Chapter 12, Problem 16CR

a.

To determine

To calculate:Thearea of trapezoidUVWX.

Answer to Problem 16CR

The area of trapezoid UVWXis$80$ .

Explanation of Solution

Given information:

SideUV=13,

Side XW= 7,

Side UX = 10.

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Area of trapezoid: $A=\frac{1}{2}\times (b_1+b_2)\times h$

Where

“ $b_1$ ”and“ $b_2$ ”are bases of trapezoid.

“h” is height of trapezoid.

Calculation:

  

Side XY can be calculated by applying Pythagoras Theorem.

In right angled triangle UYX, we get

  $\begin{array}{l}UY=UV-YV\\ YV=XW=7\\ UY=UV-XW=13-7=6\\ {\left(UX\right)}^2={\left(UY\right)}^2+{\left(YX\right)}^2\\ {\left(10\right)}^2={\left(6\right)}^2+{\left(YX\right)}^2\\ {\left(YX\right)}^2=100-36\\ {\left(YX\right)}^2=64\\ YX=\sqrt{64}=8\end{array}$

  $\begin{array}{l}{b}_1=XW=7\\ b_2=UV=13\\ h=YX=8\end{array}$

Area of trapezoid:

  $\begin{array}{l}A=\frac{1}{2}\times (b_1+b_2)\times h\\ A=\frac{1}{2}\times (XW+UV)\times YX\\ A=\frac{1}{2}\times (7+13)\times 8\\ A=\frac{1}{2}\times (20)\times 8\\ A=80\end{array}$

b.

To determine

To find: The area of triangle HEF with side HE as 10.

Answer to Problem 16CR

The area of triangle HEF is $84$ .

Explanation of Solution

Given information:

In ?HEF,

Side HE=10,

Side EF=17,

Side HF=21.

Formula used:

We can use Heron’s Formula to determine the area of a triangle when lengths of sides are given.

The semi perimeter of triangle:

  $s=\frac{a+b+c}{2}$

Where a, b and c are sides of triangle.

Area of triangle:

  $A=\sqrt{s(s-a)(s-b)(s-c)}$

Wheres is semi perimeter of triangle

Calculation:

  

HE,EF and HF are sides of triangle HEF.

  $s=\frac{HE+EF+HF}{2}=\frac{10+17+21}{2}=24$

Area of triangle HEF:

  $\begin{array}{l}A=\sqrt{s(s-a)(s-b)(s-c)}\\ A=\sqrt{24(24-10)(24-17)(24-21)}\\ A=\sqrt{24\times 14\times 7\times 3}\\ A=\sqrt{7056}\\ A=84\end{array}$

c.

To determine

To calculate: The area of inscribed quadrilateral ABCD in circle.

Answer to Problem 16CR

The area of inscribed quadrilateral ABCDis$62.60$ .

Explanation of Solution

Given information:

A quadrilateral ABCD isinscribed in circle.

Side AB = 3,

Side AD = 10,

Side DC =9,

Side BC =12.

Formula used:

Brahmagupta’sformula is used to find the area of any cyclic quadrilateral given the length of sides.

The semi perimeter of quadrilateral is

  $s=\frac{a+b+c+d}{2}$

wherea, b, c and d are sides of quadrilateral.

The area of quadrilateral is

  $A=\sqrt{(s-a)(s-b)(s-c)(s-d)}$

wherea, b, c and d are sides of quadrilateraland s is semi perimeter of quadrilateral.

Calculation:

  

In quadrilateralABCD,

By brahmagupta’sformula, we get

  $s=\frac{AB+BC+CD+AD}{2}=\frac{3+12+9+10}{2}=17$

Area of quadrilateral ABCD:

  $\begin{array}{l}A=\sqrt{(17-3)(17-12)(17-9)(17-10)}\\ A=\sqrt{14\times 5\times 8\times 7}\\ A=\sqrt{3920}\\ A=62.60\end{array}$