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Expert-verified Found in: Page 664 ### Algebra 1

Book edition Student Edition
Author(s) Carter, Cuevas, Day, Holiday, Luchin
Pages 801 pages
ISBN 9780078884801 # What is the area of the triangle given below? A $\sqrt{\mathbf{2}}\mathbf{+}\mathbf{10}\sqrt{\mathbf{5}}$B $\mathbf{17}\mathbf{+}\mathbf{5}\sqrt{\mathbf{10}}$C $\mathbf{12}\sqrt{\mathbf{2}}\mathbf{+}\mathbf{8}\sqrt{\mathbf{5}}$D $\mathbf{8}\mathbf{.5}\mathbf{+}\mathbf{2}\mathbf{.5}\sqrt{\mathbf{10}}$

The area of the given triangle is $\mathbf{8.5}\mathbf{+}\mathbf{2.5}\sqrt{\mathbf{10}}$ square units.

Option $\mathbit{D}$ is correct.

See the step by step solution

### Step by Step Solution

That is, $a\sqrt{x}+b\sqrt{x}+c\sqrt{x}=\left(a+b+c\right)\sqrt{x}$

## Step 2. State the rule of multiplication of radicals.

Two radicals can be only multiplied if the index of the terms are same. If the index of the terms are same, then the coefficient of the radicals are multiplied together and the radicands are multiplied together under a common radical symbol.

That is, $\left(a\sqrt{x}\right)\left(b\sqrt{y}\right)=\left(a×b\right)\sqrt{x×y}$

## Step 3. State the concept of area of triangle.

Area of a right angle triangle is half of the product of its perpendicular height and base.

Let the length of the perpendicular height be ‘$h$’ and the length of the base be ‘$b$’,

Then the area of the triangle ($A$) is given as:

$Area\left(A\right)=\frac{1}{2}×h×b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

## Step 4. Calculate the area.Observe the figure given below. From the figure,

The height(h) of the triangle is $2\sqrt{2}+\sqrt{5}$ units.

The base($b$) of the triangle is $3\sqrt{2}+\sqrt{5}$ units.

Substitute $2\sqrt{2}+\sqrt{5}$ as $h$ and $3\sqrt{2}+\sqrt{5}$ as $b$ in (1) to get the required area.

$\begin{array}{c}Area\left(A\right)=\frac{1}{2}×h×b\\ =\frac{1}{2}×\left(2\sqrt{2}+\sqrt{5}\right)×\left(3\sqrt{2}+\sqrt{5}\right)\\ =\frac{1}{2}\left[2\sqrt{2}\left(3\sqrt{2}+\sqrt{5}\right)+\sqrt{5}\left(3\sqrt{2}+\sqrt{5}\right)\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Using\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(a+b\right)\left(c+d\right)=a\left(c+d\right)+b\left(c+d\right)\right]\\ \\ =\frac{1}{2}\left[\left(2\sqrt{2}\right)\left(3\sqrt{2}\right)+\left(2\sqrt{2}\right)\left(\sqrt{5}\right)+\left(\sqrt{5}\right)\left(3\sqrt{2}\right)+\left(\sqrt{5}\right)\left(\sqrt{5}\right)\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Using\hspace{0.17em}\hspace{0.17em}}a\left(b+c\right)=a\left(b\right)+a\left(c\right)\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \\ =\frac{1}{2}\left[\left(2×3\right)\left(\sqrt{2}×\sqrt{2}\right)+\left(2\right)\left(\sqrt{2}×\sqrt{5}\right)+\left(3\right)\left(\sqrt{5}×\sqrt{2}\right)+{\left(\sqrt{5}\right)}^{2}\right]\\ =\frac{1}{2}\left[\left(6\right){\left(\sqrt{2}\right)}^{2}+\left(2\right)\left(\sqrt{2×5}\right)+\left(3\right)\left(\sqrt{5×2}\right)+{\left(\sqrt{5}\right)}^{2}\right]\\ =\frac{1}{2}\left[6\left(2\right)+2\left(\sqrt{10}\right)+3\left(\sqrt{10}\right)+5\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Using\hspace{0.17em}\hspace{0.17em}}{\left(\sqrt{x}\right)}^{2}=x\right]\\ \\ =\frac{1}{2}\left[12+2\sqrt{10}+3\sqrt{10}+5\right]\\ =\frac{1}{2}\left[12+5+2\sqrt{10}+3\sqrt{10}\right]\\ =\frac{1}{2}\left[17+\left(2+3\right)\sqrt{10}\right]\\ =\frac{1}{2}\left[17+5\left(\sqrt{10}\right)\right]\\ =\frac{1}{2}\left(17+5\sqrt{10}\right)\\ =\frac{1}{2}\left(17\right)+\frac{1}{2}\left(5\sqrt{10}\right)\\ =8.5+2.5\sqrt{10}\end{array}$

Therefore, the area of the given triangle is $8.5+2.5\sqrt{10}$ square units.

Option $D$ is correct. ### Want to see more solutions like these? 