S O L U T I O N :
From the provided figure and given information,
- Area or triangle = [tex] \sf \dfrac{9}{32}\; cm^2 [/tex]
- Height of the triangle = [tex] \sf \dfrac{3}{4}\; cm [/tex]
We are asked to calculate the length of base.
As it we know that,
[tex]\\ \twoheadrightarrow\quad \bf{Area_{(\Delta)} = \dfrac{1}{2} \times b \times h}\\ [/tex]
- h denotes height
- b denotes base
So,
[tex]\\ \twoheadrightarrow\quad \sf{\dfrac{9}{32}\; cm^2 = \dfrac{1}{2} \times b \times \dfrac{3}{4}\; cm}\\ [/tex]
[tex]\\ \twoheadrightarrow\quad \sf{\dfrac{9}{32}\; cm^2 = b \times \dfrac{3}{8}\; cm}\\ [/tex]
[tex]\\ \twoheadrightarrow\quad \sf{\dfrac{9}{32}\; cm^2\div \dfrac{3}{8}\; cm= b }\\ [/tex]
[tex]\\ \twoheadrightarrow\quad \sf{\dfrac{\cancel{9} }{\cancel{32} }\times \dfrac{\cancel{8} }{\cancel{3}} \; cm= b }\\ [/tex]
[tex]\\ \twoheadrightarrow\quad \bf \underline{ \dfrac{3}{4}\; cm= b }\\ [/tex]
Therefore, length of the base of the triangle is 3/4 cm.
