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The sides of triangles are 10cm, 24cm, and 26cm. 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? (π=3.14)(\pi=3.14).

A105.87

B110

C126.6

D114

Answer:

A. 105.87

Read Explanation:

sides of are trianlge 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=3cmRadius,r=3cm

Base b= 24., height h=10cm

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{24}\times{10}-\frac{1}{2}\times{3.14\times{3\times{3}}}

=>\frac{1}{2}[240-28.26]

=>105.87


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