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Yogita Ingle 4 years, 9 months ago
Let us consider two congruent circles (circles of same radius) with centers as O and O
In ΔAOB and ΔCO'D,
AOB = CO'D (Given)
OA = O'C (Radii of congruent circles)
OB = O'D (Radii of congruent circles)
ΔAOB ΔCO'D (SAS congruence rule)
AB = CD (By CPCT)
Hence, if chords of congruent circles subtend equal angles at their centers, then the chords are equal.
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Yogita Ingle 4 years, 9 months ago
When two angles are added together and they total 90 degrees, they are said to be complementary angles.
Let the angle be x
then x + 35° =90°
x = 90° - 35°
x = 55°
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Yogita Ingle 4 years, 9 months ago
{tex}Given ,\\ \\ Area \: \: of \: \: the \: \: equilateral \: \: \\ triangle \: \: is \: \: 36 \sqrt{3} \: cm {}^{2} \\ \\ To \: \: find \: \::- \: \:Height \: of \: \: the \\ equilateral \: \: triangle .\\ \\ \\ We \: \: know \: \: that ,\\ \\ Area \: of \: a \: equilateral \: \\ triangle \: \: = \frac{a {}^{2} }{4} \sqrt{3} \: \: \: \: ; where \: \: \: a \: \: \: \\ is \: \: the \: length \: of \: each \: side \: of \: the \\equlateral \: triangle.\\ \\ \\A.T.Q. \\ \\ \frac{a {}^{2} }{4} \sqrt{3} = 36 \sqrt{3} \\ \\ = > \frac{a {}^{2} }{4} = 36 \\ \\ = > a {}^{2} = 36 \times 4 \\ \\ = > a {}^{2} = (6) {}^{2} \times( 2) {}^{2} \\ \\ = > a = 6 \times 2 \\ \\ = > a = 12 \\ \\ \\ Length \: of \: each \: side \: of \: the \: \\ equilateral \: triangle \: is \: 12 \: cm{/tex}
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Yogita Ingle 4 years, 9 months ago
Steps Of Construction:
1. Draw a line segment AB = 12cm.
2. Draw a ray AX making right angle with AB.
3. Cut a line segment AD of length 18 cm from ray AX.
4. Join BD and make an angle DBY equal to ADB.
5. Let BY intersect AX at C. Join AC and BC.
Triangle ABC is the required triangle.
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