Question

Determine the value of $a$. [asy] pair w=(0,4); pair x=(0,0); pair y=(4,0); pair z=y+7/sqrt(2)*(1,1); dot(w); dot(x); dot(y); dot(z); draw(w--x--y--z--w); draw(0.15*w--0.15*w+0.15*y--0.15*y); label("$W$",w,NNW); label("$X$",x,SW); label("$Y$",y,SE); label("$Z$",z,E); label("4",(w+x)/2,W); label("4",(x+y)/2,S); label("9",(w+z)/2,NNW); label("$a$",(y+z)/2,SE); label("$135^\circ$",y,NNW); [/asy]

Answer 1

Answer:

a=7

Step-by-step explanation:

The image is rendered and attached below.

Triangle WXY is an Isosceles right triangle, since WX=XY.

First, we determine the length of WY using Pythagoras Theorem.

$WY=\sqrt{4^2+4^2}\\WY=\sqrt{32}$

Since triangle WXY is Isosceles, $\angle XYW=45^\circ$

$\angle XYZ=\angle XYW+\angle WYZ\\135^\circ=45^\circ+\angle WYZ\\\angle WYZ=135^\circ-45^\circ=90^\circ$

Therefore:

Triangle WYZ is a right triangle with WZ as the hypothenuse.

Applying Pythagoras Theorem

$WZ^2=WY^2+YZ^2\\9^2=(\sqrt{32})^2+a^2\\a^2=81-32\\a^2=49\\a^2=7^2\\ Therefore: a=7$