(1) The sum of three consecutive integers is forty-two, find 2. 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 $$ riangle 1 B C $$ is $$ (1+\sqrt{15}) \mathrm{cm}^{2} $$. Find the length $$ \overline{4 B} $$ $$ (a \sqrt{3}+b \sqrt{5}) $$.

Question

(1) The sum of three consecutive integers is forty-two, find 2. 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 $$ 	riangle 1 B C $$ is $$ (1+\sqrt{15}) \mathrm{cm}^{2} $$. Find the length $$ \overline{4 B} $$ $$ (a \sqrt{3}+b \sqrt{5}) $$.

UpStudy AI · Accepted Answer

1. Let the three consecutive integers be $$ x $$, $$ x+1 $$, and $$ x+2 $$. Their sum is

$$
x + (x+1) + (x+2) = 42.
$$

Combine like terms:

$$
3x + 3 = 42.
$$

Subtract 3 from both sides:

$$
3x = 39,
$$

so

$$
x = 13.
$$

Thus, the three integers are $$ 13 $$, $$ 14 $$, and $$ 15 $$.

2. In the given right‐angled triangle $$\triangle ABC$$, assume the right angle is at $$ C $$ so that the legs are $$ \overline{AC} $$ and $$ \overline{BC} $$. We are given 
$$
AC = \sqrt{3}+\sqrt{5}\quad\text{cm}
$$
and the area
$$
\text{Area} = 1+\sqrt{15}\quad\text{cm}^2.
$$

Since the area of a right triangle is

$$
\frac{1}{2}\times(\text{leg}_1)\times(\text{leg}_2),
$$

we have

$$
\frac{1}{2}\, AC\cdot BC = 1+\sqrt{15}.
$$

Substitute $$ AC = \sqrt{3}+\sqrt{5} $$:

$$
\frac{1}{2} (\sqrt{3}+\sqrt{5})\, BC = 1+\sqrt{15}.
$$

Solve for $$ BC $$:

$$
BC = \frac{2(1+\sqrt{15})}{\sqrt{3}+\sqrt{5}}.
$$

To simplify, multiply the numerator and denominator by the conjugate $$ \sqrt{3}-\sqrt{5} $$:

$$
BC = \frac{2(1+\sqrt{15})(\sqrt{3}-\sqrt{5})}{(\sqrt{3}+\sqrt{5})(\sqrt{3}-\sqrt{5})}.
$$

Since

$$
(\sqrt{3}+\sqrt{5})(\sqrt{3}-\sqrt{5}) = 3-5 = -2,
$$

we obtain

$$
BC = \frac{2(1+\sqrt{15})(\sqrt{3}-\sqrt{5})}{-2} = -(1+\sqrt{15})(\sqrt{3}-\sqrt{5}).
$$

Distribute the negative sign to reverse the order in the second factor:

$$
BC = (1+\sqrt{15})(\sqrt{5}-\sqrt{3}).
$$

Expand the product:

$$
(1+\sqrt{15})(\sqrt{5}-\sqrt{3}) = 1\cdot\sqrt{5} - 1\cdot\sqrt{3} + \sqrt{15}\cdot\sqrt{5} - \sqrt{15}\cdot\sqrt{3}.
$$

Simplify each term:

$$
\sqrt{15}\cdot\sqrt{5} = \sqrt{75} = 5\sqrt{3},\quad \sqrt{15}\cdot\sqrt{3} = \sqrt{45} = 3\sqrt{5}.
$$

Thus,

$$
BC = \sqrt{5} - \sqrt{3} + 5\sqrt{3} - 3\sqrt{5}.
$$

Combine like terms:

$$
BC = (5\sqrt{3} - \sqrt{3}) + (\sqrt{5} - 3\sqrt{5}) = 4\sqrt{3} - 2\sqrt{5}.
$$

So the length $$ \overline{BC} $$ is

$$
4\sqrt{3} - 2\sqrt{5}\text{ cm},
$$

which is in the form $$ a\sqrt{3}+b\sqrt{5} $$ with $$ a=4 $$ and $$ b=-2 $$.
1. The three consecutive integers are 13, 14, and 15. 2. The length of $$ \overline{BC} $$ is $$ 4\sqrt{3} - 2\sqrt{5} $$ cm.

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

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(1) The sum of three consecutive integers is forty-two, find 2. The diagram shows right angled in which the length of is . The area of is . Find the length .