Question

$\Delta AMT\sim \Delta AHE$ . In $\Delta AMT$ , MA = 6.3cm, $\angle MAT=120{}^\circ $ , AT = 4.9cm, $\dfrac{MA}{HA}=\dfrac{7}{5}$ . Construct $\Delta AHE$ .

Answer 1

Hint: Before starting the construction use the properties of similar triangles to get the sides of $\Delta AHE$.

Complete step-by-step answer:
It is given that $\Delta AMT\sim \Delta AHE$ , and we know two triangles are similar if their corresponding angles are equal and the corresponding sides are in proportion.
$\therefore \dfrac{MA}{HA}=\dfrac{AT}{AE}=\dfrac{TM}{EH}...............(i)$
$\angle MAT=\angle HAE=120{}^\circ $

Now from the question, we know $\dfrac{MA}{HA}=\dfrac{7}{5}.$
We also know MA = 6.3 cm.
$\therefore \dfrac{6.3}{HA}=\dfrac{7}{5}$
On cross-multiplication, we get
$5\times 6.3=7HA$
$\Rightarrow 31.5=7HA$
$\Rightarrow HA=4.5cm$
On substituting the value of $\dfrac{MA}{HA}$ in equation (i), we get
$\dfrac{7}{5}=\dfrac{AT}{AE}=\dfrac{TM}{EH}$
$\dfrac{AT}{AE}=\dfrac{7}{5}$

Now we know AT = 4.9 cm, putting in the above equation, we get
$\dfrac{4.9}{AE}=\dfrac{7}{5}$
On cross-multiplication, we get
$5\times 4.9=7AE$
$\Rightarrow 24.5=7AE$
$\Rightarrow AE=3.5cm$
So, for $\Delta AHE$, we have
$\angle HAE=120{}^\circ $
$AE=3.5cm$
$HA=4.5cm$
Now let us start with the construction.

We should first draw a horizontal line segment of length equal to 4.5 cm and name it as HA, followed by drawing an arc with centre as A and radius = 3.5 cm. Now draw a line at $120{}^\circ $ from the point A cutting the arc at point E as shown in the figure. Join the points H and E to get the required triangle. Finally, show the $\angle HAE$ to finish the construction.

Note: While using the relation of similarity, generally the students mix up the sides while it is clearly mentioned that the relation is valid only for the corresponding sides of similar triangles. We should also remember that we can construct a triangle with two given sides if and only if the angle between the two given lines is mentioned or there is a way to reach it in the question as we did in the above question.