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3 years ago

Please help me...

1 answer:

7 0

A system of linear equations with no solution means that the lines never meet. So as long as your two lines have the same slope and different intercepts, they will be parallel.

Ex.) y= 2x + 5

and y=2x + 10

to have a solution, the lines have to cross at a point. so literally any two lines that aren't parallel.

Ex.) y= 3x + 1

to find the solution you start by setting the equations equal to each other

2x + 5 = 3x + 1

1) subtract 2x from both sides

5= x + 1

2) subtract 1 from both sides

4 = x

now we know the x value of the point is 4. To find the y value plug 4 into one of the equations

2(4) + 5

8 + 5

so the solution to this particular system is (4,13)

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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 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 ]

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

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.

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

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

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

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

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

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Answer:

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Step-by-step explanation:

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