Answer:
1) m∠A=60°
2) [tex]BK=3\sqrt{3}[/tex]
3) The area of ABCD is 62.35
Step-by-step explanation:
1) - You can calculate the angle m∠A as following:
[tex]cos^{-1}(\alpha })=\frac{adjacent}{hypotenuse}\\adjacent=3\\hypotenuse=6\\[/tex]
- Substitute values and solve for the angle. Let's call the angle m∠A "[tex]\alpha[/tex]"
[tex]cos^{-1}(\alpha )=\frac{3}{6}\\ \alpha =60[/tex]
2) Apply the Pythagorean Theorem to find BK:
[tex]BK=\sqrt{6^{2}-3^{2}}=3\sqrt{3}[/tex]
3) - If you analize the figure, you can see that the triangle ABD and the triangle BCD are the same triangle, both are equal, therefore both have the same area.
- Then, you can calculate the area of ABD and multiply it by 2 to calculate the area of ABCD. The area of a triangle is:
[tex]Area=\frac{(base)(height)}{2}[/tex]
- The area ABD will be the sum of the area of the triangle ABK and the area of the triangle KBD.
- First, you must calculate the angle m∠D using the triangle ABD, which is a right triangle. The sum of the interior angles of a triangle is 180°. You know the angles m∠A (60°) and m∠B (90°), therefore:
[tex]60+90+D=180\\D=30[/tex]
m∠D=30°
- Now, you can calculate KD:
[tex]tan(30)=\frac{3\sqrt{3}}{KD}\\KD=9[/tex]
- The area of ABCD is:
[tex]A_{ABCD}=2(A_{ABK}+A_{KBD} )[/tex]
[tex]A_{ABCD}=2(\frac{(3)(3\sqrt{3})}{2}+\frac{(9)3\sqrt{3}}{2})\\A_{ABCD}=62.35[/tex]
