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Question
Construct a ΔABC in which AB = 5 cm. ∠B = 60° altitude CD = 3cm. Construct a ΔAQR similar to ΔABC such that side ΔAQR is 1.5 times that of the corresponding sides of ΔACB.
Solution
Given that
Construct a triangle ΔABC in which AB = 5 cm. ∠B = 60° altitude CD = 3cm and then a triangle ΔAQR similar to it whose sides are (1.5 times = 3/2) of the corresponding sides of ΔACB.
We follow the following steps to construct the given
Step of construction
Step: I- First of all we draw a line segment AB = 5cm.
Step: II- With B as centre and draw an angle ∠B = 60°.
Step: III -From point A and B construct altitude CD = 3cm, which cut the line BS at point C
Step: IV- Join AC to obtain ΔABC.
Step: V- Below AB, makes an acute angle ∠BAX = 60°.
Step: VI- Along AX, mark off five points A1, A2 and A3 such that AA1 = A1A2 = A2A3
Step: VII -Join A2B.
Step: VIII -Since we have to construct a triangle ΔAQR each of whose sides is (1.5 times = 3/2) of the corresponding sides of ΔABC.
So, we draw a line A3Q on AX from point A3 which is A3Q||A2B and meeting AB at Q.
Step: IX- From Q point draw QR || BC and meeting AC at R
Thus, ΔAQR is the required triangle, each of whose sides is (1.5 times = 3/2) of the corresponding sides of ΔABC.
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