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Answer 1 · 2018-05-18T22:20:07

Since you know the angles must add up to 180 you have a:

$90, 45, 45$ $90, 45, 45$ triangle

okay, the pathogarean theorem says $a^{2} + b^{2} = c^{2}$ $a^2 + b^2 = c^2$

we know c = 9 and we also know a=b

so: $2 a^{2} = 9^{2}$ $2a^2 = 9^2$

$a^{2} = \frac{9^{2}}{2}$ $a^2 = 9^2/2$

$a = \sqrt{\frac{9^{2}}{2}}$ $a = sqrt(9^2/2)$

$a = \frac{9}{\sqrt{2}}$ $a = 9/sqrt2$

$a = \frac{9 \sqrt{2}}{2}$ $a = (9sqrt2)/2$

$A = \frac{1}{2} b h$ $A=1/2bh$

just turn the triangle so the short sides are the base and height.

$\frac{1}{2} \cdot \frac{9 \sqrt{2}}{2} \cdot \frac{9 \sqrt{2}}{2} = 20.25$ $1/2*(9sqrt2)/2*(9sqrt2)/2 = 20.25$

Answer 2 · 2018-05-18T22:38:46

I tried this:

For problem 26 I got the following:

enter image source here

so the area will be:

$A = \frac{9 \cdot 4.5}{2} = 20.25 {ft}^{2}$ $A=(9*4.5)/2=20.25"ft"^2$

For problem 27:
enter image source here

We know that the total perimeter is 44 units so the A (or B) bit will be:

$A = \frac{44 - 20 - 4}{2} = 10$ $A=(44-20-4)/2=10$ units:
and
$C = \frac{20 - 4}{2} = 8$ $C=(20-4)/2=8$ units

From Pythagoras:

$h = \sqrt{10^{2} - 8^{2}} = 6$ $h=sqrt(10^2-8^2)=6$ units

and the area of the trapezod:

$Area = \frac{1}{2} (20 + 4) \cdot 6 = 72$ $"Area"=1/2(20+4)*6=72$ units squared