A triangle has sides 5 cm , 12 cm and 13 cm. find the length to one decimal place , of the perpendicular from the opposite vertex to the side whose length is 13 cm.
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Let's say that the triangle is ABC with AB = 12, BC = 5, CA = 13.
Now you may see that triangle is right-angled at B, as AB² + BC² = CA².
And AB ⟂ BC with CA as hypotenuse and perpendicular's foot on CA be D.
Hence, ar(ABC) = ½ AB · BC = ½ CA · BD.
⇒ BD = ( AB · BC ) ÷ CA = ( 12 × 5 ) ÷ 13 = 60/13 cm.
Now you may see that triangle is right-angled at B, as AB² + BC² = CA².
And AB ⟂ BC with CA as hypotenuse and perpendicular's foot on CA be D.
Hence, ar(ABC) = ½ AB · BC = ½ CA · BD.
⇒ BD = ( AB · BC ) ÷ CA = ( 12 × 5 ) ÷ 13 = 60/13 cm.
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