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Practice
Simple Polygon Area
Easy
Normal
Simple Polygon Area
Problem 1
The area of a rectangle is [tex]S=18cm^2[/tex] and one of its sides has a length of [tex]6cm[/tex]. Determine the length of the other side.
Solution:
Let us denote the given side with
a
and the side whose length we are trying to find with
b
. We have that [tex]S=a.b[/tex], therefore [tex]b=\frac{S}{a}=\frac{18cm^2}{6cm}=3cm[/tex].
Problem 2
Given a rectangle [tex]ABCD[/tex] with
AB=5cm
and area
S=40cm
2
, determine the length of
CD
.
Solution:
CD
and
AB
are the parallel sides in a rectangle, so they have equal lengths. Therefore,
CD=5
.
Problem 3
Given a rectangle
ABCD
, in which
AB=5cm
and its area is [tex]S=35cm^2[/tex], determine the length of
BC
.
Solution:
The area of a rectangle is calculated by the formula
S=AB.BC
, so [tex]BC=\frac{S}{AB}=\frac{35cm^2}{5cm}=7cm[/tex].
Problem 4
The area of a right triangle is [tex]S=36cm^2[/tex]. If one of its catheti has a length of
9cm
, determine the length of the other cathetus (in centimeters).
Solution:
Let us denote the catheti with
a
and
b
. We know that
a=9cm
. We also know that [tex]\frac{1}{2}a.b=S[/tex] => [tex]b=\frac{2S}{a}=\frac{2.36}{9}=8cm[/tex]
Problem 5
Let
CH
be an altitude in the acute triangle
ABC
. If
AB=8
and [tex]S_{ABC}=28[/tex], determine the length of
CH
.
Solution:
We know that [tex]S_{ABC}=\frac{1}{2}AB.CH[/tex], therefore [tex]CH=\frac{2S_{ABC}}{AB}=\frac{2.28}{8}=7[/tex].
Problem 6
In the triangle
ABC
let
AH
be the altitude to
BC
. If the area of the triangle is [tex]S=58cm^2[/tex] and
BC=4cm
, determine the length of
AH
.
Solution:
We know that [tex]S=\frac{1}{2}.AH.BC[/tex], so [tex]AH=\frac{2S}{BC}=\frac{2.58cm^2}{4cm}=29cm[/tex]
Problem 7
In triangle
ABC
,
BH
is the altitude to
AC
. If
BH=11cm
and [tex]S=44cm^2[/tex], determine the length of
AC
(in cm).
Solution:
We know that [tex]S=\frac{1}{2}.AC.BH[/tex] => [tex]AC=\frac{2S}{BH}=\frac{2.44cm^2}{11cm}=8cm[/tex].
Easy
Normal
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