Question

$\Delta AMT\sim \Delta AHE$. In $\Delta AMT,AM=6.3cm,\angle MAT={120}^o,AT=4.9cm$ and $\frac{MA}{HA}=\frac{7}{5}$. Construct both the triangles.

Answer 1

$\Delta AMT\sim \Delta AHE$.
$\frac{AM}{AH}=\frac{MT}{HE}=\frac{AT}{AE}=\frac{7}{5}$
In $\Delta AMT$, $AM=6.3cm,AT=4.9cm$ and $\angle A={120}^{\circ }$.

Draw a ray from point $A$ forming an acute angle with $AT$ on the opposite side of $M$. Divide the ray into $7$ equal parts such that $A{S}_1=S_1{S}_2=S_2{S}_3=S_3{S}_4=S_4{S}_5=S_5{S}_6=S_6=S_7$

Join $S_7$ with $T$. From $S_5$, draw a line $\parallel$ to $S_{7}T$ and let it intersect $AT$ at $E$.

Draw $EH\parallel TM$.
$\Delta AEH$ is the required similar triangle.