Question

pls answer this question....
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Answer 1

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow PQR\:is\:a\:triangle\:which\:is\:inscribed\:in\:a\:semicircle.$

$\sf\dashrightarrow PQ=PR=7cm$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow 1)measure\:of\: \angle QPR$

$\sf\dashrightarrow 2)the\:area\:of\:shaded\:region$

$\large\underline\bold{SOLUTION,}$

FINDING QR,

$\sf\therefore in \: \triangle QPR$

$\sf\large\therefore \: BY\: PYTHAGORAS\:THEOREM.$

$\sf\dashrightarrow QR^2=QP^2+PR^2$

$\sf\implies QR^2= (7)^2+(7)^2$

$\sf\implies QR^2= 49+49$

$\sf\implies QR= \sqrt{98}$

$\sf\implies QR= \sqrt{ 7\times 2}$

$\sf\implies QR=7 \sqrt{2}$

$\large{\boxed{\bf{QR= 7\sqrt{2} }}}$

$\sf\large\star \red{QUESTION.1}$

$\sf\star \purple{what \:is\: the\: measure\: of\: \angle QPR}$

ANSWER:-

$\sf\therefore in\: \triangle QPR$

$\sf\dashrightarrow using\: trigonometric\:ratios.\:we\:get,$

$\sf\therefore in\: \triangle PQR, \:$

$\sf\dashrightarrow PQ=PR$

$\sf\therefore cos \theta= \dfrac{(opposite)^2+(adjacent)^2- (hypotenuse)^2}{ 2 \times (opposite \times adjacent)}$

$\sf\implies cos \theta= \dfrac{(7)^2+(7)^2-(7\sqrt{2})^2}{2 \times 7 \times 7}$

$\sf\implies \dfrac{ 49+49-98}{98}$

$\sf\implies \dfrac{98-98}{98}$

$\sf\implies \dfrac{0}{98}$

$\sf\therefore cos \theta =0$

$\sf\therefore cos 90 \degree = 0$

$\sf\dashrightarrow \theta = 90 \degree$

$\large{\boxed{\bf{ \star\:\: \angle QPR=90 \degree \:\: \star}}}$

$\sf\therefore NOTE:-QR\:is\:a\:diameter.$

$\sf\therefore RADIUS= \dfrac{QR}{2}$

$\sf\implies RADIUS = \dfrac{7 \sqrt{2}}{2}$

$\large{\boxed{\bf{ radius\:of\:the\: semicircle= \dfrac{7 \sqrt{2}}{2}}}}$

$\sf\large\star \red{QUESTION.2}$

$\purple{\text{what is the area of shaded region. }}$

ANSWER:-

$\sf{\boxed{\bf{ \star\:\: area \:of\:shaded\:region= \dfrac{\pi r^2}{2} - \dfrac{1}{2} \times b \times h \:\: \star}}}$

$\sf\therefore \bigg( \dfrac{\pi \bigg(\dfrac{7 \sqrt{2}}{2} \bigg)^2}{2} \bigg) - \bigg( \dfrac{1}{2} \times b \times h \bigg)$

$\sf\implies \bigg( \dfrac{ \dfrac{22}{7} \times \bigg( \dfrac{7 \sqrt{2}}{2 } \bigg)^2}{2} \bigg) - \dfrac{1}{2} \times 7 \times 7$

$\sf\implies \bigg( \dfrac{ \dfrac{22}{\cancel{7}} \times \bigg( \dfrac{\cancel{49 }\times \cancel{2}}{\cancel{2} } \bigg)^2}{2} \bigg) - \dfrac{1}{2} \times 7 \times 7$

$\sf\implies \dfrac{22 \times 7}{2}- \dfrac{49}{2}$

$\sf\implies \dfrac{154-49}{2}$

$\sf\implies \dfrac{105}{2}$

$\sf\implies \cancel \dfrac{105}{2}$

$\sf\implies 52.5cm^2$

$\large{\boxed{\bf{ \star\:\: area\:of\:shaded\:region\:is\:52.5cm^2\:\: \star}}}$

$\large\underline\bold{AREA\:OF\:SHADED\:REGION\:IS\:52.5cm^2}$

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