Question

Question #1b82b

Answer 1

Tina would need a minimum of `54`in of wood for DFG, dependent on its thickness. Thicker frames would need longer lengths to accommodate the three corners that project beyond the art pieces.

Explanation:

The two triangles are `similar` because they both share a common angle, and the given sides of one are similar multiples of the given sides of the other.

Then, by the properties of similar triangles:

Since side `DF = 3*AB, and BC=3*FG, then DG = 3*AC`

So `DG = 3*7`in `= 21`in.

Perimeter of #DFG = 15+18+21 = 54in.
This is the correct answer and a correct method for the solution.

There is also a method of solving the larger triangle with a formula to find the value of the common angle using the three given sides of the smaller triangle:

`cosB = (c^2+a^2-b^2)/(2ca) = (5^2+6^2-7^2)/(2xx6*5) = 0.2`

By the way that angle is `78.5degrees`.

Then we can apply the same formula to the larger triangle:

`cosB = (c^2+a^2-b^2)/(2ca) = (15^2+18^2-b^2)/(2xx18*15) = 0.2`

`(15^2+18^2-b^2)/(2xx18*15) = 0.2`

`b = 21 = DG`

Answer checked!

Answer 2

`131.76` `"in"^2`

Explanation:

When we see the triangles, we can say that the triangles are similar to each other. The corresponding sides are in multiples of their common sides.

We can say that,

`color(orange)("AB"xx3="DF"`

`color(orange)("BC"xx3="FG"`

So,

`color(orange)("DG"=7xx3=21`

Now we know the three sides of the triangle.We can find the area of the triangle using the Heron's formula

`color(blue)("Area"=sqrt(s(s-a)(s-b)(s-c))`

Where, `a,b and c` are the sides and `s=(a+b+c)/2`

`color(blue)(s=(15+18+21)/2=27`

Let's apply the values to the formula

`rarrsqrt(27(27-15)(27-18)(27-21))`

`rarrsqrt(27(12)(9)(6))`

Split into prime factors

`rarrsqrt(underbrace(3*3)*3*underbrace(2*2)*underbrace(3*3)*underbrace(3*3)*2)`

`rarr3*3*3*2sqrt(3*2)`

`color(green)(rArr54sqrt6`

`color(green)(~~131.76`

So, the minimum area of wood required is `131.76` `"in"^2`

Hope this helps...! :)