Question

∆AMT ~ ∆ANE. In ∆AMT, AM = 6.3 cm, ∠TAM = 500, AT = 5.6 cm. = AM/AH = 7/5 Construct ∆AHE.

Answer 1

Analysis: 

As shown in the figure, 

Let A – H – M and A – E – T. 

∆AMT ~ ∆AHE … [Given] 

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

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

But, AM/AH = 7/5  ....(ii)   [Given ]

∴ AM/AH = MT/HE = AT/AJ = 7/5 ....[Form (i) and (ii)]

∴ Sides of AAMT are longer than corresponding sides of ∆AHE. 

∴ The length of side AH will be equal to 5 parts out of 7 equal parts of side AM. 

So, if we construct AAMT, 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: 

i. Draw ∆AMT of given measure. Draw ray AB making an acute angle with side AM.

ii. Taking convenient distance on the compass, mark 7 points A₁, A₂, A₃, A₄,A₅, A₆ and A₇, such that 

AA₁ = A₁A₂ = A₂ A₃ = A₃A₄ = A₄A₅= A₅ A₆= A₆ A₇ .

iii. Join A₇M. Draw line parallel to A₇M through A₅ to intersects seg AM at H. 

iv. 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.