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प्रश्न
∆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.
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उत्तर
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:
- Draw ∆AMT such that AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm.
- Draw ray AB making an acute angle with side AM.
- 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.
- Join A7M. Draw line parallel to A7M through A5 to intersects seg AM at H.
- 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.
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