User:Afly

About Afly

Some of my asymptote creations

$[asy] pair A=(0,0); pair B=(2,0); pair C=(1,sqrt(3)); draw(A--B--C); draw(A--C--B); [/asy]$

$\text{That is a equilateral triangle with side length 2!}$

$\text{The height is } \sqrt{2^2-1^2} = \sqrt{3}$

$\text{The area is } 2 \cdot \sqrt{3} \cdot \frac{1}{2} = \boxed{\sqrt{3}}$

$\dots$

$[asy] pair A=(0,0); pair B=(0,2); pair C=(1,1); draw(A--B--C); draw(A--C--B); [/asy]$

$\text{Homework: calculate the area}$

ANSWER: LOOK ONLY WHEN THE PROBLEM IS COMPLETED >>> THIS IS ONLY TO CHECK YOUR ANSWER

$\text{This is an isosceles right triangle. It has side length }\sqrt{2}$

$\text{It has an height of } \frac{\sqrt{2}}{\sqrt{2}} = 1 \text{ and a base of } \sqrt{2} \cdot \sqrt{2} = 2$

$\text{The area is } 1 \cdot 2 \cdot \frac{1}{2} = \boxed{1}$ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ $[asy] pair A=(0,0); pair B=(0,3); pair C=(4,0); draw(A--B--C--A,black+linewidth(1)); pair D=(-3,0); pair E=(-3,3); draw(A--B--E--D--A,red+linewidth(1)); pair F=(0,-4); pair G=(4,-4); draw(A--C--G--F--A,orange+linewidth(1)); pair H=(7,4); pair I=(3,7); draw(B--C--H--I--B,yellow+linewidth(1)); draw(A--B--C--A,black+linewidth(1)); [/asy]$ $\text{The pythagorean theorem: red plus orange equals yellow}$ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ $[asy] pair A=(0,0); pair B=(10,0); pair C=(6,8); pair D=(2,8); pair E=(30/7,40/7); draw(A--B--C--D--A,black+linewidth(5)); fill(A--E--D--cycle,red); fill(B--E--C--cycle,yellow); dot(A); dot(B); dot(C); dot(D); dot(E); label(A,'A',SW); label(B,'B',SE); label(C,'C',NE); label(D,'D',NW); label(E,'E',S); draw(A--E--C,blue+linewidth(2)); draw(B--E--D,blue+linewidth(2)); [/asy]$ $\text{Given that } AB \parallel DC \text{ , prove that } A _ \triangle AED = A _ \triangle BEC .$

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User:Afly

About Afly

Some of my asymptote creations