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Ask QuestionPosted by Zarfa Umer 6 years, 10 months ago
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Sia ? 6 years, 4 months ago
Area of a remaining portion of the square = Area of square - (4 {tex}\times{/tex} Area of a quadrant + Area of a circle)
{tex}= 4 \times 4 - \left[ 4 \times \frac { 90 } { 360 } \times \frac { 22 } { 7 } \times 1 ^ { 2 } + \frac { 22 } { 7 } \times \left( \frac { 2 } { 2 } \right) ^ { 2 } \right]{/tex}
{tex}= 16 - 2 \times \frac { 22 } { 7 }{/tex}
{tex}= \frac { 68 } { 7 } \mathrm { cm } ^ { 2 }{/tex}
Posted by Tushir Family 6 years, 10 months ago
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Sia ? 6 years, 4 months ago
Let the larger number be x,
Then, (smaller number)2 = 8(larger number)
= 8x {tex}\Rightarrow{/tex} smaller number {tex}= \sqrt { 8 x }{/tex}
According to the question, x2 - 8x = 180
{tex}\Rightarrow x ^ { 2 } - 8 x - 180 = 0{/tex}
Using the quadratic formula, {tex}= \frac { - b \pm \sqrt { b ^ { 2 } - 4 a c } } { 2 a }{/tex}
we get {tex}= \frac { - ( - 8 ) \pm \sqrt { ( - 8 ) ^ { 2 } - 4 ( 1 ) ( - 180 ) } } { 2 ( 1 ) }{/tex}
{tex}= \frac { 8 \pm \sqrt { 64 + 720 } } { 2 } = \frac { 8 \pm \sqrt { 784 } } { 2 } = \frac { 8 \pm 28 } { 2 }{/tex}
{tex}= \frac { 8 + 28 } { 2 } , \frac { 8 - 28 } { 2 } \Rightarrow x = 18 , - 10{/tex}
x = -10 is inadmissible
Then smaller number {tex}= \sqrt { 8 ( - 10 ) } = \sqrt { - 80 }{/tex}
which does not exist.
{tex}\therefore x = 18 \therefore \sqrt { 8 x } = \sqrt { 8 \times 18 } = \sqrt { 144 } = \pm 12{/tex}
Hence, the two numbers are 18, 12 or 18, -12.
Posted by Insha . 6 years, 10 months ago
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Lustig Yashu? 6 years, 10 months ago
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Sia ? 6 years, 4 months ago
In order to draw the pair of tangents, we follow the following steps.
Steps of construction
STEP I Take a point O on the plane of the paper and draw a circle of radius OA = 5 cm.
STEP II Produce OA to B such that OA = AB = 5 cm.
STEP III Taking A as the centre draw a circle of radius AO = AB = 5 cm. Suppose it cuts the circle drawn in step I at P and Q.
STEP IV Join BP and BQ to get the desired tangents.
Justification: In OAP, we have
OA = OP = 5 cm (= Radius)
Also, AP = 5 cm (= Radius of circle with centre A)
{tex}\therefore{/tex} {tex}\Delta O A P{/tex} is equilateral. {tex}\Rightarrow \angle P A O = 60 ^ { \circ } \Rightarrow \angle B A P = 120 ^ { \circ }{/tex}
In {tex}\Delta B A P{/tex}, we have
BA = AP and {tex}\angle B A P{/tex} = 120°
{tex}\angle A B P = \angle A P B = 30 ^ { \circ }{/tex}
{tex}\Rightarrow \quad \angle P B Q = 60 ^ { \circ }{/tex}
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