Question

in figure RP : P K = 3:2 then find A(ΔTRP):A(ΔTPK)

Answer 1

Answer:Extend the segment RK to line l and draw perpendicular from point T to the line l at M as shown in the figure

TM becomes height of ∆TRP and ∆TPK

Area of triangle = (1/2) × base × height

A(∆TRP) = (1/2) × RP × TM

A(∆TPK) = (1/2) × PK × TM

∴ using (i)

Hence A(∆TRP): A(∆TPK) = 3:2

Step-by-step explanation:Extend the segment RK to line l and draw perpendicular from point T to the line l at M as shown in the figure

TM becomes height of ∆TRP and ∆TPK

Area of triangle = (1/2) × base × height

A(∆TRP) = (1/2) × RP × TM

A(∆TPK) = (1/2) × PK × TM

∴ using (i)

Hence A(∆TRP): A(∆TPK) = 3:2

Answer 2

The required ratio is 9:4.

Step-by-step explanation:

Since we have given that

RP : PK = 3:2

So, we can rewrite it as $\dfrac{RP}{PK}=\dfrac{3}{2}$

We need to find the ratio of Ar(ΔTRP) : Ar(ΔTPK)

As we know the Area similarity formula:

So, we get that

$\dfrac{Ar(\Delta TRP)}{Ar(\Delta TPK)}=\dfrac{RP^2}{PK^2}=\dfrac{3^2}{2^2}=\dfrac{9}{4}$

Hence, the required ratio is 9:4.

# learn more:

In figure RP : P K = 5:3 o 3 then A(ΔTRP):A(ΔTPK)

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