a. Answer Thearea of triangle ABCis 6 . Explanation Given information: In ΔABC Side A B = 3 , Side B C = 4 , Side A C = 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 + c 2 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 = A B + B C + C A 2 = 3 + 4 + 5 2 = 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 × 1 A = 36 A = 6 b. To determine To find: The area of triangle DEF . Answer The area of triangle DEF is 2 5 . Explanation Given information: In ΔDEF Side D E = 3 , Side E F = 3 , Side D F = 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 + c 2 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 = D E + E F + D F 2 = 3 + 3 + 4 2 = 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 × 1 A = 20 A = 2 5 c. To determine To find: The area of triangle LMN . Answer The area of triangle LMN is 10 2 . Explanation Given information: In ΔLMN Side L M = 5 , Side M N = 6 , Side L N = 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 + c 2 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 = L M + M N + L N 2 = 5 + 6 + 9 2 = 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 × 1 A = 200 A = 10 2 d. To determine To find: The area of triangle OPQ . Answer The area of triangle OPQ is 6 3 . Explanation Given information: In ΔOPQ Side O P = 3 , Side P Q = 7 , Side O Q = 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 + c 2 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 = O P + P Q + O Q 2 = 3 + 7 + 8 2 = 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 × 1 A = 108 A = 6 3 e. To determine To find: The area of triangle XYZ . Answer The area of triangle XYZ is 60 . Explanation Given information: In ΔXYZ Side X Y = 8 , Side Y Z = 15 , Side X Z = 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 + c 2 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 = X Y + Y Z + X Z 2 = 8 + 15 + 17 2 = 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 × 3 A = 3600 A = 60 f. To determine To find: The area of triangle UVW . Answer The area of triangle UVW is 84 . Explanation Given information: In ΔUVW Side U V = 13 , Side V W = 14 , Side U W = 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 + c 2 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 = U V + V W + U W 2 = 13 + 14 + 15 2 = 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 × 6 A = 7056 A = 84

91st Edition

ISBN: 9780866099653

Author: Richard Rhoad, George Milauskas, Robert Whipple

Publisher: McDougal Littell

See similar textbooks

Videos

Question

Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).

Chapter 11.8, Problem 1PSA

To determine

To calculate:Thearea of triangle ABC .

Expert Solution

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:

Geometry For Enjoyment And Challenge, Chapter 11.8, Problem 1PSA , additional homework tip 1

AB , BC and CA are sides of triangle ABC .

$s=AB+BC+CA2=3+4+52=6$

Area of triangle ABC :

$A=s(s−a)(s−b)(s−c)A=6(6−3)(6−4)(6−5)A=6×3×2×1A=36A=6$

To determine

To find: The area of triangle DEF .

Expert Solution

Answer to Problem 1PSA

The area of triangle DEF is $25$ .

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:

Geometry For Enjoyment And Challenge, Chapter 11.8, Problem 1PSA , additional homework tip 2

DE , EF and DF are sides of triangle DEF .

$s=DE+EF+DF2=3+3+42=5$

Area of triangle DEF :

$A=s(s−a)(s−b)(s−c)A=5(5−3)(5−3)(5−4)A=5×2×2×1A=20A=25$

To determine

To find: The area of triangle LMN .

Expert Solution

Answer to Problem 1PSA

The area of triangle LMN is $102$ .

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:

Geometry For Enjoyment And Challenge, Chapter 11.8, Problem 1PSA , additional homework tip 3

LM , MN and LN are sides of triangle LMN .

$s=LM+MN+LN2=5+6+92=10$

Area of triangle LMN :

$A=s(s−a)(s−b)(s−c)A=10(10−5)(10−6)(10−9)A=10×5×4×1A=200A=102$

To determine

To find: The area of triangle OPQ .

Expert Solution

Answer to Problem 1PSA

The area of triangle OPQ is $63$ .

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:

Geometry For Enjoyment And Challenge, Chapter 11.8, Problem 1PSA , additional homework tip 4

LM , MN and LN are sides of triangle LMN .

$s=OP+PQ+OQ2=3+7+82=9$

Area of triangle LMN :

$A=s(s−a)(s−b)(s−c)A=9(9−3)(9−7)(9−8)A=9×6×2×1A=108A=63$

To determine

To find: The area of triangle XYZ .

Expert Solution

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:

Geometry For Enjoyment And Challenge, Chapter 11.8, Problem 1PSA , additional homework tip 5

XY , YZ and XZ are sides of triangle XYZ .

$s=XY+YZ+XZ2=8+15+172=20$

Area of triangle XYZ :

$A=s(s−a)(s−b)(s−c)A=20(20−8)(20−15)(20−17)A=20×12×5×3A=3600A=60$

To determine

To find: The area of triangle UVW .

Expert Solution

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:

Geometry For Enjoyment And Challenge, Chapter 11.8, Problem 1PSA , additional homework tip 6

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 11.7, Problem 22PSC

Chapter 11.8, Problem 2PSA

Chapter 11 Solutions

Geometry For Enjoyment And Challenge

Ch. 11.1 - Prob. 1PSA Ch. 11.1 - Prob. 2PSA Ch. 11.1 - Prob. 3PSA Ch. 11.1 - Prob. 4PSA Ch. 11.1 - Prob. 5PSA Ch. 11.1 - Prob. 6PSA Ch. 11.1 - Prob. 7PSA Ch. 11.1 - Prob. 8PSB Ch. 11.1 - Prob. 9PSB Ch. 11.1 - Prob. 10PSB

Ch. 11.1 - Prob. 11PSB Ch. 11.1 - Prob. 12PSB Ch. 11.1 - Prob. 13PSB Ch. 11.1 - Prob. 14PSB Ch. 11.1 - Prob. 15PSB Ch. 11.1 - Prob. 16PSC Ch. 11.1 - Prob. 17PSC Ch. 11.2 - Prob. 1PSA Ch. 11.2 - Prob. 2PSA Ch. 11.2 - Prob. 3PSA Ch. 11.2 - Prob. 4PSA Ch. 11.2 - Prob. 5PSA Ch. 11.2 - Prob. 6PSA Ch. 11.2 - Prob. 7PSA Ch. 11.2 - Prob. 8PSA Ch. 11.2 - Prob. 9PSA Ch. 11.2 - Prob. 10PSA Ch. 11.2 - Prob. 11PSA Ch. 11.2 - Prob. 12PSA Ch. 11.2 - Prob. 13PSB Ch. 11.2 - Prob. 14PSB Ch. 11.2 - Prob. 15PSB Ch. 11.2 - Prob. 16PSB Ch. 11.2 - Prob. 17PSB Ch. 11.2 - Prob. 18PSB Ch. 11.2 - Prob. 19PSB Ch. 11.2 - Prob. 20PSB Ch. 11.2 - Prob. 21PSB Ch. 11.2 - Prob. 22PSB Ch. 11.2 - Prob. 23PSB Ch. 11.2 - Prob. 24PSB Ch. 11.2 - Prob. 25PSB Ch. 11.2 - Prob. 26PSC Ch. 11.2 - Prob. 27PSC Ch. 11.2 - Prob. 28PSC Ch. 11.2 - Prob. 29PSC Ch. 11.2 - Prob. 30PSC Ch. 11.2 - Prob. 31PSC Ch. 11.2 - Prob. 32PSC Ch. 11.2 - Prob. 33PSC Ch. 11.3 - Prob. 1PSA Ch. 11.3 - Prob. 2PSA Ch. 11.3 - Prob. 3PSA Ch. 11.3 - Prob. 4PSA Ch. 11.3 - Prob. 5PSA Ch. 11.3 - Prob. 6PSA Ch. 11.3 - Prob. 7PSB Ch. 11.3 - Prob. 8PSB Ch. 11.3 - Prob. 9PSB Ch. 11.3 - Prob. 10PSB Ch. 11.3 - Prob. 11PSB Ch. 11.3 - Prob. 12PSB Ch. 11.3 - Prob. 13PSB Ch. 11.3 - Prob. 14PSC Ch. 11.3 - Prob. 15PSC Ch. 11.3 - Prob. 16PSC Ch. 11.3 - Prob. 17PSC Ch. 11.3 - Prob. 18PSC Ch. 11.3 - Prob. 19PSC Ch. 11.3 - Prob. 20PSC Ch. 11.4 - Prob. 1PSA Ch. 11.4 - Prob. 2PSA Ch. 11.4 - Prob. 3PSA Ch. 11.4 - Prob. 4PSB Ch. 11.4 - Prob. 5PSB Ch. 11.4 - Prob. 6PSB Ch. 11.4 - Prob. 7PSB Ch. 11.4 - Prob. 8PSB Ch. 11.4 - Prob. 9PSB Ch. 11.4 - Prob. 10PSC Ch. 11.4 - Prob. 11PSC Ch. 11.4 - Prob. 12PSC Ch. 11.4 - Prob. 13PSC Ch. 11.5 - Prob. 1PSA Ch. 11.5 - Prob. 2PSA Ch. 11.5 - Prob. 3PSA Ch. 11.5 - Prob. 4PSA Ch. 11.5 - Prob. 5PSA Ch. 11.5 - Prob. 6PSA Ch. 11.5 - Prob. 7PSA Ch. 11.5 - Prob. 8PSA Ch. 11.5 - Prob. 9PSA Ch. 11.5 - Prob. 10PSA Ch. 11.5 - Prob. 11PSA Ch. 11.5 - Prob. 12PSB Ch. 11.5 - Prob. 13PSB Ch. 11.5 - Prob. 14PSB Ch. 11.5 - Prob. 15PSB Ch. 11.5 - Prob. 16PSB Ch. 11.5 - Prob. 17PSB Ch. 11.5 - Prob. 18PSB Ch. 11.5 - Prob. 19PSC Ch. 11.5 - Prob. 20PSC Ch. 11.5 - Prob. 21PSC Ch. 11.5 - Prob. 22PSC Ch. 11.5 - Prob. 23PSC Ch. 11.5 - Prob. 24PSC Ch. 11.5 - Prob. 25PSC Ch. 11.5 - Prob. 26PSC Ch. 11.6 - Prob. 1PSA Ch. 11.6 - Prob. 2PSA Ch. 11.6 - Prob. 3PSA Ch. 11.6 - Prob. 4PSA Ch. 11.6 - Prob. 5PSA Ch. 11.6 - Prob. 6PSA Ch. 11.6 - Prob. 7PSA Ch. 11.6 - Prob. 8PSB Ch. 11.6 - Prob. 9PSB Ch. 11.6 - Prob. 10PSB Ch. 11.6 - Prob. 11PSB Ch. 11.6 - Prob. 12PSB Ch. 11.6 - Prob. 13PSB Ch. 11.6 - Prob. 14PSB Ch. 11.6 - Prob. 15PSB Ch. 11.6 - Prob. 16PSB Ch. 11.6 - Prob. 17PSB Ch. 11.6 - Prob. 18PSC Ch. 11.6 - Prob. 19PSC Ch. 11.6 - Prob. 20PSC Ch. 11.6 - Prob. 21PSC Ch. 11.6 - Prob. 22PSC Ch. 11.6 - Prob. 23PSC Ch. 11.7 - Prob. 1PSA Ch. 11.7 - Prob. 2PSA Ch. 11.7 - Prob. 3PSA Ch. 11.7 - Prob. 4PSA Ch. 11.7 - Prob. 5PSA Ch. 11.7 - Prob. 6PSA Ch. 11.7 - Prob. 7PSA Ch. 11.7 - Prob. 8PSA Ch. 11.7 - Prob. 9PSB Ch. 11.7 - Prob. 10PSB Ch. 11.7 - Prob. 11PSB Ch. 11.7 - Prob. 12PSB Ch. 11.7 - Prob. 13PSB Ch. 11.7 - Prob. 14PSB Ch. 11.7 - Prob. 15PSB Ch. 11.7 - Prob. 16PSB Ch. 11.7 - Prob. 17PSB Ch. 11.7 - Prob. 18PSC Ch. 11.7 - Prob. 19PSC Ch. 11.7 - Prob. 20PSC Ch. 11.7 - Prob. 21PSC Ch. 11.7 - Prob. 22PSC Ch. 11.8 - Prob. 1PSA Ch. 11.8 - Prob. 2PSA Ch. 11.8 - Prob. 3PSA Ch. 11.8 - Prob. 4PSB Ch. 11.8 - Prob. 5PSB Ch. 11.8 - Prob. 6PSB Ch. 11.8 - Prob. 7PSB Ch. 11.8 - Prob. 8PSB Ch. 11.8 - Prob. 9PSB Ch. 11.8 - Prob. 10PSB Ch. 11.8 - Prob. 11PSC Ch. 11.8 - Prob. 12PSC Ch. 11.8 - Prob. 13PSC Ch. 11 - Prob. 1RP Ch. 11 - Prob. 2RP Ch. 11 - Prob. 3RP Ch. 11 - Prob. 4RP Ch. 11 - Prob. 5RP Ch. 11 - Prob. 6RP Ch. 11 - Prob. 7RP Ch. 11 - Prob. 8RP Ch. 11 - Prob. 9RP Ch. 11 - Prob. 10RP Ch. 11 - Prob. 11RP Ch. 11 - Prob. 12RP Ch. 11 - Prob. 13RP Ch. 11 - Prob. 14RP Ch. 11 - Prob. 15RP Ch. 11 - Prob. 16RP Ch. 11 - Prob. 17RP Ch. 11 - Prob. 18RP Ch. 11 - Prob. 19RP Ch. 11 - Prob. 20RP Ch. 11 - Prob. 21RP Ch. 11 - Prob. 22RP Ch. 11 - Prob. 23RP Ch. 11 - Prob. 24RP Ch. 11 - Prob. 25RP Ch. 11 - Prob. 26RP Ch. 11 - Prob. 27RP Ch. 11 - Prob. 28RP Ch. 11 - Prob. 29RP Ch. 11 - Prob. 30RP Ch. 11 - Prob. 31RP Ch. 11 - Prob. 32RP Ch. 11 - Prob. 33RP Ch. 11 - Prob. 34RP Ch. 11 - Prob. 35RP Ch. 11 - Prob. 36RP Ch. 11 - Prob. 37RP Ch. 11 - Prob. 38RP Ch. 11 - Prob. 39RP Ch. 11 - Prob. 40RP Ch. 11 - Prob. 41RP Ch. 11 - Prob. 42RP Ch. 11 - Prob. 43RP Ch. 11 - Prob. 44RP Ch. 11 - Prob. 45RP

Additional Math Textbook Solutions

Find more solutions based on key concepts

In parts (a) - (f), use the following figure. a. Solve f( x )=50 . b Solve f( x )=80 . c. Solve f( x )=0 . d. S...

Precalculus

In Exercises 9–22, write the function in the form y = f(u) and u = g(x). Then find dy/dx as a function of x. 15...

University Calculus: Early Transcendentals (4th Edition)

ASSESSMENT Each of the following sequences is either arithmetic or geometric. Identify the sequences 230,193+82...

A Problem Solving Approach To Mathematics For Elementary School Teachers (13th Edition)

Fill in each blanks so that the resulting statement is true. Any set of ordered pairs is called a/an _______. T...

College Algebra (7th Edition)

SAT Test. In Exercises 15–20, assume that random guesses are made for eight multiple choice questions on an SAT...

Elementary Statistics (13th Edition)

The equivalent expression of x(y+z) by using the commutative property.

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

Knowledge Booster

$Geometry$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, geometry and related others by exploring similar questions and additional content below.