4. In the figure, triangle APM and triangle BQM are equal triangles. PQ = 10 cm, PA = 4 cm. So what is AM? How much is AB?
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As ∆AQM and ∆BQM are equal
therefore, PM=MQ
hence PM=5cm and MQ=5cm
In ∆APM by Pythagoras theorem
PM²=PA²+AM²
5²=4²+AM²
25=16+AM²
25-16=AM²
9=AM²
AM=√9
AM=3cm
Again as ∆APM and ∆BQM are equal
Therefore AM=MB =3cm
AM+MB = AB
3+3=AB
AB=9cm
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Step-by-step explanation:
Draw a line passing through P.
expand BQ such that it will intersect the line passing through P at 'O' making a angle of 90° at O
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