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The area of the region bounded by the lines x73a+yb=4,x=0\frac{x}{7 \sqrt{3} a}+\frac{y}{b}=4, x=0 and y=0y=0 is :

Solution

✅ Correct Option: 1

Let's find the intercepts of the line:

x-intercept (when y=0y = 0):

x73a+0b=4\dfrac{x}{7 \sqrt{3} a}+\dfrac{0}{b}=4

x73a=4\dfrac{x}{7 \sqrt{3} a}=4

x=283ax = 28\sqrt{3}a

y-intercept (when x=0x = 0):

073a+yb=4\dfrac{0}{7 \sqrt{3} a}+\dfrac{y}{b}=4

yb=4\dfrac{y}{b}=4

y=4by = 4b


The triangle has vertices at (0,0)(0,0), (283a,0)(28\sqrt{3}a, 0), and (0,4b)(0, 4b).

Area of the triangle =12×base×height= \dfrac{1}{2} \times \text{base} \times \text{height}

=12×283a×4b= \dfrac{1}{2} \times 28\sqrt{3}a \times 4b

=12×1123ab= \dfrac{1}{2} \times 112\sqrt{3}ab

=563ab= 56\sqrt{3}ab square units

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