CUET Mathematics 2024 - The area of the region bounded by the lines x7 sqrt(3) a+y/b=4, x=0 and y=0 is : | PYQs + Solutions

CUET Mathematics 2024

Calculus

Application of Integrals

Medium

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

56\sqrt{3} \ ab

56a

\frac{ab}{2}

3ab

✅ Correct Option: 1

Let's find the intercepts of the line:

x-intercept (when

y = 0

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

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

x = 28\sqrt{3}a

y-intercept (when

x = 0

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

\dfrac{y}{b}=4

y = 4b

The triangle has vertices at

(0,0)

(28\sqrt{3}a, 0)

, and

(0, 4b)

Area of the triangle

= \dfrac{1}{2} \times \text{base} \times \text{height}

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

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

= 56\sqrt{3}ab

square units