Question

∆AMT ~ ∆AHE. In ∆AMT, AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm, `(AM)/(AH) = 7/5`. Construct ∆AHE.

∆AMT ~ ∆AHE. In ∆AMT, AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm, `(AM)/(AH) = 7/5`, then construct △AMT and ΔAHE.

Answer 1

Analysis:

As shown in the figure,

Let A – H – M as well as points A – E – T be collinear.

∆AMT ~ ∆AHE,

∴ ∠TAM ≅ ∠EAH   ...[Corresponding angles of similar triangles]

`(AM)/(AH) = (MT)/(HE) = (AT)/(AE)`   ...(i) [Corresponding sides of similar triangles]

∴ `(AM)/(AH) = 7/5`   ...(ii) [Given]

`(AM)/(AH) = (MT)/(HE) = (AT)/(AE) = 7/5`   ...[From (i) and (ii)]

∴ Sides of ∆AHE are smaller than sides of ∆AMT.

∴ If seg AH will be equal to 5 parts out of 7 equal parts of side AM.

So, if we construct ∆AMT, point H will be on side AM, at a distance equal to 5 parts from A.

Now, point E is the point of intersection of ray AT and a line through H, parallel to MT.

∆AHE is the required triangle similar to ∆AMT.

Steps of construction:

1. Draw ∆AMT such that AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm.
2. Draw ray AB making an acute angle with side AM.
3. Taking convenient distance on the compass, mark 7 points A₁, A₂, A₃, A₄, A₅, A₆ and A₇, such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
4. Join A₇M. Draw line parallel to A₇M through A₅ to intersects seg AM at H.
5. Draw a line parallel to side TM through H. Name the point of intersection of this line and seg AT as E.
   ∆AHE is the required triangle similar to ∆AMT.
   

Here, ∆AMT ~ ∆AHE.