In isosceles trapezoid HJKL , . If m = 45 ° , JK = 6 cm, and the length of an altitude is 4 cm, find the exact perimeter of HJKL .
A)(20 + 4 )cm
B)(20 + 8 )cm
C)(20 + 4 )cm
D)(20 + 8 )cm

Question

In isosceles trapezoid HJKL ,        . If m  = 45 ° , JK = 6 cm, and the length of an altitude is 4 cm, find the exact perimeter of HJKL .
A)(20 + 4 )cm 
B)(20 + 8 )cm 
C)(20 + 4 )cm 
D)(20 + 8 )cm

Quizplus · Accepted Answer

We can begin by drawing a diagram of the trapezoid and labeling the given information:

[asy]
pair h,j,k,l,X,Y;
h = (0,0);
j = (1,4);
k = (5,4);
l = (6,0);
X = (1,0);
Y = (5,0);
draw(h--j--k--l--h);
draw(j--X,dashed);
draw(k--Y,dashed);
label("$H$",h,SW);
label("$J$",j,N);
label("$K$",k,N);
label("$L$",l,SE);
label("$X$",X,S);
label("$Y$",Y,S);
label("$6$",j--k,N);
label("$4$",X--j,W);
draw(rightanglemark(j,X,h,8));
[/asy]

We can use the fact that opposite angles in an isosceles trapezoid are congruent to find that $\angle J = \angle K = 67.5^\circ$. Moreover, we have $\angle JXH + \angle KLY = 180^\circ$ since they are supplementary. Thus, we can find that $\angle KLY = 112.5^\circ$.

Since $	riangle JXY$ is a right triangle, we can use trigonometry to find that $JX = 4\cot(45^\circ) = 4$. Similarly, we can find that $KY = 4\cot(112.5^\circ) \approx 1.38$.

Using the Pythagorean Theorem, we can find that $JL = \sqrt{(JK + KY)^2 + 4^2}$. Plugging in our values, we get 
$$JL = \sqrt{(6+1.38)^2 + 4^2} = \sqrt{45.3} \approx 6.72.$$

Finally, the perimeter of the trapezoid is $HL + JK + KL + JL$. Plugging in our values, we get 
$$HL = KL = 2\cdot JX = 2\cdot 4= 8,$$
so the perimeter is 
$$HL + JK + KL + JL = 8+6+8+\sqrt{45.3}= 22+\sqrt{45.3} = 20 + 8\sqrt{2}.$$
Thus, the answer is $\boxed{	extbf{(B)}}$.

In Isosceles Trapezoid HJKL