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ABC is aright triangled at B …

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ABC is aright triangled at B .AB=28cm and BC=21cm, with AC as diameter a semicircle is drawn and with BC as radius a quadrant is drawn.find the area of shaded part

Posted by Abhishek Kandari 6 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

In {tex}\triangle{/tex}ABC, {tex}\angle{/tex}B = 90^o
AC² = AB² + BC²
AC² = 28²+ 21²
AC²= 784 + 441
AC² = 1225
AC = 35 cm
Area of shaded region = Area of triangle + Area of semi-circle with diameter AC - area of quadrant with radius BC
{tex}\;=\;\frac12\times(21\times28)+\frac12\times\frac{22}7\times\frac{35}2\times\frac{35}2-\frac14\times\frac{22}7\times21\times21{/tex}
{tex}\;=\;(7\times28)+\;11\times\frac52\times\frac{35}2-\frac12\times\frac{11}1\times3\times21{/tex}
{tex}= 294 + 481.25 - 346.5{/tex}
{tex}= 775.25 - 346.5{/tex}
{tex}= 428.75 \;cm^2.{/tex}

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