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?
+2446 
This is a sketch of the given info. ¯AD is an altitude and ¯AC≅¯AB, and the triangle is right.
Because ¯AC≅¯AB by the given info, the isosceles triangle theorem tells us that the angles opposite of their sides are congruent, so ∠B≅∠C. We can now figure out m∠Bandm∠C.
| m∠B+m∠C+m∠BAC=180∘ | Now, substitute in the known values. |
| m∠B+m∠C+90=180 | m∠B=m∠C |
| m∠B+m∠B+90=180 | Now, solve for both angles. |
| 2m∠B+90=180 | |
| m∠B+45=90 | |
| m∠B=45∘ | |
| m∠B=m∠C=45∘ | |
Since an altitude always bisects the angle of an isosceles triangle, m∠CAD=12∗90=45∘.
Since m∠CAD=m∠C=45∘, by the inverse of the isosceles triangle theorem, AD=DC=4√2
The entire base, however, is 2 times this length, so BC=2∗4√2=8√2. The altitude is 4√2. Now, let's use the area formula for a triangle.
| 12bh=12(8√2)(4√2) | Now, simplify here. |
| 12(32∗(√2)2) | |
| 12(32∗2) | The 1/2 and 2 cancel out here. |
| 32 | This is the area of the triangle. |
The height of the triangle =4sqrt(2)
Area =[4sqrt(2)]^2 / 2 =16 sq. units - the area of triangle
+2446 
I know why the guest got this question incorrect. Let's assume that m∠ADC=90∘, thus making it a right triangle. Let's make the assumption, too, that AD=DC, thus making it an isosceles triangle.
| (4√2)22 | This utilizing the triangle area formula. |
| 16∗(√2)22 | |
| 8∗√22 | |
| 8∗2 | |
| 16units2 | |
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.
+2446
This is a sketch of the given info. ¯AD is an altitude and ¯AC≅¯AB, and the triangle is right.
Because ¯AC≅¯AB by the given info, the isosceles triangle theorem tells us that the angles opposite of their sides are congruent, so ∠B≅∠C. We can now figure out m∠Bandm∠C.
| m∠B+m∠C+m∠BAC=180∘ | Now, substitute in the known values. |
| m∠B+m∠C+90=180 | m∠B=m∠C |
| m∠B+m∠B+90=180 | Now, solve for both angles. |
| 2m∠B+90=180 | |
| m∠B+45=90 | |
| m∠B=45∘ | |
| m∠B=m∠C=45∘ | |
Since an altitude always bisects the angle of an isosceles triangle, m∠CAD=12∗90=45∘.
Since m∠CAD=m∠C=45∘, by the inverse of the isosceles triangle theorem, AD=DC=4√2
The entire base, however, is 2 times this length, so BC=2∗4√2=8√2. The altitude is 4√2. Now, let's use the area formula for a triangle.
| 12bh=12(8√2)(4√2) | Now, simplify here. |
| 12(32∗(√2)2) | |
| 12(32∗2) | The 1/2 and 2 cancel out here. |
| 32 | This is the area of the triangle. |
