In a right-angled triangle $$ ABC $$, the altitude $$ AB $$ is 5 cm, and the base $$ BC $$ is 12 cm. $$ P $$ and $$ Q $$ are two points on $$ BC $$ such that the areas of $$ riangle ABP $$, $$ riangle ABQ $$, and $$ riangle ABC $$ are in arithmetic progression. If the area of $$ riangle ABC $$ is 1.5 times the area of $$ riangle ABP $$, the length of $$ PQ $$, in cm, is:

Question

In a right-angled triangle $$ ABC $$, the altitude $$ AB $$ is 5 cm, and the base $$ BC $$ is 12 cm. $$ P $$ and $$ Q $$ are two points on $$ BC $$ such that the areas of $$ 	riangle ABP $$, $$ 	riangle ABQ $$, and $$ 	riangle ABC $$ are in arithmetic progression. If the area of $$ 	riangle ABC $$ is 1.5 times the area of $$ 	riangle ABP $$, the length of $$ PQ $$, in cm, is:

Accepted Answer

![Screenshot 2025-03-30 20.26.31.png](https://res.cloudinary.com/aptidude/image/upload/v1743346600/aptidude_questions/gnvliwhzznorlk6yi2im.png)

#### Step 1: Calculate the area of $$ 	riangle ABC $$
The area of a triangle is given by:  
$$ 	ext{Area} = \frac{1}{2} 	imes 	ext{Base} 	imes 	ext{Height}. $$
For $$ 	riangle ABC $$:  
$$ 	ext{Area} = \frac{1}{2} 	imes 12 	imes 5 = 30~	ext{cm}^2. $$

#### Step 2: Relationship between areas
It is given that the area of $$ 	riangle ABC $$ is **1.5 times** the area of $$ 	riangle ABP $$. Let the area of $$ 	riangle ABP = x $$. Then:  
$$ 1.5x = 30. $$
Solve for $$ x $$:  
$$ x = \frac{30}{1.5} = 20~	ext{cm}^2. $$
Thus, the area of $$ 	riangle ABP = 20~	ext{cm}^2. $$

#### Step 3: Find length of segment BP
The area of a triangle is also given by:  
$$ 	ext{Area} = \frac{1}{2} 	imes AB 	imes BP. $$
For $$ 	riangle ABP $$:  
$$ 20 = \frac{1}{2} 	imes 5 	imes BP. $$
Solve for $$ BP $$:  
$$ BP = \frac{20}{\frac{1}{2} 	imes 5} = 8~	ext{cm}. $$

#### Step 4: Relationship between areas in arithmetic progression
The areas of triangles are in arithmetic progression:  
$$ Ar(	riangle ABP), Ar(	riangle ABQ), Ar(	riangle ABC). $$
Let the area of $$ 	riangle ABQ = y $$. Since the areas are in arithmetic progression:  
$$ y - x = (30 - y). $$
Substitute for $$ x = 20~	ext{cm}^2$$:  
$$ y - 20 = (30 - y). $$
Solve for $$ y $$:  
$$ 2y = 50. $$
$$ y = 25~	ext{cm}^2. $$
Thus, the area of $$ 	riangle ABQ = 25~	ext{cm}^2. $$

#### Step 5: Find length of segment BQ
For triangle $$ ABQ $$:  
$$ Ar(	riangle ABQ) = \frac{1}{2} 	imes AB 	imes BQ. $$
Substitute values:  
$$ 25 = \frac{1}{2} 	imes 5 	imes BQ. $$
Solve for $$ BQ $$:  
$$ BQ = \frac{25}{\frac{1}{2} 	imes 5} = 10~	ext{cm}. $$

#### Step 6: Find length of segment PQ
The length of segment PQ is given by:  
$$ PQ = BQ - BP. $$
Substitute values:  
$$ PQ = 10 - 8 = 2~	ext{cm}. $$

#### Final Answer:
The length of segment **PQ** is **2 cm**.