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In a certain isosceles right triangle, the altitude to the hypotenuse has length 4 times the square root of 2. What is the area of the triangle?

Guest Nov 28, 2017

Best Answer 

 #4
avatar+1493 
+2

 

This is a sketch of the given info. \(\overline{AD}\) is an altitude and \(\overline{AC}\cong\overline{AB}\), and the triangle is right.

 

Because \(\overline{AC}\cong\overline{AB}\) by the given info, the isosceles triangle theorem tells us that the angles opposite of their sides are congruent, so \(\angle B\cong\angle C\). We can now figure out \(m\angle B\quad\text{and}\quad m\angle C\).

 

\(m\angle B+m\angle C+m\angle BAC=180^{\circ}\)Now, substitute in the known values.
\(m\angle B+m\angle C+90=180\)\(m\angle B=m\angle C\)
\(m\angle B+m\angle B+90=180\)Now, solve for both angles. 
\(2m\angle B+90=180\) 
\(m\angle B+45=90\) 
\(m\angle B=45^{\circ}\) 
\(m\angle B=m\angle C=45^{\circ}\) 
  

 

Since an altitude always bisects the angle of an isosceles triangle, \(m\angle CAD=\frac{1}{2}*90=45^{\circ}\)

 

Since \(m\angle CAD=m\angle C=45^{\circ}\), by the inverse of the isosceles triangle theorem, \(AD=DC=4\sqrt{2}\)

 

The entire base, however, is 2 times this length, so \(BC=2*4\sqrt{2}=8\sqrt{2}\). The altitude is \(4\sqrt{2}\). Now, let's use the area formula for a triangle.

 

\(\frac{1}{2}bh=\frac{1}{2}(8\sqrt{2})(4\sqrt{2})\)Now, simplify here.
\(\frac{1}{2}(32*\left(\sqrt{2}\right)^2)\) 
\(\frac{1}{2}(32*2)\)The 1/2 and 2 cancel out here.
\(32\)This is the area of the triangle.
  
TheXSquaredFactor  Nov 28, 2017
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5+0 Answers

 #1
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+1

The height of the triangle =4sqrt(2)

Area =[4sqrt(2)]^2 / 2 =16 sq. units - the area of triangle

Guest Nov 28, 2017
edited by Guest  Nov 28, 2017
 #2
avatar
+1

is the answer for sure 8 square units? for sure?

Guest Nov 28, 2017
 #3
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+1

this is an important problem. anyone else get the same answer?

Guest Nov 28, 2017
 #5
avatar+1493 
+2

 

I know why the guest got this question incorrect. Let's assume that \(m\angle ADC=90^{\circ}\), thus making it a right triangle. Let's make the assumption, too, that \(AD=DC\), thus making it an isosceles triangle. 

 

\(\frac{\left(4\sqrt{2}\right)^2}{2}\)This utilizing the triangle area formula.
\(\frac{16*(\sqrt{2})^2}{2}\) 
\(8*\sqrt{2}^2\) 
\(8*2\) 
\(16units^2\) 
  

 

However, there is a key word. That word would happen to be "altitude." An altitude is a perpendicular height that extends from a vertex to the opposite side. Notice how the height of this triangle does not do this. Therefore, the original diagram does not fit the original criteria. 

TheXSquaredFactor  Nov 28, 2017
 #4
avatar+1493 
+2
Best Answer

 

This is a sketch of the given info. \(\overline{AD}\) is an altitude and \(\overline{AC}\cong\overline{AB}\), and the triangle is right.

 

Because \(\overline{AC}\cong\overline{AB}\) by the given info, the isosceles triangle theorem tells us that the angles opposite of their sides are congruent, so \(\angle B\cong\angle C\). We can now figure out \(m\angle B\quad\text{and}\quad m\angle C\).

 

\(m\angle B+m\angle C+m\angle BAC=180^{\circ}\)Now, substitute in the known values.
\(m\angle B+m\angle C+90=180\)\(m\angle B=m\angle C\)
\(m\angle B+m\angle B+90=180\)Now, solve for both angles. 
\(2m\angle B+90=180\) 
\(m\angle B+45=90\) 
\(m\angle B=45^{\circ}\) 
\(m\angle B=m\angle C=45^{\circ}\) 
  

 

Since an altitude always bisects the angle of an isosceles triangle, \(m\angle CAD=\frac{1}{2}*90=45^{\circ}\)

 

Since \(m\angle CAD=m\angle C=45^{\circ}\), by the inverse of the isosceles triangle theorem, \(AD=DC=4\sqrt{2}\)

 

The entire base, however, is 2 times this length, so \(BC=2*4\sqrt{2}=8\sqrt{2}\). The altitude is \(4\sqrt{2}\). Now, let's use the area formula for a triangle.

 

\(\frac{1}{2}bh=\frac{1}{2}(8\sqrt{2})(4\sqrt{2})\)Now, simplify here.
\(\frac{1}{2}(32*\left(\sqrt{2}\right)^2)\) 
\(\frac{1}{2}(32*2)\)The 1/2 and 2 cancel out here.
\(32\)This is the area of the triangle.
  
TheXSquaredFactor  Nov 28, 2017

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