Answer:
Parallelograms I, II, and IV
Step-by-step:
Area of parallelograms:
I. A=3*5=15 units squared
II. A=5*3=15 units squared
III. A=4*4=16 units squared
IV. A=5*3=15 units squared
So, parallelograms I, II, and IV have the same area of 15 units squared.
Answer:
Step-by-step explanation:
it would be 7000 because the 7 is greater than five
Answer:
If EFGH is an isosceles trapezoid,
if
EH=4x-27+
FG=x%2B9+
EG=3y%2B19
FH=11y-21
then EH=FG and EG=FH
set up equations:
4x-27=x%2B9 .......solve for x
4x-x=27%2B9
3x=36
x=36%2F3
3y%2B19=11y-21.......solve for y
11%2B19=11y-3y
30=8y
y=30%2F8
y=15%2F4
y=3.75
so,x=12 and y=3.75
you can also check the length of equal sides and equal diagonals:
EH=4%2A12-27+=21
FG=12%2B9+=21
EG=3%283.75%29%2B19=30.25
FH=11%283.75%29-21=30.25x=12
Step-by-step explanation:
1. You must apply the following formula:
LA=<span>1/2 ×(Perimeter of the base)(Slant length)</span>
<span>
</span>
<span> 2. The slant length is (By </span><span> the
Pythagorean Theorem):</span>
<span> a</span><span>^2=b^2+c^2</span>
<span> </span><span>a= b^2+c^2</span>
<span> a</span><span>=</span><span>√(</span><span>1^2+3^2)</span>
<span> </span><span>a=√10</span>
<span>
</span>
<span> 3. </span><span>The p</span><span>erimeter of the base is:</span>
<span>
</span>
<span> </span><span>P=2+2+2+2</span>
<span> </span><span>P=8</span>
<span>
</span>
<span> 4. </span><span>Then, you have:</span>
<span>
</span>
<span> LA=</span><span>1/2 ×(Perimeter of the base)(Slant Length)</span>
<span> LA=</span><span>1/2 ×(8)(</span><span>√</span><span>10)</span>
<span> LA=</span><span>(</span><span>8√10)/2</span>
<span> LA=</span><span>4√10</span>