Mathematics High School

12. In the given figure, RS is parallel to PQ, If RS = 3 cm, PQ = 6 cm and ar(∆TRS) = 15cm³, then ar (∆TPQ) = ? (a) 70 cm² (b) 58 cm² (c) 60 cm² (d) 64 cm²​

12. In The Given Figure, RS Is Parallel To PQ, If RS = 3 Cm, PQ = 6 Cm And Ar(TRS) = 15cm, Then Ar (TPQ)

Answers

Answer 1

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

In triangle TPQ,

RS || PQ,

RS = 3 cm,

PQ = 6 cm,

ar(∆ TRS) = 15 sq. cm

As it is given that, RS || PQ

So, it means

⇛∠TRS = ∠TPQ [ Corresponding angles ]

⇛ ∠TSR = ∠TPQ [ Corresponding angles ]

[tex]\rm\implies \: \triangle TPQ \: \sim \: \triangle TRS \: \: \: \: \: \: \{AA \}[/tex]

Now, We know

Area Ratio Theorem,

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

[tex]\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{ar( \triangle \: TRS)} = \dfrac{ {PQ}^{2} }{ {RS}^{2} } [/tex]

[tex]\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = \dfrac{ {6}^{2} }{ {3}^{2} } [/tex]

[tex]\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = \dfrac{36 }{9} [/tex]

[tex]\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = 4 [/tex]

[tex]\rm\implies \:ar( \triangle \: TPQ) = 60 \: {cm}^{2} [/tex]

12. In The Given Figure, RS Is Parallel To PQ, If RS = 3 Cm, PQ = 6 Cm And Ar(TRS) = 15cm, Then Ar (TPQ)

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Step-by-step explanation:

For some common ratio r, the first of the "geometric means" will be -4r, the next will be -4r^2, and so on. The 5th will be -4r^5, and finally, -62500 will be -4r^6.

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recheck by filling value

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