Question

Angle ABC is isosceles with AB =AC = 10 cm, BC =12 cm, AD perpendicular BC and AD =8cm. find the area of ANGLE ABC. also, find the length of CE​

Answer 1

Explanation:

Since triangle ABC is isosceles, the angles at A and C will be the same, so I will call them theta. Since E is the midpoint of E and AB=12cm, it follows that AE = AB/2 = 6cm. If we draw a line from E to AC that forms a right angle to AC and name that point I, then trigonometry tells us than AI = 6*cos(theta) and EI = 6*sin(theta).

If we draw another line from point B to line AC forming another right angle at AC, and name the point of intersection G, then AG = 12*cos(theta), and also GC = 12*cos(theta), so AC = AG + GC = 24*cos(theta). CD = AC = 24*cos(theta), so the distance AD would be the sum of those two, or 48*cos(theta). Line segment ID, is then AD - AI = 42*cos(theta).

Then, we need to draw a third new line, from point F to point H, again forming a right angle where the line intersects AD. Applying trigonometry to this right triangle, we see that FH/CH = tan(theta). Also, note that FHD is also a right triangle, which is similar to EID, the triangle that we formed earlier, so

FH/HD = EI/ID

Note that HD = HC + CD, and plug in values that we found earlier:

[CH*tan(alpha)]/[CH + 24*cos(alpha)] = [6*sin(alpha)]/[42*cos(alpha)]

Noting that tan(alpha) = sin(alpha)/cos(alpha), and solving for CH yields:

CH = 4*cos(alpha)

Since FCH is still a right triangle, CH = FC* cos(alpha), so substitution tells us that FC = 4. The answer is 4.