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IrinaVladis [17]
3 years ago
12

Factor out the greatest common factor 15x^2-21x ^=exponent

Mathematics
1 answer:
gavmur [86]3 years ago
4 0

Answer:

3x

Step-by-step explanation:

3x(5x-7)

brainliest!

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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) 5
Gnesinka [82]

\large\underline{\sf{Solution-}}

Given that,

In <u>triangle TPQ, </u>

  • RS || PQ,

  • RS = 3 cm,

  • PQ = 6 cm,

  • ar(∆ TRS) = 15 sq. cm

As it is given that, <u>RS || PQ</u>

So, it means

⇛∠TRS = ∠TPQ [ Corresponding angles ]

⇛ ∠TSR = ∠TPQ [ Corresponding angles ]

\rm\implies \: \triangle TPQ \:  \sim \: \triangle TRS \:  \:  \:  \:  \:  \:  \{AA \}

<u>Now, We know </u>

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.

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

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

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

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = 4

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

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