side of an equilateral
DABC is 8cm. Find the
sides of an equilatesal
PQR whose area is
twice the area of the
first.
Answers
Answer:
Given :-
Side of equilateral triangle = 8 cm
We know that all sides of equilateral triangle are equals !
We have to find the Side of triangle whose area is twice the area of 1st
\boxed{\sf\ \ Area\ of \ equilateral\ \triangle= \dfrac{\sqrt{3}}{4}a^2}
Area of equilateral △=
4
3
a
2
Where a = side
\begin{gathered}\longrightarrow\sf Area = \dfrac{\sqrt{3}}{4}\times (8)^2\\ \\ \longrightarrow\sf Area = \dfrac{\sqrt{3}}{\cancel{4}}\times \cancel{64}\\ \\ \longrightarrow\sf Area = \sqrt{3}\times 16\\ \\ \longrightarrow\sf Area \ of \ equilateral \triangle = 16\sqrt{3}cm^2\end{gathered}
⟶Area=
4
3
×(8)
2
⟶Area=
4
3
×
64
⟶Area=
3
×16
⟶Area of equilateral△=16
3
cm
2
Now the Area of that triangle whose area is twice the first
\begin{gathered}\longrightarrow\sf Area \ of \ \triangle_2= 2\times 16\sqrt{3}\\ \\ \longrightarrow\sf Area \ of \ \triangle_2= 32\sqrt{3}cm^2\end{gathered}
⟶Area of △
2
=2×16
3
⟶Area of △
2
=32
3
cm
2
Now the side of triangle 2
\begin{gathered}\longrightarrow\sf \dfrac{\sqrt{3}}{4}a^2= 32\sqrt{3}\\ \\ \longrightarrow\sf a^2= 32\cancel{\sqrt{3}}\times \dfrac{4}{\cancel{\sqrt{3}}}\\ \\ \longrightarrow\sf a^2= 128\\ \\ \longrightarrow\sf a= \sqrt{2\times 4\times 4\times 4}\\ \\ \longrightarrow\sf a= 8\sqrt{2}cm\end{gathered}
⟶
4
3
a
2
=32
3
⟶a
2
=32
3
×
3
4
⟶a
2
=128
⟶a=
2×4×4×4
⟶a=8
2
cm
\boxed{\sf \dag\ \ Side\ of \ triangle_2= 8\sqrt{2}cm}
† Side of triangle
2
=8
2
cm
Step-by-step explanation:
pl
Answer:
do u use I.n.s.t.a if u
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