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Answer :
Steps of construction by the given data are:
Step 1. Draw a \(\triangle \) ABC with sides as AB=5 cm, BC= 7 cm, AC= 6cm.
Step 2. At point B, draw an acute angle CBX below BC.
Step 4. Join \(B_5\) and C.
Step 5.Draw \(B_7C'\) parallel to \(B_5C\), where C` is a point on extended line BC.
Step 6. Draw A'C' parallel to AC, where A' is a point on extended line BA.
Hence,A’BC’ is the required triangle.
JUSTIFICATION:
In \(\triangle \) ABC and \(\triangle \) A’BC’,
AC || A’C’
\(\frac{AB}{A’B}=\frac{BC}{BC’}\)
(By the basic Proportionality Theorem)...(i)
In \(ΔBB_5C \)and \(ΔBB_7C’\),
\(B_5C || B_7C’\)
\(\frac{BC}{BC’}=\frac{BB_5}{BB_7}=\frac{5}7\) (By the basic Proportionality Theorem)...(ii)
Equating (i) and (ii) \(\frac{AB}{A’B}=\frac{BB_5}{BB_7} \Rightarrow \frac{AB}{A’B}=\frac{5}7 \)
\(\therefore A’B=\frac{7}5AB\)
\(\therefore \) Sides of new triangle is \(\frac{7}5 \)times the corresponding sides of the first triangle.