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Answer :
Steps to construct figure by given data are as follows:
Step 1: Construct an isosceles triangle ABC with BC=8cm and altitude AD=4cm.
Step 2:Draw a ray BX, making an acute angle with BC.
Step 3: Locate 3 points on BX, such that BP=PQ=QR.
Step 4: Join QC.
Step 5: Through R, draw a line RC parallel to OC, meeting produced line BC at C’.
Step 6: Through C, draw a line CA parallel to CA, meeting the produced line BA at A’.
Thus, \(\triangle \) A’BC’ is the required isosceles triangle
JUSTIFICATION:
In \(\triangle \) ABC and \(\triangle \) A’BC’, we have
\(\triangle ACB=\triangle A’C’B \)(corresponding angles)
\(∠B=∠ B \)(common)
\(∴ ΔABC∼ΔA’BC’ \) (By AA similarity)
\(∴ \frac{AB}{A’B}=\frac{AC}{A’C’}=\frac{BC}{BC’} \)
But, \(\frac{BC}{BC’}=\frac{BQ}{BR}=\frac{2}3 ∴ \frac{BC}{BC’}=\frac{2}3 \)
\(⇒ \frac{A’B}{AB}=\frac{A’C’}{AC}=\frac{B’C}{BC}=\frac{3}2.\)
Hence, we get the required triangle whose sides are \(\frac{3}{2} \)
, i.e., \(1\frac{1}{2}\) times of the corresponding sides of the isosceles \(\therefore \) ABC.