JKL}$ (by AA similarity or the property of parallel lines in a triangle), we have the proportion $\frac{GK}{JK}=\frac{GH}{JL}$. Given $GK = 6$, $JK=6 + 4=10$, and $JL = 15$. Let $GH=x$, then $\frac{6}{10}=\frac{x}{15}$. Cross - multiply: $10x=6\times15$, so $10x = 90$, and $x = 9$.

Step3: Solve problem 12.

Since $DE\parallel BC$, we have $\frac{AD}{AB}=\frac{DE}{BC}$. Given $AD = 10$, $AB=10 + 40=50$, and $BC = 30$. Let $DE=x$, then $\frac{10}{50}=\frac{x}{30}$. Cross - multiply: $50x=10\times30$, so $x = 6$.

Step4: Solve problem 13.

Since $BD\parallel AE$, we have $\frac{CB}{CA}=\frac{CD}{CE}$. Let $CA=x + 32$, $CB = 32$, $CE=24 + 9=33$, and $CD = 24$. Then $\frac{32}{x + 32}=\frac{24}{33}$. Cross - multiply: $32\times33=24\times(x + 32)$. So $1056=24x+768$. Subtract 768 from both sides: $24x=1056 - 768=288$, and $x = 12$.

Step5: Solve problem 14.

Since the two triangles are similar (by AA similarity or parallel - line property), we have $\frac{x}{6}=\frac{x + 1}{9}$. Cross - multiply: $9x=6(x + 1)$. Expand: $9x=6x+6$. Subtract $6x$ from both sides: $3x=6$, and $x = 2$.

Step6: Solve problem 15.

Since the two triangles are similar, we have $\frac{x + 4}{2x}=\frac{9}{12}$. Cross - multiply: $12(x + 4)=9\times2x$. Expand: $12x+48 = 18x$. Subtract $12x$ from both sides: $6x=48$, and $x = 8$.

Step7: Solve problem 16.

Since the two triangles are similar, we have $\frac{4}{x}=\frac{6}{5}$. Cross - multiply: $6x=4\times5$, so $6x = 20$, and $x=\frac{10}{3}$.

For the proportion - completion problems:

  1. $\frac{RS}{SJ}=\frac{JR}{KJ}$
  2. $\frac{QL}{PM}=\frac{SQ}{SM}$
  3. $\frac{KL}{LW}=\frac{KM}{MW}$
  4. $\frac{KQ}{QP}=\frac{KS}{SP}$
  5. $\frac{PT}{PJ}=\frac{TQ}{KQ}$
  6. $\frac{LJ}{KP}=\frac{LQ}{KQ}$
  7. $\frac{c}{a}=\frac{b}{d}$
  8. $\frac{d}{a}=\frac{e}{b}$
  9. $\frac{f}{d + e}=\frac{c}{a + b}$
  10. $\frac{a + b + c}{a}=\frac{d+e + f}{d}$

Answer:

  1. SJ
  2. SM
  3. KM
  4. SP
  5. PJ
  6. LJ
  7. b
  8. b
  9. a + b
  10. $\frac{d+e + f}{d}$
  11. 9
  12. 6
  13. 12
  14. 2
  15. 8
  16. $\frac{10}{3}$