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let $b$ and $c$ be the base and hypotenuse of the right triangle, respectively

The area of the triangle is given by the formula: $A=\dfrac{1}{2}bh$ where $b$ = base and $h$ = height

Substituting, we get

$84=\dfrac{1}{2}(7)(b)$ $\large \implies$ $b=24$

By pythagorean theorem, we have

$c^2=7^2+24^2$ $\large \implies$ $c^2=625$ $\large \implies$ $c=\sqrt{625}$ $\large \implies$ $c=25$

Finally the perimeter is

$P=7+24+25=56~m$

From the picture, we can say

$7x=84\times 2$

$x=24$

Using Pythagoras

${ 7 }^{ 2 }+24^{ 2 }=49+576=625$

$Hypoteneuse=\sqrt { 625 } =25$

Therefore the perimeter is $7+24+25=\boxed { 56 }$

Write a solution. Area = x*7/2=84 where x=length of other leg. x=24; hypotenuse=(7^2+24^2)^(1/2) = 625^(1/2) = 25. Perimeter=7+24+25 = 56.

Because the area of the triangle is 84 units, 74=1/2bh. bh=168. 7b=168 b=24. From the basic Pythagorean triple 7+24+25 one gets 56 units as the perimeter.

let $b$ be the base and $h$ be the hypotenuse of the right triangle

The area of a triangle is given by, $A=\dfrac{1}{2} \times base \times height$ .

Substitute the given values,

$84=\dfrac{1}{2}(b)(7)$

$b=24$

Applying pythagorean theorem,

$h=\sqrt{7^2+24^2}=\sqrt{625}=25$

Therefore,

$P=24+7+25=56~m$

84= 1/2 b 7 84*2/7=b b=24. using pythagoras theorem: h^2=7^2 +24^2 h^2=625 h=25 perimeter= sum of all sides. ie 7+24+25=56

(1/2) * 7x = 84 7x = 2 * 84 7x = 168 x = 168/7 = 24 by Pythagoras 7^2+ 24^2 = H^2 H = 625 ^ (1/2) H = 25 perimeter = 24+25+7 = 56

Using the formula of area of right triangle, we can find the length of the other

side of the right angle to be 24

Using the Pythagoras theorem, we can find the length of the hypotenuse to be 25

So the perimeter = 7 + 24 + 25 = 56