Dear student Given: area of △=8 cm2Also, area of △=12×base×height Assuming that it is a right triangle=12×b×h where b=base,h=height⇒8=12×b×h ⇒b×h=16⇒b=16h ⇒b∝1h, where constant of proportionality =16 Regards

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Dear student

Given : area of △ = 8 {cm}^{2} Also, area of △ = \frac{1}{2} \times base \times height [A s s u m i n g t h a t i t i s a r i g h t t r i a n g l e] = \frac{1}{2} \times b \times h [where b = base, h = height] \Rightarrow 8 = \frac{1}{2} \times b \times h \Rightarrow b \times h = 16 \Rightarrow b = \frac{16}{h}

\Rightarrow b \propto \frac{1}{h}, where constant of proportionality = 16

Regards

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