Question

In the figure the equilateral triangles of side 1 cm joined .what is the distance from A to C ?​

Answer 1

Answer:

2or3 is answer ok it is an right

Answer 2

Given:

In the figure the equilateral triangles of side 1 cm joined .what is the distance from A to C ?​

Solution:

Join point A to C,

Refer the image given below,

Since,  triangle ADM, DMC and BCM are equilateral,

Hence $\[\angle MDC = \angle DCM = 60^\circ \]$

Observe that ADCM is a parallelogram,

Therefore, $\[\angle DCA = 30^\circ \]$

Take triangle DOC,

$\[\begin{array}{l} \Rightarrow \angle DOC = 180^\circ - \angle MDC - \angle DCA\\ \Rightarrow \angle DOC = 180^\circ - 90^\circ - 30^\circ \\ \Rightarrow \angle DOC = 90^\circ \end{array}\]$

It shows that diagonal DM and AC interests at  $\[90^\circ \]$.

Hence DCAM is rhombus,

Now equate the area of rhombus DCM and sum of area of triangles ADM and DCM,

$\[\begin{array}{l} \Rightarrow \frac{1}{2} \times DM \times AC = 2 \times \frac{{\sqrt 3 }}{4} \times {1^2}\\ \Rightarrow \frac{1}{2} \times 1 \times AC = \frac{{\sqrt 3 }}{2} \times {1^2}\\ \Rightarrow AC = \sqrt 3 \,cm\end{array}\]$

Hence the distance from A to C is $\[\sqrt 3 \,cm\]$ or 1.732 cm.