Question

13. Each side of an equilateral triangle
measures 2v3 cm. The length of its
altitude is.​

Answer 1

$\underline{\underline{\textsf{\maltese\:\:Question :}}}$

Each side of an equilateral triangle measures 2√3 cm. The length of its altitude is ?

$\underline{\underline{\textsf{\maltese\:\: Diagram :}}}$

$\setlength{\unitlength}{2cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.95,3.1){ \bf A }\put(0.9,-0.2){ \bf B }\put(5,-0.2){ \bf C } \put(4.3,1.5){{ 2\sqrt{3}\sf{cm} }} \put(1,1.5){{ 2\sqrt{3}\sf{cm} }} \put(2.7,-.5){{ 2\sqrt{3}\sf{cm} }} \qbezier(3,3)(3,3)(3,0) \put(2.89,-0.2){ \bf M }\put(3.1,0.3){\small90 ^\circ }\put(3.01,.001){\framebox(0.2,0.2)} \put(2.8,.001){\framebox(0.2,0.2)}\put(2.7,0.3){\small90 ^\circ } \put(1.3,0.2){\small60 ^\circ } \put(4.5,0.2){\small60 ^\circ } \put(3.03,2.5){\small30 ^\circ } \put(2.75,2.5){\small30 ^\circ } \end{picture}$

$\setlength{\unitlength}{2 cm}\begin{picture}(0,0)\thicklines \qbezier(3,3)(3,3)(3,0) \qbezier(3,0)(3,0)(5,0) \qbezier(3,3)(3,3)(5,0) \put(3.01,.001){\framebox(0.2,0.2)}\put(2.85,3.2){ \bf A } \put(2.85,-0.4){ \bf M } \put(5.2,-0.3){ \bf C }\put(3.7,-.2){\small \sqrt3 cm}\put(4.2,1.5){\bf 2\sqrt{3}\sf{cm} }}\put(4.5,.1){60 ^\circ } \put(3.1,0.3){\small90 ^\circ }\put(3.1,2.3){30 ^\circ }\end{picture}$

$\underline{\underline{\textsf{\maltese\:\: Solution :}}}$

In ∆AMC

MC = $\sf \frac{BC}{2}$

MC = $\dfrac{2\sqrt{3}}{2}$

MC = √3

Applying Pythagoras Theorem in ∆AMC

Hypotenuse² = Base² + Perpendicular ²

⇒ AC² = MC² + AM²

⇒ (2√3 cm)² = (√3 cm)² + AM²

⇒ 12cm² = 3cm² + AM²

⇒ 12cm² - 3cm² = AM²

⇒ 9cm² = AM²

⇒ √9 cm² = AM

⇒ 3cm = AM

⇒ AM = 3cm

∴ The length of its altitude is 3cm.