Chapter 11, Problem 1RP

a.

To determine

To calculate:Thearea of rectangle ABCD with base BC as $12$ and height AB as $7$ .

Answer to Problem 1RP

Thearea of rectangle is $84$ .

Explanation of Solution

Given information:

In rectangle ABCD, Side AB = $7$ ,

Side BC = $12$ .

Formula used:

Area of rectangle: $A=l\times w$

l = length of rectangle

w= width of triangle

Calculation:

  

Area of rectangle:

  $\begin{array}{l}A(\square ABCD)=BC\times AB\\ A(\square ABCD)=12\times 7=84\end{array}$

b.

To determine

To find: The area of triangle PQRwith base QR as $12$ and height PD as $7$ .

Answer to Problem 1RP

The area of triangle is $42$ .

Explanation of Solution

Given information:

In trianglePQR, Side PD = $7$ ,

Side QR = $12$ .

Formula used:

Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

Area of triangle:

  $\begin{array}{l}A(△PQR)=\frac{1}{2}\times QR\times PD\\ A(△PQR)=\frac{1}{2}\times 12\times 7=42\end{array}$

c.

To determine

To calculate: The area of parallelogram PQRSwith base QR as $15$ and height PQ as $5$ .

Answer to Problem 1RP

The area of rectangle is $75$ .

Explanation of Solution

Given information:

In rectangle PQRS, Side PQ = $5$ ,

Side QR = $15$ .

Formula used:

Area of parallelogram: $A=l\times w$

l = length of parallelogram

w= width of parallelogram

Calculation:

  

Area of parallelogram:

  $\begin{array}{l}A(▱PQRS)=QR\times PQ\\ A(▱PQRS)=15\times 5=75\end{array}$

d.

To determine

To find: The area of trapezoid UVWXwith basesUXas $3$ , VW as $10$ and heightUAas $8$ .

Answer to Problem 1RP

The area of trapezoid UVWX is $52$ .

Explanation of Solution

Given information:

A trapezoid UVWXwith bases UX as $3$ , VW as $10$ and heightUAas $8$ .

Formula used:

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

b = base of trapezoid

h = height of trapezoid

Calculation:

  

Area of Trapezoid:

  $\begin{array}{l}A(UVWX)=\frac{1}{2}\times \left(UX+VW\right)\times h\\ A(UVWX)=\frac{1}{2}\times \left(3+10\right)\times 8=52\end{array}$

e.

To determine

To find: The area of kite KITEwith diagonals IE as $5$ and KT as $8$ .

Answer to Problem 1RP

The area of kite KITEis $20$ .

Explanation of Solution

Given information:

A kite KITEwith diagonals IE as $5$ and KT as $8$ .

Formula used:

Area of kite $=\frac{1}{2}(d_1\times d_2)$

d = diagonal of kite.

Calculation:

  

Area of kite:

  $\begin{array}{l}A(KITE)=\frac{1}{2}\times IE\times KT\\ A(KITE)=\frac{1}{2}\times 5\times 8=20\end{array}$

f.

To determine

To find: The area of trapezoid UVWXwith median MN as $4$ and height UAas $2$ .

Answer to Problem 1RP

The area of trapezoid UVWX is $8$ .

Explanation of Solution

Given information:

A trapezoid UVWXwith median MN as $4$ and height UAas $2$ .

Formula used:

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

b = base of trapezoid

h = height of trapezoid Median of trapezoid:

Median $=\frac{1}{2}\times (b_1+b_2)$

b = base of trapezoid

Calculation:

  

A median or mid-segment of a trapezoid is the line segment connecting the midpoints of the two non-parallel sides of a trapezoid.

Median $\overline{MN}=\frac{1}{2}\times (UV+XW)$

  $\overline{MN}$ =4.

  $\begin{array}{l}A(UVWX)=\frac{1}{2}\times \left(UX+VW\right)\times h\\ A(UVWX)=4\times 2=8\end{array}$