Question

Help please! find the area of the isoceles triangle

Answer 1

Answer:

This is a long answer. I got 82.5

Step-by-step explanation:

Area of isoceles triangle is

$\frac{1}{2} (b)(h)$

where b is the base and h is height.

Let draw a altuide going through Point D that split side FE into 2 equal lines. Let call that point that is equidistant from FE, H.

Since it is a altitude,it forms a right angle. So angle H=90.

Angle H is equidistant from F and E so

FH=11

EH=11.

The height is still unknown.

We can use pythagorean theorem to find side h but we need to know the slanted side or side DF to use the theorem.

Using triangle DFH, we know that angle H is 90 and angle F is 34. So using triangle interior rule,Angle D equal 56.

• We know side FH=11
• We know Angle D equal 56
• We are trying to find side DF
• We know angle H equal 90

We can use law of sines to find side DF

$\frac{fh}{ \sin(d) } = \frac{df}{ \sin(h) }$

Plug in the numbers

$\frac{11}{ \sin(56) } = \frac{df}{ \sin(90) }$

sin of 90 =1 so

$\frac{11}{ \sin(56) } = df$

Side df is about 13.3 inches.

Since we know our slanted side is 13.3 we can set up our pythagorean theorem equation,

$(fh) {}^{2} + (dh) {}^{2} = df {}^{2}$

$(11) {}^{2} +( dh) {}^{2} = (13.3) {}^{2}$

$121 + dh {}^{2} = 176.89$

$(dh) {}^{2} = 55.89$

$\sqrt{55.89} = 7.5$

is approximately 7.5 so dh=7.5 approximately.

Now using base times height times 1/2 multiply them out

$\frac{1}{2} (22)(7.5)$

$82.5$