(A) The perimeter of an equilateral triangle is given by $P = 3a$, where $a$ is the side length.Given $P = 24$, we have $3a = 24$, which implies $a =

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(A) The perimeter of an equilateral triangle is given by $P = 3a$, where $a$ is the side length.Given $P = 24$, we have $3a = 24$, which implies $a = 8$.The altitude $h$ of an equilateral triangle is given by the formula $h = \frac{\sqrt{3}}{2} a$.Substituting $a = 8$ into the formula, we get $h = \frac{\sqrt{3}}{2} \times 8 = 4 \sqrt{3}$.Therefore, the length of the altitude is $4 \sqrt{3}$.

Class 10 Mathematics Triangles Similarity of Triangles and Basic Proportionality Theorem

Medium

The perimeter of equilateral $\Delta ABC$ is $24$. Find the length of its altitude. (in $\sqrt{3}$)

A
$4$
B
$5$
C
$6$
D
$8$

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Triangles

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Similarity of Triangles and Basic Proportionality Theorem

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AD}$ is a median. Prove that $AC^{2} = 4AD^{2} - 3AB^{2}$.

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The perimeter of equilateral $\Delta ABC$ is $24$. Find the length of its altitude. (in $\sqrt{3}$)

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AD}$ is a median. Prove that $AC^{2} = AD^{2} + 3BD^{2}$.

The diagonals of convex quadrilateral $PQRS$ intersect at right angles. Prove that $PQ^{2} + RS^{2} = PS^{2} + QR^{2}$.

In the figure,$PS$ is the bisector of $\angle QPR$ of $\Delta PQR$. Prove that $\frac{QS}{SR} = \frac{PQ}{PR}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AD}$ is a median. Prove that $AC^{2} = 4AD^{2} - 3AB^{2}$.

In $\Delta ABC$,$m\angle A + m\angle C = m\angle B$. If $AB = 7$ and $BC = 24$,then $AC = \ldots$

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