Geometry For Enjoyment And Challenge
Geometry For Enjoyment And Challenge
91st Edition
ISBN: 9780866099653
Author: Richard Rhoad, George Milauskas, Robert Whipple
Publisher: McDougal Littell
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Chapter 11, Problem 28RP

a.

To determine

To find: The ratio of theareas of region I to that of region II.

a.

Expert Solution
Check Mark

Answer to Problem 28RP

The ratio of the area of region I to that of region II is 16:81.

Explanation of Solution

Given Information:

Sides of the quadrilateral are

  a=8,b=9,c=18andd=10

Formula used:

Ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Calculation:

Sides of the quadrilateral ABCD are

  AB=8,BC=9,CD=18andDA=10

Let the diagonals intersect at point O

  Geometry For Enjoyment And Challenge, Chapter 11, Problem 28RP , additional homework tip  1

In ΔAOBandΔCOD,

  AOB=COD … (Vertically opposite angles)

  OAB=OCD … (Alternate angles)

  OBA=ODC … (Alternate angles)

  ΔAOBΔCOD … (By AAA similarity test)

As we know that, the ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

  ar(ΔAOB)ar(ΔCOD)=AB2CD2ar(ΔAOB)ar(ΔCOD)=82182ar(ΔAOB)ar(ΔCOD)=1681

Hence, ratio of the area of region I to that of region II is 16:81.

b.

To determine

To find: The ratio of the areas of triangle I to that of triangle II.

b.

Expert Solution
Check Mark

Answer to Problem 28RP

The ratio of the area of region I to that of region II is 49:81.

Explanation of Solution

Given Information:

Sides of the triangle I are

  a=6andb=7

Sides of the triangle II are

  c=9andd=15

Formula used:

Ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Calculation:

Sides of the triangle I are (a)=6and(b)=7

Sides of the triangle II are (c)=9and(d)=15

Let the triangles intersect at point O

  Geometry For Enjoyment And Challenge, Chapter 11, Problem 28RP , additional homework tip  2

In ΔAOBandΔCOD,

  AOB=COD … (Vertically opposite angles)

  OAB=OCD … (Alternate angles)

  OBA=ODC … (Alternate angles)

  ΔAOBΔCOD … (By AAA similarity test)

As we know that, the ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

  ar(ΔAOB)ar(ΔCOD)=OB2OD2ar(ΔAOB)ar(ΔCOD)=7292ar(ΔAOB)ar(ΔCOD)=4981

Hence, ratio of the area of region I to that of region II is 49:81.

c.

To determine

To calculate: The ratio of the areas of the two triangles.

c.

Expert Solution
Check Mark

Answer to Problem 28RP

The ratio of the area of region I to that of region II is 1:!.

Explanation of Solution

Given Information:

Sides of the triangle I are (a)=6and(b)=10

One side of the triangle II is (c)=10

Two angles of triangle I are equal to the corresponding angles of the triangle II.

Formula used:

Ratio of areas of two congruent triangle is equal to 1:1.

Calculation:

Sides of the triangle I are (a)=6and(b)=10

One side of the triangle II is (c)=10

  Geometry For Enjoyment And Challenge, Chapter 11, Problem 28RP , additional homework tip  3

In ΔABEandΔBCD,

  BAE=DBC … (Given)

  AEB=BCD … (Given)

  AE=BC=10 … (Given)

  ΔBAEΔDBC … (By ASA congruency test)

As the triangles are congruent, so the triangles will have equal areas.

  ar(ΔBAE)=ar(ΔDBC)ar(ΔBAE)ar(ΔDBC)=11

Hence, the ratio of the area of region I to that of region II is 1:1.

Chapter 11 Solutions

Geometry For Enjoyment And Challenge

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

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