Question

The right angled triangle of base 60cm and height 61 cm. At each vertex of the triangle, circles of radius 3cm are drawn. What is the area of the triangle in sqcm, excluding the portion enclosed by circles? $(\pi=3.14)$.

• A. 1815.87
• B. 2000
• C. 1596.7
• D. 1728.5

Answer 1

Answer:

A. 1815.87

Area of Remaining portion = Area of $\triangle$ - Area of Sector

Area of Sector = $\frac{\theta}{360^o}\times{\pi{r^2}}$

$Radius,r=3cm$

Base b= 60., height h=61cm

Area of remaining portion = $\frac{1}{2}{bh}-\frac{\theta_1}{360^o}\times{\pi{r^2}}+\frac{\theta_2}{360^o}\times{\pi{r^2}}+\frac{\theta_3}{360^o}\times{\pi{r^2}}$

=>\frac{1}{2}{bh}-\frac{\pi{r^2}}{360^o}[\theta_1+\theta_2+\theta_3]

Sum of interior angles of an triangle is $180^o$

$Here, \theta_1+\theta_2+\theta_3=180^o$

=>\frac{1}{2}{bh}-\frac{180^o}{360^o}{\pi{r^2}}

=>\frac{1}{2}\times{60}\times{61}-\frac{1}{2}\times{3.14\times{3\times{3}}}

=>\frac{1}{2}[3660-28.26]

=>1815.87 Cm^2