The sum of three consecutive integers is forty-two, find the The diagram shows right angled $$ riangle A B C $$ in which the length of $$ \overline{A C} $$ is $$ (\sqrt{3}+\sqrt{5}) \mathrm{cm} $$. The area of $$ riangle A B C $$ is $$ (1+\sqrt{15}) \mathrm{cm}^{2} $$. Find the length $$ \overline{A B} $$ in the form

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The sum of three consecutive integers is forty-two, find the The diagram shows right angled $$ 	riangle A B C $$ in which the length of $$ \overline{A C} $$ is $$ (\sqrt{3}+\sqrt{5}) \mathrm{cm} $$. The area of $$ 	riangle A B C $$ is $$ (1+\sqrt{15}) \mathrm{cm}^{2} $$. Find the length $$ \overline{A B} $$ in the form

UpStudy AI · Accepted Answer

**1. Sum of Three Consecutive Integers**

Let the integers be $$ n $$, $$ n+1 $$, and $$ n+2 $$. Their sum is
$$
n + (n+1) + (n+2) = 3n + 3.
$$
We are given that the sum is 42, so
$$
3n + 3 = 42.
$$
Subtract 3 from both sides:
$$
3n = 39,
$$
and then divide by 3:
$$
n = 13.
$$
Thus, the three consecutive integers are
$$
13,\quad 14,\quad 15.
$$

---

**2. Right Triangle $$\triangle ABC$$ with $$ \angle A = 90^\circ $$**

The diagram shows a right triangle in which the leg $$\overline{AC}$$ is given by
$$
\sqrt{3}+\sqrt{5}\ \text{cm},
$$
and the area is
$$
1+\sqrt{15}\ \text{cm}^2.
$$
Assume that the right angle is at $$ A $$; hence, $$\overline{AB}$$ and $$\overline{AC}$$ are the perpendicular sides. Let
$$
\overline{AB} = x.
$$
Then the area is given by
$$
\frac{1}{2}\, x\, (\sqrt{3}+\sqrt{5}) = 1+\sqrt{15}.
$$
Multiply both sides by 2 to eliminate the fraction:
$$
x(\sqrt{3}+\sqrt{5}) = 2(1+\sqrt{15}).
$$
Solve for $$ x $$:
$$
x = \frac{2(1+\sqrt{15})}{\sqrt{3}+\sqrt{5}}.
$$

To simplify, rationalize the denominator by multiplying the numerator and denominator by $$\sqrt{5}-\sqrt{3}$$:
$$
x = \frac{2(1+\sqrt{15})(\sqrt{5}-\sqrt{3})}{(\sqrt{3}+\sqrt{5})(\sqrt{5}-\sqrt{3})}.
$$
Since
$$
(\sqrt{3}+\sqrt{5})(\sqrt{5}-\sqrt{3}) = (\sqrt{5})^2 - (\sqrt{3})^2 = 5-3 = 2,
$$
this simplifies to
$$
x = \frac{2(1+\sqrt{15})(\sqrt{5}-\sqrt{3})}{2} = (1+\sqrt{15})(\sqrt{5}-\sqrt{3}).
$$

Thus, the length of $$\overline{AB}$$ is
$$
(\sqrt{5}-\sqrt{3})(1+\sqrt{15})\ \text{cm}.
$$
The three consecutive integers are 13, 14, and 15. The length of $$\overline{AB}$$ is $$(\sqrt{5} - \sqrt{3})(1 + \sqrt{15})$$ cm.

The sum of three consecutive integers is forty-two, find the
The diagram shows right angled in which the
length of is . The area of is
. Find the length in the form

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The sum of three consecutive integers is forty-two, find the The diagram shows right angled in which the length of is . The area of is . Find the length in the form