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2 years ago

Pythagoras ' Theorem - Solving Complex Problems

Mathematics

2 answers:

Snezhnost [94]2 years ago

7 0

Answer:

1.you don't need to calculate 25²,7²,30² and 16² because you have a formula a²-b²=(a+b)(a-b)

OP²=MO²-MP²

OP²=25²-7²

OP²=(25+7)(25-7)

=32(8)

=256

OP=√256

OP=16 cm

NP²=NO²-OP²

NP²=30²-16²

=(30+16)(30-16)

=46(14)

=144

NP=√144

=12 cm

2.draw a straight line from the angle C and name that line with CE

CE=AD and AE=CD (because AECD is a rechtangular)

So EB=10-7 =3 cm

BC²=EC²+EB²

=4²+3²

=16+9

=25

BC=√25

=5 cm

for area of the square? i don't understand this question...

Pepsi [2]2 years ago

5 0

Answer:

see explanation

Step-by-step explanation:

using Pythagoras' identity in right triangle MOP to find OP

OP² + MP² = MO²

OP² + 7² = 25²

OP² + 49 = 625 ( subtract 49 from both sides )

OP² = 576 ( take square root of both sides )

OP = $\sqrt{576}$ = 24

Using Pythagoras' identity in right triangle NOP

PN² + OP² = NO²

PN² + 576 = 30²

PN² + 576 = 900 ( subtract 576 from both sides )

PN² = 324 ( take square root of both sides )

PN = $\sqrt{324}$ = 18 cm

----------------------------------------------------------------------------

Consider a vertical line from C to AB parallel to AD

then using Pythagoras' identity in the formed right triangle

BC² = AD² + (10 - 7)²

BC² = 4² + 3² = 16 + 9 = 25 ( take square root of both sides )

BC = $\sqrt{25}$ = 5

then perimeter (P) of ABCD is

P = AB + BC + CD + AD = 10 + 5 + 7 + 4 = 26 cm

The perimeter of the square is therefore 26 cm

with side s = 26 ÷ 4 = 6.5 cm

the area (A) of the square is calculated as

A = s² = 6.5² = 42.25 cm²

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Answer:

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Step-by-step explanation:

Simplify left side by adding 11x+x

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Get x alone on left side by subtracting 5 on both sides

12x=5

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x=5/12

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3 years ago

Prove each statement that is true and ﬁnd a counterexample for each statement that is false. Assume all sets are subsets of a un

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Answer with Step-by-step explanation:

We are given that A, B and C are subsets of universal set U.

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$A\cap (B-C)=(A\cap B)-(A\cap C)$

Proof:

Let x $\in A\cap (B-C)$

Then $x\in A$ and $x\in(B-C)$

When $x\in ( B-C)$ then $x\in B$ but $x\notin C$

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Therefor, $x\in A\cap (B-C)$

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3 years ago

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Answer:

Solution given:

we have

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as rate for 1 print tickets =$1.05/tickets

for t

t=$30+$1.05*t

is a required expression.

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3 years ago

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Answer:

B. Perimeter of a square and

C. Side length of a square

Step-by-step explanation:

if n= side length of square then

Area of square is $n^{2}$
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Thus,

Perimeter of square can be expressed as

$2\sqrt{2}$ ×diagonal length of a square

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$\frac{1}{\sqrt{2} }$ ×diagonal length of a square

but Area of square is

$\frac{1}{\sqrt{2} }$ ×n×diagonal length of a square

As a Result, Area of square is also dependent of the value n, wheras in other cases it is a proportion of diagonal length of a square

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Pythagoras ' Theorem - Solving Complex Problems​

Pythagoras ' Theorem - Solving Complex Problems