In a right-angled triangle $\triangle A B C$, the altitude $A B$ is $5 \mathrm{~cm}$, and the base $B C$ is $12 \mathrm{~cm}$. $P$ and $Q$ are two points on $B C$ such that the areas of $\triangle \mathrm{ABP}, \triangle \mathrm{ABQ}$ and $\triangle \mathrm{ABC}$ are in arithmetic progression. If the area of $\triangle \mathrm{ABC}$ is $1.5$ times the area of $\triangle \mathrm{ABP}$, the length of PQ in cm , is

We have a right-angled triangle ABC where AB (altitude) = 5 cm, BC (base) = 12 cm, and P and Q are points on BC. The areas of triangles ABP, ABQ, and ABC are in arithmetic progression, with area of triangle ABC = 1.5 × area of triangle ABP.

All three triangles ABP, ABQ, and ABC share the same height AB = 5 cm. When triangles have the same height, their areas are directly proportional to their bases.

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Area of triangle ABC = $\dfrac{1}{2} \times 12 \times 5 = 30$ cm²

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Since area of ABC is 1.5 times area of ABP:

Area of triangle ABP = $\dfrac{30}{1.5} = 20$ cm²

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Since area of triangle ABP = $\dfrac{1}{2} \times BP \times 5$:

$20 = \dfrac{1}{2} \times BP \times 5$

$20 = 2.5 \times BP$

$BP = 8$ cm

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The areas form an arithmetic progression: 20, x, 30

In an arithmetic progression, the middle term is the average of the first and last terms:

Area of ABQ = $\dfrac{20 + 30}{2} = 25$ cm²

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Since area of triangle ABQ = $\dfrac{1}{2} \times BQ \times 5$:

$25 = \dfrac{1}{2} \times BQ \times 5$

$25 = 2.5 \times BQ$

$BQ = 10$ cm

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Since both P and Q are on BC:

$PQ = BQ - BP = 10 - 8 = 2$ cm

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Therefore, PQ = 2 cm.