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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Answer:
26 cm²
Step-by-step explanation:
The area of the rectangle whose dimensions are shown at the right and bottom is ...
(6 cm)(7 cm) = 42 cm²
The figure is smaller than that by the area of the space whose dimensions are shown at the right and in the middle left:
(4 cm)(4 cm) = 16 cm²
The figure area is then the difference ...
42 cm² - 16 cm² = 26 cm²
_____
<em>Alternate solution</em>
Draw a diagonal line between the lower right inside corner and the lower right outside corner. This divides the figure into two trapezoids.
The trapezoid at lower left has bases 7 and 4 cm, and height 6-4 = 2 cm. Its area is ...
A = (1/2)(b1 +b2)h = (1/2)(7 + 4)(2) = 11 . . . . cm²
The trapezoid at upper right has bases 6 cm and 4 cm and height 3 cm. Its area is ...
A = (1/2)(b1 +b2)h = (1/2)(6 + 4)(3) = 15 . . . . cm²
Then the area of the figure is the sum of the areas of these trapezoids, so is ...
11 cm² + 15 cm² = 26 cm²
_____
<em>Comment on other alternate solutions</em>
There are many other ways you can find the area of this figure. It can be divided into rectangles, triangles, or other figures of your choice. The appropriate area formulas should be used, and the resulting partial areas added or subtracted as required.
You can also let a geometry program find the area for you. (It is 26 cm².)
