a.
To calculate:Thearea of
a.
![Check Mark](/static/check-mark.png)
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=a+b+c2
a , b and c are sides of triangle
Area of triangle:
A=√s(s−a)(s−b)(s−c)
Calculation:
AB , BC and CA are sides of triangle ABC .
s=AB+BC+CA2=3+4+52=6
Area of triangle ABC :
A=√s(s−a)(s−b)(s−c)A=√6(6−3)(6−4)(6−5)A=√6×3×2×1A=√36A=6
b.
To find: The area of triangle DEF .
b.
![Check Mark](/static/check-mark.png)
Answer to Problem 1PSA
The area of triangle DEF is 2√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=a+b+c2
a , b and c are sides of triangle
Area of triangle:
A=√s(s−a)(s−b)(s−c)
Calculation:
DE , EF and DF are sides of triangle DEF .
s=DE+EF+DF2=3+3+42=5
Area of triangle DEF :
A=√s(s−a)(s−b)(s−c)A=√5(5−3)(5−3)(5−4)A=√5×2×2×1A=√20A=2√5
c.
To find: The area of triangle LMN .
c.
![Check Mark](/static/check-mark.png)
Answer to Problem 1PSA
The area of triangle LMN is 10√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=a+b+c2
a , b and c are sides of triangle
Area of triangle:
A=√s(s−a)(s−b)(s−c)
Calculation:
LM , MN and LN are sides of triangle LMN .
s=LM+MN+LN2=5+6+92=10
Area of triangle LMN :
A=√s(s−a)(s−b)(s−c)A=√10(10−5)(10−6)(10−9)A=√10×5×4×1A=√200A=10√2
d.
To find: The area of triangle OPQ .
d.
![Check Mark](/static/check-mark.png)
Answer to Problem 1PSA
The area of triangle OPQ is 6√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=a+b+c2
a , b and c are sides of triangle
Area of triangle:
A=√s(s−a)(s−b)(s−c)
Calculation:
LM , MN and LN are sides of triangle LMN .
s=OP+PQ+OQ2=3+7+82=9
Area of triangle LMN :
A=√s(s−a)(s−b)(s−c)A=√9(9−3)(9−7)(9−8)A=√9×6×2×1A=√108A=6√3
e.
To find: The area of triangle XYZ .
e.
![Check Mark](/static/check-mark.png)
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=a+b+c2
a , b and c are sides of triangle
Area of triangle:
A=√s(s−a)(s−b)(s−c)
Calculation:
XY , YZ and XZ are sides of triangle XYZ .
s=XY+YZ+XZ2=8+15+172=20
Area of triangle XYZ :
A=√s(s−a)(s−b)(s−c)A=√20(20−8)(20−15)(20−17)A=√20×12×5×3A=√3600A=60
f.
To find: The area of triangle UVW .
f.
![Check Mark](/static/check-mark.png)
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=a+b+c2
a , b and c are sides of triangle
Area of triangle:
A=√s(s−a)(s−b)(s−c)
Calculation:
UV , VW and UW are sides of triangle UVW .
s=UV+VW+UW2=13+14+152=21
Area of triangle UVW :
A=√s(s−a)(s−b)(s−c)A=√21(21−13)(21−14)(21−15)A=√21×8×7×6A=√7056A=84
Chapter 11 Solutions
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