Answer:
The statement, one fifth of the square of a number is written as
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P is the centroid of △JLN, LO⊥JN, ∠NJK≅∠JNM, OP=8, OJ=15. Find the perimeter of △JLN.
1 answer:
The complete question in the attached figure
we know that
The Centroid of a Triangle is the point where the three medians of the triangle intersect.
<span>Each median divides the triangle into two smaller triangles of equal area.
<span>and
the centroid divides each median into two segments whose lengths are in the ratio 2:1, with the longest one nearest the vertex
therefore
triangles LOJ and LON are similar
ON=OJ=15 units
then
NJ=15+15-------> NJ=30 units
Let
x--------> distance OL
so
OP=x/3
OP=8
x/3=8--------> x=8*3---------> x=24 units
OL=24 units
in the right triangle JLO
OJ=15 units
OL=24 units
JL=?
applying the Pythagorean Theorem
JL</span></span>²=OJ²+OL²--------> JL²=15²+24²------> JL=28.30 units
<span>
perimeter of triangle JLN
P=JL*2+NJ-------> P=2*28.30+30------> P=86.60 units------> P=86.6 units
the answer is
</span><span>the perimeter of △JLN is </span><span>86.6 units
</span>
we know that
The Centroid of a Triangle is the point where the three medians of the triangle intersect.
<span>Each median divides the triangle into two smaller triangles of equal area.
<span>and
the centroid divides each median into two segments whose lengths are in the ratio 2:1, with the longest one nearest the vertex
therefore
triangles LOJ and LON are similar
ON=OJ=15 units
then
NJ=15+15-------> NJ=30 units
Let
x--------> distance OL
so
OP=x/3
OP=8
x/3=8--------> x=8*3---------> x=24 units
OL=24 units
in the right triangle JLO
OJ=15 units
OL=24 units
JL=?
applying the Pythagorean Theorem
JL</span></span>²=OJ²+OL²--------> JL²=15²+24²------> JL=28.30 units
<span>
perimeter of triangle JLN
P=JL*2+NJ-------> P=2*28.30+30------> P=86.60 units------> P=86.6 units
the answer is
</span><span>the perimeter of △JLN is </span><span>86.6 units
</span>
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