Alliterated Area Again

Let $a$ be the side of the triangle and $b$ the side of the square. We know that $3a=4b$ .

Now, the area of the triangle is given by $\dfrac{a^2\sqrt{3}}{4}$ , so $\dfrac{a^2\sqrt{3}}{4}=16\sqrt{3} \Longrightarrow \dfrac{a^2}{4}=16 \Longrightarrow a^2=64$ . Hence, $a=8$ .

We have: $3(8)=4b \Longrightarrow b=\dfrac{3(8)}{4}=6$ .

Finally, the area of the square is $b^2$ , so the area is $6^2=\boxed{36}$ .

Area of Triangle = $(1/2) * Base * Height$

Side of Triangle = X

Height = ( $sqrt{3}$ /2 ) * X

Area = (1/2) * Height * X

Area = $sqrt{3}$ /4 * $X^2$

Solve = $sqrt{3}$ /4 * $X^2$ ­=16 * $sqrt{3}$

we get X= 8

$3 * 8$ = $4 * Side of Square$

Side of Square = 6

$6^2$ = 36 square units.

The area of triangle with side $a$ is given by:

$A_{\triangle} = \dfrac{1}{2} a \left( \dfrac{\sqrt{3}a}{2} \right) = \dfrac{\sqrt{3}a^2}{4}$ .

Since $A_\triangle = 16\sqrt(3)$

$\Rightarrow \dfrac{\sqrt{3}a^2}{4} = 16\sqrt(3)\quad \Rightarrow a^2 = 64$

Now the area of square, $A_\square = \left( \dfrac{3a}{4} \right)^2 = \dfrac{9\times 64}{16} = \boxed{36}$

Area of the triangle : x . 1/2 . x . √3 . 1/2 = 16 . √3

x . x = 64

x = 8

Perimeter of the triangle : 3 . x = 24

3 . 8 = 24

Length of side of the square : 24 . 1/4 = 6

Area of the square : 6 . 6 = 36

Let the side of the equilateral triangle be a.

Put the heron formula of triangle to find area.

Area of triangle = √(s(s-a)(s-b)(s-c)) = 16√3

Where a b c are the sides of triangle since triangle is equilateral then all sides are a.

s=(a+b+c)/2

In this case s = 3a/2

Now

Putting the values

16√3 = √(s〖(s-a)〗^3 )

Now putting value of s in terms of a

16√3 = √(3a/2〖(3a/2-a)〗^3 )

16√3 = √(3a/2 〖(3a/2-a)〗^3 )

16√3 = √(3a/2 〖(a⁄2)〗^3 )

16√3= √(〖3a〗^4/16)

16√3 = a^2/4 √3

a^2 = 16√3 x 4/√3

a = 8

now perimeter of triangle and square is equal

3a = 4x where x = side of square

3a = 8 x 3 = 24

Therefore 24 = 4x

x = 6

area of square = x^2

= 6 x 6 = 36

Area of square is 36

sqrt A root3/4=16 root3 =>sqrt A=16 4 =>A= 4 2= 8

3A=4X =>X=3A/4= 3*8/4=6 =>sqrt X= 36

Let the side of the square be x and that of triangle be y. A/Q 3y=4x. ----------(1) now area of an equilateral triangle is sqrt(3)/4 (side) (side); there fore the value of x can be calculated by using the above formula. it is given that area of triangle is 16sqrt(3); so from this we calculate that y=8; now putting the values in eq 1; we get x=3 8/4; x=6; area of square = (side) (side); hence area=6*6=36;

From equlateral triangle of base 2a say,we can find height of triangle that is from pependicular from vertex to base ,will be underroot 3 a.so area of triangle is underroot 3 X a^2( given as 16 underroot 3..This means perimeter of square is 3a,so each side is 3/4 a.( therefore, side of square is 6 since16 underroot 3 = 9a^2)) so with this area of square is 6 x6=36 square

Area of equilateral triangle =√3/4 a^2=16√3 Therefore a=8 Perimeter of triangle 24=4 side of square Therefore side of square = 6 Area = 6*6=36

Area of triangle is √3/4 a a so equate it to 16√3 we get a as 8 so perimeter equal to 8*3=24equate it to perimeter of square so we get side equal to 6 now area of square=36

The relationship between the height of an equilateral triangle and its side length is important. Taking one of the two right angled triangles that make up this equilateral triangle, we can label the longer leg as $h$ , the shorter leg as $\frac{1}{2}b$ and the hypotenuse as $b$ .

By Pythagoras' theorem:

$(\frac{1}{2}b^{2}) + h^{2} = b^{2}$

$h^{2} = \frac{3}{4}b^{2}$

$h = \frac{\sqrt(3)}{2}b$

The area of a triangle is equal to half times the base times height:

$A_{T} = \frac{1}{2}bh = \frac{1}{2}b(\frac{\sqrt{3}}{2}b) = \frac{\sqrt{3}}{4}b^{2})$

From the information given in the question:

$\frac{\sqrt{3}}{4}b^{2} = 16\sqrt{3}$

$b^{2} = 64$

$b = 8$ or $b = -8$

Reject the latter as a length must be positive

The perimeter of the square is the same as the perimeter of the triangle.

$P_{T} = 8*3 = 24$

$P_{S} = 24$

The side length of the square is the perimeter divided by 4

$L_{S} = 6$

The area of the square is the side length squared

$A_{S} = \boxed{36}$

Area of equilateral triangle= √3÷4×(side)²=16√3 So . side=8cm Perimeter of equilateral triangle = 24 Hence . perimeter of square= 24 So , side of square = 6cm Hence, area of square = side×side = 36 cm²

From the data for area of the triabgle, we get the side of the triangle as 8. Which translates to a perimeter of 24. Therefore the side of the square has to be 6 ( p=4s) .

Area of the square therefore is s^2 I.e. 36

Sea "a" el costado del cuadrado y "b" el costado del triangulo:

Perímetros iguales: 3b = 4a ...(E1)

Área del triangulo= base x altura/2=b x (b x raíz(3)/2)/2 = 16 x raíz(3)

b=8 ...(E2) -->(E1): a=3 x 8/4=6;

Área del cuadrado= a x a = 6 x 6=36.

Let the equilateral triangle be of side length l.

Using the sine formula for areas of triangles,

16√3 = (l^2 sin60˚)/2

Solving this gives l = 8,

Then the length of a side of the square is l*3/4 = 6.

The area of the square is 6^2 = 36.

Let the side of the triangle be t and the side of the square be s

So, we know that 3t=4s

Therefore, t=4/3s

Now, given $\sqrt {3}$ / 4 $t^{2}$ = 16 $\sqrt {3}$

So, t=8

And, t=4/4s , so s=3/4 X 8, i.e, 6

So, Area of square=* $s^{2}$ = 36 *

Because 3a=4b the perimeters must be multiples of 12. Having a perimeter of 12 does not satisfy the second requirement that the area of the equilateral triangle must be 16 x sqrt3. (One can verify this using Heron's Formula.) However a perimeter of 24 satisfies both requirements. 24/4 equals 6, therefore the area of the square is 6 x 6+ 36 units.

16√3=(√3/4)x² x²=64 x=8 So perimeter of triangle = 24 So each side of square = 6 Area of square = 6 × 6 = 3

1)letX be the side of the□and Y for △. you know that 4x=3y.
2)the area of △is ½Y*L(l is the triangle height )
3)L=√(¾Y²) .so a for△ is √(3/4Y²)=16√3
4)Y=8..X=6 ..a for□=36

16 sqrt3 = (sqrt3 /4) a^2 a^2 = 64 a= 8(length of triangle) area of square = [(8*3)/4]^2 =36

A=1/2 a(\sqrt{3a}\frac{2})=\sqrt{3a^{2}}\frac{4} then; A=16\sqrt{3} => \sqrt{3a^{2}}\frac{4}=16\sqrt{3a}=>a^{2}=64 Now The area of a Square is A = (\sqrt{3a}\frac{2})^2=9*64\frac{16}=\boxed{36}

herons formula s=a+b+c+/2

area=sqroot(s(s-a)(s-b)(s-c)) here equilateral triangle so.......s=3c/2.......... solving,,,,,,,,,,,=(16^2) 3=3c/2 1/8*c^3 ....c=8......... perimetr=24.................. square side....=6 ...........area 36

Solution Image

Area of equilateral triangle = √3 / 4 * a^2, where a is the side of triangle Therefore, 16√3 = √3 / 4 * a^2 a = 8

Since perimeters are same, let 'b' be the side of the square hence,
3a = 4b b = 6

Area = b^2 = 36

just guess.common sense goes along way.

if T is the side of the triangle, and S is the side of the square then we get that 3T=4S, also we knoe that the area of a triangle is base height/2, and this equals to 16sqr(3), so we use pythagoras to get that T/2 T/2 sqr(3)=16 sqr(3), so we get that T=4, now we substitue with 3T=4S, which would be S=3T/4, substituting we get 3*8/4 which is S=6, now 6^2=36 WHICH IS THE AREA OF THE SQUARE

let a is the side of equilateral triangle and q is the side of square then, perimeter of Triangle will be 3a and the perimeter of square will be 4q. Now, Area of triangle = sqrt {s (s-a) (s-b) (s-c)} where, s=(a+b+c)/2 Here a=b=c, so s=3a/2 and (s-a) = a/2 , putting these values in above formula we get area of triangle (a^2) (3^1/2)/4 = 16 (3^1/2) given. a=8. In question it is also given that 3a=4q therefore q=6 and area of square = 6*6 =36 Ans.

let 2x be the length of one of the sides of the triangle. our area is 1/2(2x)(x(sqrt(3))), or x^2(sqrt(3))=16sqrt(3). Therefore, x^2 = 16, or x = 4 (since length is positive). The side of the triangle is 8 and the perimeter is 24. Since the square has the same perimeter, the length of one sides is 24/4, or 6. The area, it follows, is 6^2, or our answer of 36.

sqr((3)/4)a^2=16sqr(3)=>a=8=>3a=24=>3a/4=24/4=>x=6=>x^2=36 is the answer

(sqrt(3)/4) ((side)^2)= 16 sqrt(3)==> giving side of equil triangle as 8 units.==>perimeter =8*3=24 units. but this 24 equals perimeter of the given square giving side of square as (24)/4= 6 units... Hence area of square = 6^2= 36 sq. units... :)

Let x be the side of the triangle and y the side of the square.

3x=4y --------------a

and (sqroot of 3* x^2 )/4=16sq.root3

therefor x=8

a implies y=6 and

hence area of square is 6^2=36

S(Triangle) = sqrt(p (p-a) (p-b) (p-c)) p = (a+b+c)/2 a=b=c(equilateral) S(Triangle) = [a^2 sqrt(3)]/4 16 sqrt(3) = [a^2 sqrt(3)]/4 a = 8

a+b+c = 4x (triangle perimeter = square perimeter) 3a = 4x x = 6

x^2 = 36 (square area)

Let Lt, Ls, Pt, Ps, At, As, rispectively:

the side of the equilater triangle, the side of the square, the

perimeter of the triangle, the perimeter of the square, the

area of the triangle, the area of the square; we have:

At = 16√3; Pt = Ps, As = ? (question);

Pt = Ps, => 3Lt = 4Ls;

At = 16√3 = (1/2)Lt^2√3/2 = √3Lt^2/4;

Lt^2/4 = 16; Lt^2 = 64; Lt = 8;

Ls = (3/4)Lt = 3*8/4 = 6; As = 6^2 = 36