Construction of a Similar Triangle, When One Vertex Is Common
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Question
If you are asked to construct △APQ ∼△ABC with the scale factor 35, which of the following are correct? In △ABC,AB=4cm, BC=3cm and ∠ABC=90∘.
A
∠AQP=∠ACB
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B
∠AA3P=∠AA5P
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C
AP:PB=3:2
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D
5PQ = 3BC
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Solution
The correct options are A∠AQP=∠ACB CAP:PB=3:2 D5PQ = 3BC The following steps will give you the information on how to construct the similar triangle to △ABC. Step 1: Draw a line AB = 4 cms. Step2: Draw a line BC = 3 cm, perpendicular to AB passing through B. Step 3: Join AC. Step 4: Draw a ray AX, making an acute angle with line AB. Step 5: Mark 5 points A1,A2,A3,A4 and A5 such that A1A2=A2A3=A3A4=A4A5.
Step 6: Join BA5. Step 7: Draw a line parallel to BA5 passing through A3 by making an angle equal to ∠AA5B, intersecting AB at the point P.APAB=35.(This is the given scale factor of the smaller triangle which is the ratio of corresponding sides.) Step 8: Draw a line parallel to BC passing through P, intersecting AC at Q. Now we have constructed the triangle △APQ ∼△ABC. By basic proportionality theorem, The corresponding angles are equal. Therefore in ΔABC and ΔAPQ,∠BAC=∠PAQ,∠ABC=∠APQand∠ACB=∠PQA. And the corresponding sides are proportional. Therefore in ΔABCandΔAPQ,APAB=PQBC=AQAC We know that APAB=35 Therefore, PQBC=35 5PQ = 3BC. From construction we have, A3P||A5B.∠AA3P=∠AA5B, because they are corresponding angles of parallel lines.Now ∠AA3P≠∠AA5P as it is equal to ∠AA5B. Consider triangle AA5B, the line A3P being parallel to A5B cuts the sides AB and AA5 in the same proportion. i.e. AA3A3A5=APPB APPB=32 Therefore A3P will divide the line AB in the ratio 3:2.