Chapter 11.8, Problem 1PSA

a.

To determine

To calculate:Thearea of triangle ABC .

Answer to Problem 1PSA

Thearea of triangle ABCis $6$ .

Explanation of Solution

Given information:

In ΔABC

Side $AB=3,$

Side $BC=4,$

Side $AC=5.$

Formula used:

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

  $s=\frac{a+b+c}{2}$

a , b and c are sides of triangle

Area of triangle:

  $A=\sqrt{s(s-a)(s-b)(s-c)}$

Calculation:

  

AB , BC and CA are sides of triangle ABC .

  $s=\frac{AB+BC+CA}{2}=\frac{3+4+5}{2}=6$

Area of triangle ABC :

  $\begin{array}{l}A=\sqrt{s(s-a)(s-b)(s-c)}\\ A=\sqrt{6(6-3)(6-4)(6-5)}\\ A=\sqrt{6\times 3\times 2\times 1}\\ A=\sqrt{36}\\ A=6\end{array}$

b.

To determine

To find: The area of triangle DEF .

Answer to Problem 1PSA

The area of triangle DEF is $2\sqrt{5}$ .

Explanation of Solution

Given information:

In ΔDEF

Side $DE=3,$

Side $EF=3,$

Side $DF=4.$

Formula used:

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

  $s=\frac{a+b+c}{2}$

a , b and c are sides of triangle

Area of triangle:

  $A=\sqrt{s(s-a)(s-b)(s-c)}$

Calculation:

  

DE , EF and DF are sides of triangle DEF .

  $s=\frac{DE+EF+DF}{2}=\frac{3+3+4}{2}=5$

Area of triangle DEF :

  $\begin{array}{l}A=\sqrt{s(s-a)(s-b)(s-c)}\\ A=\sqrt{5(5-3)(5-3)(5-4)}\\ A=\sqrt{5\times 2\times 2\times 1}\\ A=\sqrt{20}\\ A=2\sqrt{5}\end{array}$

c.

To determine

To find: The area of triangle LMN .

Answer to Problem 1PSA

The area of triangle LMN is $10\sqrt{2}$ .

Explanation of Solution

Given information:

In ΔLMN

Side $LM=5,$

Side $MN=6,$

Side $LN=9.$

Formula used:

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

  $s=\frac{a+b+c}{2}$

a , b and c are sides of triangle

Area of triangle:

  $A=\sqrt{s(s-a)(s-b)(s-c)}$

Calculation:

  

LM , MN and LN are sides of triangle LMN .

  $s=\frac{LM+MN+LN}{2}=\frac{5+6+9}{2}=10$

Area of triangle LMN :

  $\begin{array}{l}A=\sqrt{s(s-a)(s-b)(s-c)}\\ A=\sqrt{10(10-5)(10-6)(10-9)}\\ A=\sqrt{10\times 5\times 4\times 1}\\ A=\sqrt{200}\\ A=10\sqrt{2}\end{array}$

d.

To determine

To find: The area of triangle OPQ .

Answer to Problem 1PSA

The area of triangle OPQ is $6\sqrt{3}$ .

Explanation of Solution

Given information:

In ΔOPQ

Side $OP=3,$

Side $PQ=7,$

Side $OQ=8.$

Formula used:

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

  $s=\frac{a+b+c}{2}$

a , b and c are sides of triangle

Area of triangle:

  $A=\sqrt{s(s-a)(s-b)(s-c)}$

Calculation:

  

LM , MN and LN are sides of triangle LMN .

  $s=\frac{OP+PQ+OQ}{2}=\frac{3+7+8}{2}=9$

Area of triangle LMN :

  $\begin{array}{l}A=\sqrt{s(s-a)(s-b)(s-c)}\\ A=\sqrt{9(9-3)(9-7)(9-8)}\\ A=\sqrt{9\times 6\times 2\times 1}\\ A=\sqrt{108}\\ A=6\sqrt{3}\end{array}$

e.

To determine

To find: The area of triangle XYZ .

Answer to Problem 1PSA

The area of triangle XYZ is $60$ .

Explanation of Solution

Given information:

In ΔXYZ

Side $XY=8,$

Side $YZ=15,$

Side $XZ=17.$

Formula used:

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

  $s=\frac{a+b+c}{2}$

a , b and c are sides of triangle

Area of triangle:

  $A=\sqrt{s(s-a)(s-b)(s-c)}$

Calculation:

  

XY , YZ and XZ are sides of triangle XYZ .

  $s=\frac{XY+YZ+XZ}{2}=\frac{8+15+17}{2}=20$

Area of triangle XYZ :

  $\begin{array}{l}A=\sqrt{s(s-a)(s-b)(s-c)}\\ A=\sqrt{20(20-8)(20-15)(20-17)}\\ A=\sqrt{20\times 12\times 5\times 3}\\ A=\sqrt{3600}\\ A=60\end{array}$

f.

To determine

To find: The area of triangle UVW .

Answer to Problem 1PSA

The area of triangle UVW is $84$ .

Explanation of Solution

Given information:

In ΔUVW

Side $UV=13,$

Side $VW=14,$

Side $UW=15.$

Formula used:

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

  $s=\frac{a+b+c}{2}$

a , b and c are sides of triangle

Area of triangle:

  $A=\sqrt{s(s-a)(s-b)(s-c)}$

Calculation:

  

UV , VW and UW are sides of triangle UVW .

  $s=\frac{UV+VW+UW}{2}=\frac{13+14+15}{2}=21$

Area of triangle UVW :

  $\begin{array}{l}A=\sqrt{s(s-a)(s-b)(s-c)}\\ A=\sqrt{21(21-13)(21-14)(21-15)}\\ A=\sqrt{21\times 8\times 7\times 6}\\ A=\sqrt{7056}\\ A=84\end{array}$