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# help urgent!

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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

#4
+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
Sort:

#1
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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
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is the answer for sure 8 square units? for sure?

Guest Nov 28, 2017
#3
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this is an important problem. anyone else get the same answer?

Guest Nov 28, 2017
#5
+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
+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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