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# HELP! PLZ

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A pentagon is drawn by placing an isosceles right triangle on top of a square as pictured. What percent of the area of the pentagon is the area of the right triangle?

Guest Nov 27, 2017

#1
+5587
+3

Let's call the length of each leg of the right triangle  " t " .

And let's call the side length of the square  " s " .

The triangle is an isosceles right triangle, so

t2 + t2  =  s2

2t2  =  s2

And...

area of triangle  =  (1/2)(t * t)   =  t2/2

area of pentagon  =  area of square + area of triangle

area of pentagon  =          s2             +         t2/2                Substitute  2t2  in for  s2 .

area of pentagon  =          2t2            +         t2/2

So...

$$\frac{\text{area of triangle}}{\text{area of pentagon}}=\frac{\frac{t^2}{2}}{2t^2+\frac{t^2}{2}}=\frac{\frac12}{2+\frac12}=\frac12\cdot\frac25=\frac15=\frac{20}{100}$$

The area of the triangle is  20%  of the area of the pentagon.

hectictar  Nov 28, 2017
Sort:

#1
+5587
+3

Let's call the length of each leg of the right triangle  " t " .

And let's call the side length of the square  " s " .

The triangle is an isosceles right triangle, so

t2 + t2  =  s2

2t2  =  s2

And...

area of triangle  =  (1/2)(t * t)   =  t2/2

area of pentagon  =  area of square + area of triangle

area of pentagon  =          s2             +         t2/2                Substitute  2t2  in for  s2 .

area of pentagon  =          2t2            +         t2/2

So...

$$\frac{\text{area of triangle}}{\text{area of pentagon}}=\frac{\frac{t^2}{2}}{2t^2+\frac{t^2}{2}}=\frac{\frac12}{2+\frac12}=\frac12\cdot\frac25=\frac15=\frac{20}{100}$$

The area of the triangle is  20%  of the area of the pentagon.

hectictar  Nov 28, 2017
#2
+2

Let the sides of the isosceles triangle =1, then:

Hypotenuse=sqrt(2)

Area of triangle = [1 x 1] / 2 =1/2

Area of the square =sqrt(2)^2 =2

Area of pentagon =2 + 1/2 =2 1/2 sq. units.

{1/2} / {2 1/2} =1/5 - ratio of area of triangle to the pentagon.

Guest Nov 28, 2017

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