Construct a triangle ABP such that AB=6cm angle ABP=30`and BP=4cm.Complete rhombus ABCD such that P is equidistant from AB and BC. Locate the point Q on the line BP such that Q is equidistant from. An andBsuch
Asked by tharaganesh74 | 1st Mar, 2020, 09:38: AM
- Construct ∆ABP
- P is equidistant from AB and BC, P lies on the bisector of ∠ABC. ∠ABP = 30°, therefore, ∠ABR = 60°, mark BC = 5 cm from BR.
- Complete rhombus ABCD.
- Since Q is equidistant from A and B, draw perpendicular bisector of AB. The point of intersection of the right bisector of AB and the line BP is the required point Q
Answered by Arun | 1st Mar, 2020, 10:47: AM
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