1. p and q are points on ab and bc respectively of δ abc. pq || ac. ac = 18 cm, ap = 4 cm and pb = 12 cm. find pq.
2. the sides of a quadrilateral are 12, 18, 20, and 16. the longest side of a similar quadrilateral is 6. find the remaining sides.
3. the corresponding sides of two similar figures are 12\ and 21\. find the ratio of their perimeters.
4. a triangle has sides of 6\, 4\ and 7\. the shortest side of a similar triangle is 12\. find the perimeter.
5. in δ abc, p is on ab and q is on ac. ∠apq = ∠c. if ab = 40, ac = 32, and cq = 6, find pb.
6. in a 30° - 60° - 90° triangle, the shortest side is 15\. a similar triangle is four times as large. find the longest side of the similar triangle.
7. a field has the dimensions 375 yd, 425 yd, 275 yd and 300 yd. a plot plan is drawn to scale where 1\ represents 25 yd. find the dimensions of the plot plan.
8. in δ abc, points p and q are on ab and ac respectively. ∠ aqp = ∠ b. if ab = 24\, aq = 8\, cq = 22\, and pq = 12\, find ap, pb, and bc.
9. lines ab and cd intersect at e. ac is drawn parallel to bd. if ae = 9, ec = 15, and be = 12, find de.
10. in δ abc, d is on ab and e is on ac. de || bc. ac = 20 cm, ae = 6 cm, bc = 24 cm, and ad = 8 cm. find de and ab.

Question

Accepted Answer

### Problem 1
# Explanation:
## Step1: Identify similar triangles
Since $ PQ \parallel AC $, by the Basic Proportionality Theorem (Thales' theorem), $ \triangle BPQ \sim \triangle BAC $. So the ratio of corresponding sides is equal.
First, find $ AB = AP + PB = 4 + 12 = 16 $ cm.
## Step2: Set up the proportion
The ratio of $ PB $ to $ AB $ is equal to the ratio of $ PQ $ to $ AC $. So $ \frac{PB}{AB}=\frac{PQ}{AC} $.
Substitute the values: $ \frac{12}{16}=\frac{PQ}{18} $.
## Step3: Solve for $ PQ $
Cross - multiply: $ 16\times PQ = 12\times18 $.
$ 16PQ = 216 $.
$ PQ=\frac{216}{16}=\frac{27}{2} = 13.5 $ cm.
# Answer:
$ 13.5 $ cm

### Problem 2
# Explanation:
## Step1: Find the scale factor
The sides of the first quadrilateral are $ 12',18',20',16' $. The longest side is $ 20' $. The longest side of the similar quadrilateral is $ 6' $. The scale factor $ k=\frac{6}{20}=\frac{3}{10} $.
## Step2: Find the remaining sides
For the side of length $ 12' $: New length $ = 12\times\frac{3}{10}=3.6' $.
For the side of length $ 18' $: New length $ = 18\times\frac{3}{10}=5.4' $.
For the side of length $ 16' $: New length $ = 16\times\frac{3}{10}=4.8' $.
# Answer:
$ 3.6',5.4',4.8' $

### Problem 3
# Explanation:
## Step1: Recall the property of similar figures
For similar figures, the ratio of their perimeters is equal to the ratio of their corresponding sides.
## Step2: Find the ratio of corresponding sides
The corresponding sides are $ 12'' $ and $ 21'' $. The ratio of the sides is $ \frac{12}{21}=\frac{4}{7} $.
Since the ratio of perimeters is equal to the ratio of corresponding sides, the ratio of their perimeters is $ \frac{4}{7} $.
# Answer:
$ \frac{4}{7} $

### Problem 4
# Explanation:
## Step1: Find the scale factor
The sides of the first triangle are $ 6'',4'',7'' $. The shortest side is $ 4'' $. The shortest side of the similar triangle is $ 12'' $. The scale factor $ k = \frac{12}{4}=3 $.
## Step2: Find the perimeter of the first triangle
Perimeter of the first triangle $ P_1=6 + 4+7=17'' $.
## Step3: Find the perimeter of the similar triangle
Since the ratio of perimeters of similar triangles is equal to the scale factor, the perimeter of the similar triangle $ P_2=k\times P_1 $.
$ P_2 = 3\times17 = 51'' $.
# Answer:
$ 51'' $

### Problem 5
# Explanation:
## Step1: Identify similar triangles
Given $ \angle APQ=\angle C $ and $ \angle A=\angle A $ (common angle), so $ \triangle APQ\sim\triangle ACB $.
## Step2: Set up the proportion
Let $ AP = x $, then $ PB=40 - x $. $ AQ=AC - CQ=32 - 6 = 26' $.
From the similarity of triangles, $ \frac{AP}{AC}=\frac{AQ}{AB} $.
Substitute the values: $ \frac{x}{32}=\frac{26}{40} $.
## Step3: Solve for $ AP $
Cross - multiply: $ 40x=32\times26 $.
$ 40x = 832 $.
$ x=\frac{832}{40}=20.8' $.
## Step4: Find $ PB $
$ PB = 40 - 20.8 = 19.2' $.
# Answer:
$ 19.2' $

### Problem 6
# Explanation:
## Step1: Recall the properties of $ 30^{\circ}-60^{\circ}-90^{\circ} $ triangle
In a $ 30^{\circ}-60^{\circ}-90^{\circ} $ triangle, the sides are in the ratio $ 1:\sqrt{3}:2 $, where the shortest side (opposite $ 30^{\circ} $) is $ x $, the longer leg (opposite $ 60^{\circ} $) is $ x\sqrt{3} $ and the hypotenuse (longest side) is $ 2x $.
## Step2: Find the scale factor
The shortest side of the first triangle is $ 15'' $. The similar triangle is four times as large, so the scale factor $ k = 4 $.
## Step3: Find the longest side of the similar triangle
The longest side of the first triangle is $ 2\times15 = 30'' $.
The longest side of the similar triangle is $ 30\times4=120'' $.
# Answer:
$ 120'' $

### Problem 7
# Explanation:
## Step1: Use the scale
The scale is $ 1'' $ represents $ 25 $ yd. So to find the length on the plot plan, we divide each dimension by $ 25 $.
## Step2: Calculate each dimension
For $ 375 $ yd: $ \frac{375}{25}=15'' $.
For $ 425 $ yd: $ \frac{425}{25}=17'' $.
For $ 275 $ yd: $ \frac{275}{25}=11'' $.
For $ 300 $ yd: $ \frac{300}{25}=12'' $.
# Answer:
$ 15'',17'',11'',12'' $

### Problem 8
# Explanation:
## Step1: Identify similar triangles
Given $ \angle AQP=\angle B $ and $ \angle A=\angle A $ (common angle), so $ \triangle AQP\sim\triangle ABC $.
## Step2: Find $ AP $
Let $ AP = x $, $ AB = 24'' $, $ AQ = 8'' $, $ CQ = 22'' $, so $ AC=AQ + CQ=8 + 22 = 30'' $, $ PQ = 12'' $.
From similarity, $ \frac{AP}{AC}=\frac{AQ}{AB}=\frac{PQ}{BC} $.
First, use $ \frac{AP}{30}=\frac{8}{24} $.
Cross - multiply: $ 24AP=30\times8 $.
$ 24AP = 240 $.
$ AP = 10'' $.
## Step3: Find $ PB $
$ PB=AB - AP=24 - 10 = 14'' $.
## Step4: Find $ BC $
Using $ \frac{8}{24}=\frac{12}{BC} $.
Cross - multiply: $ 8BC=24\times12 $.
$ 8BC = 288 $.
$ BC = 36'' $.
# Answer:
$ AP = 10'' $, $ PB = 14'' $, $ BC = 36'' $

### Problem 9
# Explanation:
## Step1: Identify similar triangles
Since $ AC\parallel BD $, $ \triangle AEC\sim\triangle BED $.
## Step2: Set up the proportion
We have $ \frac{AE}{BE}=\frac{EC}{DE} $.
Substitute $ AE = 9' $, $ BE = 12' $, $ EC = 15' $.
So $ \frac{9}{12}=\frac{15}{DE} $.
## Step3: Solve for $ DE $
Cross - multiply: $ 9\times DE=12\times15 $.
$ 9DE = 180 $.
$ DE = 20' $.
# Answer:
$ 20' $

### Problem 10
# Explanation:
## Step1: Identify similar triangles
Since $ DE\parallel BC $, $ \triangle ADE\sim\triangle ABC $.
## Step2: Find $ DE $
The ratio of $ AE $ to $ AC $ is $ \frac{AE}{AC}=\frac{6}{20}=\frac{3}{10} $.
Since the ratio of corresponding sides of similar triangles is equal, $ \frac{DE}{BC}=\frac{AE}{AC} $.
Substitute $ BC = 24 $ cm: $ \frac{DE}{24}=\frac{3}{10} $.
Cross - multiply: $ 10DE=24\times3 $.
$ 10DE = 72 $.
$ DE = 7.2 $ cm.
## Step3: Find $ AB $
The ratio of $ AD $ to $ AB $ is equal to the ratio of $ AE $ to $ AC $. Let $ AB = y $.
$ \frac{AD}{AB}=\frac{AE}{AC} $, $ \frac{8}{y}=\frac{6}{20} $.
Cross - multiply: $ 6y=8\times20 $.
$ 6y = 160 $.
$ y=\frac{160}{6}=\frac{80}{3}\approx26.67 $ cm.
# Answer:
$ DE = 7.2 $ cm, $ AB=\frac{80}{3}\approx26.67 $ cm

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QUESTION IMAGE

Question

Explanation:

Problem 1

Step1: Identify similar triangles

Step2: Set up the proportion

Step3: Solve for \( PQ \)

Step1: Find the scale factor

Step2: Find the remaining sides

Step1: Recall the property of similar figures

Step2: Find the ratio of corresponding sides

Answer:

Problem 2