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3 years ago

What is the perimeter of a square whose diagonal is 3 square root 2? Show all work on how you got your answer

Mathematics

1 answer:

Alika [10]3 years ago

6 0

Answer:

Step-by-step explanation:

Given: Diagonal of square= $3\sqrt{2}$

To find the perimeter of square, we need to find the length of sides of square.

∴ Using the formula of diagonal to find side of square.

Formula; $Diagonal= s\sqrt{2}$

Where, s is side of square.

⇒ $3\sqrt{2} = s\sqrt{2}$

Dividing both side by √2

⇒ $s= \frac{3\sqrt{2} }{\sqrt{2} }$

∴ $s= 3$

Hence, Length of side of square is 3.

Now, finding the perimeter of square.

Formula; $Perimeter= 4s$

⇒ $Perimeter= 4\times 3$

∴ $Perimeter= 12$

Hence, Perimeter of square is 12.

You might be interested in

Which are the solutions of the quadratic equation x2 =-5-3

alexdok [17]

Answer: x = 2 • ± √2 = ± 2.8284

Step-by-step explanation:

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 8 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares.

Equation at the end of step 1 :

x2 - 8 = 0

Step 2 :

Solving a Single Variable Equation :

2.1 Solve : x2-8 = 0

Add 8 to both sides of the equation :

x2 = 8

When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:

x = ± √ 8

Can √ 8 be simplified ?

Yes! The prime factorization of 8 is

2•2•2

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 8 = √ 2•2•2 =

± 2 • √ 2

The equation has two real solutions

These solutions are x = 2 • ± √2 = ± 2.8284

Two solutions were found :

x = 2 • ± √2 = ± 2.8284

Not sure what you need help with, but I hope I helped you somehow.

5 0

3 years ago

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$