आकृति में, 14, cm की त्रिज्याएँ लेकर तथा P , | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Given that, radii of each arc ( $r$ ) $=14\, cm$

Now, area of the sector with central $\angle P=\frac{\angle P}{360^{\circ}} 	imes \pi r^{2}$

$

Vedclass · Accepted Answer

$308$

Vedclass · Answer

Given that, radii of each arc ( $r$ ) $=14\, cm$

Now, area of the sector with central $\angle P=\frac{\angle P}{360^{\circ}} 	imes \pi r^{2}$

$=\frac{\angle P}{360^{\circ}} 	imes \pi 	imes(14)^{2} \,cm ^{2}$

[$\because$ area of any sector with central angle $	heta$ and radius $\left.r=\frac{\pi^{2}}{360^{\circ}} 	imes 	hetaight]$

Area of the sector with central angle $=\frac{\angle Q}{360^{\circ}} 	imes \pi r^{2}=\frac{\angle Q}{360^{\circ}} 	imes \pi 	imes(14)^{2}\,cm ^{2}$

and area of the sector with central angle $R=\frac{\angle R}{360^{\circ}} 	imes \pi r^{2}=\frac{\angle R}{360^{\circ}} 	imes \pi 	imes(14)^{2} \,cm ^{2}$

Therefore, sum of the areas (in $cm ^{2}$ ) of three sectors

$=\frac{\angle P}{360^{\circ}} 	imes \pi 	imes(14)^{2}+\frac{\angle 	heta}{360^{\circ}} 	imes \pi 	imes(14)^{2}+\frac{\angle R}{360^{\circ}} 	imes \pi 	imes(14)^{2}$

$=\frac{\angle P+\angle Q+\angle R}{360} 	imes 196 	imes \pi=\frac{180^{\circ}}{360^{\circ}} 	imes 196 \pi \,cm ^{2}$

[since, sum of all interior angles in any triangle is $\left.180^{\circ}ight]$

$=98 \pi \,cm ^{2}=98 	imes \frac{22}{7}$

$=14 	imes 22=308 \,cm ^{2}$

Hence, the required area of the shaded region is $308 cm ^{2}$.

Similar Questions

क्या यह कहना सत्य है कि व्यास $p \,cm$ वाले एक वृत्त के अंतर्गत वर्ग का क्षेत्रफल $p^{2} \,cm ^{2}$ है? क्यों?

व्यासों $36\, cm$ और $20 \,cm$ वाले दो वृत्तों की परिधियों के योग के बराबर परिधि वाले एक वृत्त की त्रिज्या है ($cm$ में)