Chapter 7.4, Problem 29CYU

To determine

To find: The area of the given triangle.

Answer to Problem 29CYU

  $\frac{29}{2}+3\sqrt{15}$

Explanation of Solution

In the given figure length of base and height of the triangle is given.

Concept Used:

• Area of triangle $=$ $\frac{1}{2}bh$ , where $b$ and $h$ are the base and height of the given triangle respectively. (given in the question)
• Distributive law: It states that the product of sum of two number and a third number is equal to the sum of the products of third number and the two numbers separately. For example $a\left(b+c\right)=ab+ac$ and vice versa.

Calculation:

Area of the given triangle $=\frac{1}{2}bh$

  $\begin{array}{l}=\frac{1}{2}\left(2\sqrt{3}+\sqrt{5}\right)\left(4\sqrt{3}+\sqrt{5}\right)\\ =\frac{1}{2}\left(2\sqrt{3}·4\sqrt{3}+2\sqrt{3}·\sqrt{5}+\sqrt{5}·4\sqrt{3}+\sqrt{5}·\sqrt{5}\right)\\ =\frac{1}{2}\left(2·4·\sqrt{3\times 3}+2·\sqrt{3\times 5}+4·\sqrt{3\times 5}+\sqrt{5\times 5}\right)\\ =\frac{1}{2}\left(8·3+2\sqrt{15}+4\sqrt{15}+5\right)\\ =\frac{1}{2}\left(24+\sqrt{15}\left(2+4\right)+5\right)\\ =\frac{1}{2}\left(24+\sqrt{15}\times 6+5\right)\\ =\frac{1}{2}\left(24+5+6\sqrt{15}\right)\\ =\frac{1}{2}\left(29+6\sqrt{15}\right)\\ =\frac{29}{2}+\frac{6}{2}\sqrt{15}\\ =\frac{29}{2}+3\sqrt{15}\end{array}$

Thus, Area of the given triangle $=\left(\frac{29}{2}+3\sqrt{15}\right)$ square units.