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Sia ? 6 years, 6 months ago
Let the radius of the circle be r cm.
Its circumference {tex} = 2\pi r\;cm{/tex}
As per the problem,
The circumference of the circle exceeds its diameter by 18 cm
{tex} \Rightarrow {/tex} Circumference – Diameter =180 cm
{tex} \Rightarrow 2\pi r - 2r = 180{/tex}
{tex} \Rightarrow \left( {2 \times \frac{{22}}{7} \times r} \right) - 2r = 180{/tex}
{tex} \Rightarrow \left( {\frac{{44r - 14r}}{7}} \right) = 180{/tex}
{tex} \Rightarrow \frac{{30r}}{7} = 180{/tex}
{tex} \Rightarrow r = \frac{{180 \times 7}}{{30}}{/tex}
{tex}= \frac{{420}}{{10}}{/tex}
= 42 cm
Hence the radius of the circle is 42 cm.
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Sia ? 6 years, 5 months ago
To construct: To construct a triangle ABC with side BC = 7 cm, {tex}\angle B = 45^\circ {/tex} and {tex}\angle A = 105^\circ {/tex} and then a triangle similar to it whose sides are {tex}\frac{4}{3}{/tex} of the corresponding sides of the first triangle ABC.
Steps of construction:
- Draw a triangle ABC with side BC = 7 cm, {tex}\angle B = 45^\circ {/tex}, {tex}\angle A= 105^\circ {/tex}and {tex}\angle C= 30^\circ {/tex}
- From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
- Locate 4 points B1, B2, B3 and B4 on BX such that BB1 = B1 B2 = B2 B3 = B3B.
- Join B3 C and draw a line through the point B4, draw a line parallel to B3 C intersecting BC at the point C'.
- Draw a line through C' parallel to the line CA to intersect BA at A'.
Then, A'BC' is the required triangle.
Justification :
{tex}\because {B_4}C'||{B_3}C{/tex} [By construction]
{tex}\therefore \triangle B{B_4}C' \sim \triangle B{B_3}C{/tex} [AA similarity]
{tex}\therefore \frac{{B{B_4}}}{{B{B_3}}} = \frac{{BC'}}{{BC}}{/tex} [By Basic Proportionality Theorem]
But {tex}\frac{{B{B_4}}}{{B{B_3}}} = \frac{4}{3}{/tex} [By construction]
Therefore, {tex}\frac{{BC'}}{{BC}} = \frac{4}{3}{/tex} ............... (i)
{tex}\because CA||C'A'{/tex} [By construction]
{tex}\therefore \triangle BC'A \sim \triangle BCA{/tex} [AA similarity]
{tex}\therefore \frac{{A'B}}{{AB}} = \frac{{A'C'}}{{AC}} = \frac{{BC'}}{{BC}} = \frac{4}{3}{/tex} [From eq. (i)]
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Prajjwal Srivastava 7 years, 9 months ago
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