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d1i1m1o1n [39]
2 years ago
6

Write an equivalent expression for each question

Mathematics
1 answer:
Mandarinka [93]2 years ago
4 0

Answer:

A. 16 - 4 x   = 4(4 - x)

B. 4 x + 8  = 4( x + 2)

C. -8 b - 24   = -8(b + 3)

D. 54 + 9 x  = 9 (x  + 6)

Step-by-step explanation:

Here, the given expressions are:

A. 16 - 4 x

Now, 16 and 4 have 4 as their COMMON FACTOR.

⇒Taking out 4 as the factor, we get:  16 - 4 x = 4 ( 4 - x)

Hence, the equivalent expression of 16 - 4 x = 4 ( 4 - x)

B. 4x + 8

Now, 4 and 8 have 4 as their COMMON FACTOR.

⇒Taking out 4 as the factor, we get:  4 x  + 8 = 4 ( x + 2)

Hence, the equivalent expression of 4 x  + 8 = 4 ( x + 2)

C. -8b - 24

Now, -8 and -24 have (-8) as their COMMON FACTOR.

⇒Taking out (-8) as the factor, we get:  -8b - 24  = (-8) ( b+ 3)

Hence, the equivalent expression of -8b - 24  = (-8) ( b+ 3)

D. 54 + 9x

Now, 54 and 9  have (9) as their COMMON FACTOR.

⇒Taking out (9) as the factor, we get:  54 + 9 x = 9( 6 + x)

Hence, the equivalent expression of 54 + 9 x = 9( 6 + x)

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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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