Question

$△$AMT$~$$△$AHE, In $△$AMT, MA = 6.3 cm, AT = 4.9 cm,$\angle$MAT = 120$°$ and $\frac{\mathrm{MA}}{\mathrm{HA}}=\frac{7}{5}$
Construct $△$AHE.

Answer 1

Steps of construction:
1. Draw a line segment AM = 6.3 cm.
2. Draw an angle $\angle$MAX = 120$°$.
3. With A as the centre, draw an arc of radius 4.9 cm that cuts the ray AX at vertex T. Join MT.
4. Divide the line segment AM in the seven equal parts and take a point H on AM such that AM:AH=7:5 i.e., AH:HM=5 : 2
5. Draw EH parallel to TM.

Thus, $∆$AHE is the required triangle, which is similar to $△$AMT.