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The sides of triangles are 3cm, 4cm, and 5cm. At each vertex of the triangle, circles of radius 6 cm are drawn. What is the area of the triangle in sqcm, excluding the portion enclosed by circles? (π=3.14)(\pi=3.14).

A5 sqcm

B8 sqcm

C5.76 sqcm

D5.72 sqcm

Answer:

D. 5.72 sqcm

Read Explanation:

sides of are triangle values are triplets so it is an right angled triangle.

image.png

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

Area of Sector = θ360o×πr2\frac{\theta}{360^o}\times{\pi{r^2}}

Radius,r=1cmRadius, r=1 cm

Base b= 4., height h=3cm

Area of remaining portion = 12bhθ1360o×πr2+θ2360o×πr2+θ3360o×πr2\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 180o180^o

Here,θ1+θ2+θ3=180oHere, \theta_1+\theta_2+\theta_3=180^o

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

=>\frac{1}{2}\times{3}\times{4}-\frac{1}{2}\times{3.14\times{1\times{1}}}

=>\frac{1}{2}[6-3.14]

=>5.72 sqcm


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