Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 8 - Section 8.1 - Area and Initial Postulates - Exercises - Page 348: 6

Answer

The areas of the four resulting smaller triangles is one fourth of the area of the given rhombus.

Work Step by Step

Draw diagonals MP and QN. the intersection of diagonals is named as O. Let us compare $\triangle$ OPN and $\triangle$ OQP we know in rhombus diagonals are perpendicular bisectors to each other So OP = OM and OQ = ON $\angle$ PON = $\angle$ POQ = $\angle$ MON = $\angle$ MOQ = 90$^{\circ}$ In $\triangle$ OPN and $\triangle$ OQP PN =QP(in rhombus opposite sides are equal) OP = OP (common) $\angle$ PON = $\angle$ POQ = 90$^{\circ}$ By SAS rule $\triangle$ OPN is congruent to $\triangle$ OQP Therefore area of $\triangle$ OPN = area of $\triangle$ OQP Since In rhombus sides are equal Similarly area of $\triangle$ OMN = area of $\triangle$ OMQ area of $\triangle$ PON = area of $\triangle$ OMN area of $\triangle$ OQP = area of $\triangle$ OQM so area of $\triangle$ OQP = area of $\triangle$ OPN = area of $\triangle$ OQM = area of $\triangle$ OQN Therefore area of MNPQ = area of $\triangle$ OQP + area of $\triangle$ OPN + area of $\triangle$ OQM + area of $\triangle$ OQN. Therefore area of MNPQ = 4 * area of $\triangle$ OQP area of $\triangle$ OQP = $\frac{area of MNPQ}{4}$. Therefore area of smaller triangle is one fourth of the area of rhombus MNPQ.
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