$$ riangle C A B $$ is right-angled at $$ A $$, $$ E \in \overline{\mathrm{BC}} $$ such that $$ \overline{\mathrm{AE}} \perp \overline{\mathrm{BC}} $$, D is the midpoint of $$ \overline{\mathrm{BC}}, \mathrm{AE}=9.6 \mathrm{~cm} $$. and $$ \mathrm{BC}=20 \mathrm{~cm} $$. Find : The length of each of $$ \overline{\mathrm{AB}} $$ and $$ \overline{\mathrm{AC}} $$

Question

$$ 	riangle C A B $$ is right-angled at $$ A $$, $$ E \in \overline{\mathrm{BC}} $$ such that $$ \overline{\mathrm{AE}} \perp \overline{\mathrm{BC}} $$, D is the midpoint of $$ \overline{\mathrm{BC}}, \mathrm{AE}=9.6 \mathrm{~cm} $$. and $$ \mathrm{BC}=20 \mathrm{~cm} $$. Find : The length of each of $$ \overline{\mathrm{AB}} $$ and $$ \overline{\mathrm{AC}} $$

UpStudy AI · Accepted Answer

We start with a right triangle $$\triangle CAB$$ with the right angle at $$A$$. Let the legs be $$\overline{AB}$$ and $$\overline{AC}$$ and the hypotenuse be $$\overline{BC}$$ with
$$
BC = 20\text{ cm}.
$$

An altitude from $$A$$ to $$\overline{BC}$$ meets $$BC$$ at $$E$$ and has length
$$
AE = 9.6\text{ cm}.
$$

In any right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments (say, $$BE$$ and $$EC$$) which satisfy
$$
AE^2 = BE \times EC.
$$
Let
$$
BE = x \quad \text{and} \quad EC = 20-x.
$$
Then
$$
9.6^2 = x(20-x).
$$
Calculating,
$$
92.16 = x(20-x).
$$
This gives the quadratic equation
$$
x^2 -20x + 92.16 = 0.
$$

The discriminant of this quadratic is
$$
\Delta = 20^2 - 4(1)(92.16) = 400 - 368.64 = 31.36,
$$
so
$$
\sqrt{\Delta} = 5.6.
$$

Thus the solutions are
$$
x = \frac{20 \pm 5.6}{2}.
$$
This leads to:
$$
x = \frac{25.6}{2} = 12.8 \quad \text{or} \quad x = \frac{14.4}{2} = 7.2.
$$

It does not matter which segment is labeled $$BE$$ or $$EC$$ because the legs will simply swap. The key extra property in a right triangle is that each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to the leg. That is, we have:
$$
AB^2 = BC \times BE \quad \text{and} \quad AC^2 = BC \times EC.
$$

Taking the case $$BE = 7.2$$ and $$EC = 12.8$$, we obtain:
$$
AB^2 = 20 \times 7.2 = 144 \quad \Longrightarrow \quad AB = \sqrt{144} = 12\text{ cm},
$$
$$
AC^2 = 20 \times 12.8 = 256 \quad \Longrightarrow \quad AC = \sqrt{256} = 16\text{ cm}.
$$

Thus, the lengths of the legs are:
$$
\overline{AB} = 12\text{ cm} \quad \text{and} \quad \overline{AC} = 16\text{ cm}.
$$

Alternatively, if the segments are interchanged (i.e. $$BE=12.8$$ and $$EC=7.2$$), then the leg lengths would be swapped, but the triangle would still have legs of 12 cm and 16 cm.
The lengths of the legs are $$ \overline{AB} = 12\text{ cm} $$ and $$ \overline{AC} = 16\text{ cm} $$.

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Geometry Questions