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Preeti Dabral 1 year, 11 months ago
Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm. Let the radius of the circle be r cm. Draw OP {tex}\perp{/tex} AB and OQ {tex}\perp{/tex} CD. Since AB || CD and OP {tex}\perp{/tex} AB, OQ {tex}\perp{/tex} CD. Therefore, points O, Q, and P are collinear.
Clearly, PQ = 3 cm
Let OQ = x cm. Then, OP = (x + 3) cm
Applying Pythagoras theorem in right triangles OAP and OCQ, we obtain
OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2
{tex}\Rightarrow{/tex} r2 = (x + 3)2 + 32 and r2 = x2 + 62 [{tex}\because{/tex} AP = {tex}\frac{1}{2}{/tex} AB = 3 cm and CQ = {tex}\frac{1}{2}{/tex}CD = 6 cm]
{tex}\Rightarrow{/tex} (x + 3)2 + 32 = x2 + 62 [On equating the values of r2]
{tex}\Rightarrow{/tex}x2 + 6x + 9 + 9 = x2 + 36 {tex}\Rightarrow{/tex} 6x = 18 {tex}\Rightarrow{/tex} x = 3
Putting the value of x in r2 = x2 + 62, we get
r2 = 32 + 62 = 45 {tex}\Rightarrow{/tex} r = {tex}\sqrt{45}{/tex} cm = 6.7 cm
Hence, the radius of the circle is 6.7 cm.
Posted by Khushee Bhogayta 1 year, 11 months ago
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Preeti Dabral 1 year, 11 months ago
Here, ∠PQR=100∘
Take a point S in the major arc. Join PS and RS.
{tex}\because{/tex} PQRS is a cyclic quadrilateral.
{tex}\therefore \angle \mathrm { PQR } + \angle \mathrm { PSQ } = 180 ^ { \circ }{/tex}
|The sum of either pair of opposite angles of a cyclic quadrilateral is 180o
{tex}\Rightarrow 100 ^ { \circ } + \angle P S R = 180 ^ { \circ }{/tex}
{tex}\Rightarrow \angle P S R = 180 ^ { \circ } - 100 ^ { \circ }{/tex}
{tex}\Rightarrow \angle P S R = 80 ^ { \circ }{/tex} ......... (1)
Now, {tex}\angle \mathrm { POR } = 2 \angle \mathrm { PSR }{/tex}
|The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle
{tex}= 2 \times 80 ^ { \circ } = 160 ^ { \circ }{/tex} ........ (2) |Using (1)
In {tex}\triangle O P R{/tex}
{tex}\because O P = O R{/tex} |Radii of a circle
{tex}\therefore \angle \mathrm { OPR } = \angle \mathrm { ORP }{/tex} ....... (3)
|Angles opposite to equal sides of a triangle are equal
In {tex}\triangle O P R{/tex}
{tex}\angle O P R + \angle O R P + \angle P O R = 180 ^ { \circ }{/tex} | Sum of all the angles of a triangle is 180o
{tex}\Rightarrow \angle \mathrm { OPR } + \angle \mathrm { OPR } + 160 ^ { \circ } = 180 ^ { \circ }{/tex} |Using (2) and (1)
{tex}\Rightarrow 2 \angle O P R + 160 ^ { \circ } = 180 ^ { \circ }{/tex}
{tex}\Rightarrow 2 \angle O P R = 180 ^ { \circ } - 160 ^ { \circ } = 20 ^ { \circ }{/tex}
{tex}\Rightarrow \angle \mathrm { OPR } = 10 ^ { \circ }{/tex}
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Preeti Dabral 1 year, 11 months ago
Inner diameter of bowl = 10.5 cm
{tex}\therefore {/tex} Inner radius of bowl {tex}\left( r \right)=\frac{10.5}{2}{/tex} = 5.25 cm
Now, Inner surface area of bowl = {tex}2\pi {{r}^{2}}{/tex}
{tex}=2\times \frac{22}{7}\times 5.25\times 5.25{/tex}
{tex}=2\times \frac{22}{7}\times \frac{21}{4}\times \frac{21}{4}{/tex}
={tex}\frac{693}{4}c{{m}^{2}}{/tex}
{tex}\because {/tex} Cost of tin-plating per{tex} 100\text{ }c{{m}^{2}}{/tex} = ₹ 16
{tex}\therefore {/tex} Cost of tin-plating per{tex} 1\text{ }c{{m}^{2}}=\frac{16}{100}{/tex}
{tex}\therefore {/tex} Cost of tin-plating per {tex}\frac{693}{4}c{{m}^{2}}=\frac{16}{100}\times \frac{693}{4}{/tex} = ₹27.72
Posted by Tanisha Singh 1 year, 11 months ago
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Preeti Dabral 1 year, 11 months ago
- (b) 120°
- (a) 60°
- (c) 180°
- (d) {tex}\angle{/tex}q
- (a) 60°
Posted by Jatin Aggarwal 9Th D 1 year, 11 months ago
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Anvi 8A Arnav Singh11 1 year, 11 months ago
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