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Ask QuestionPosted by Vicky Singh 5 years, 11 months ago
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Posted by Khushi Sharma 5 years, 11 months ago
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Raj Singh 5 years, 11 months ago
Posted by Harjeet Kaur 5 years, 11 months ago
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Yogita Ingle 5 years, 11 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.
Posted by Rohit Sharma 5 years, 11 months ago
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Posted by Tanmay Sonawane 5 years, 11 months ago
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Khushi Sharma 5 years, 11 months ago
Kritika Dhiman 5 years, 11 months ago
Posted by Tanmay Sonawane 5 years, 11 months ago
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Utsav Kumar 5 years, 11 months ago
Posted by Arjun Singh 5 years, 11 months ago
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Yogita Ingle 5 years, 11 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°
Posted by Akanksha Jha 5 years, 11 months ago
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Posted by Akanksha Jha 5 years, 11 months ago
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Yogita Ingle 5 years, 11 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}
Posted by Alia Imtiaz 5 years, 11 months ago
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Ruchi Bansal 5 years, 11 months ago
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Yogita Ingle 5 years, 11 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.
Posted by Vikaram Thakur 5 years, 11 months ago
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Posted by Nikki Poojar 5 years, 11 months ago
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Utsav Kumar 5 years, 11 months ago
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Khushi Sharma 5 years, 11 months ago
Yogita Ingle 5 years, 11 months ago
RHS (Right angle- Hypotenuse-Side)
If the hypotenuse and a side of a right- angled triangle is equivalent to the hypotenuse and a side of the second right- angled triangle, the right triangles are said to be congruent by RHS rule.
In above figure, hypotenuse XZ = RT and side YZ=ST, hence triangle XYZ ≅ triangle RST.
Posted by Mahima Agrawal 5 years, 11 months ago
- 2 answers
Yogita Ingle 5 years, 11 months ago
2+8-3(9+5-3*6)/2
= 2+8-3(9+5-18)/2
= 2+8-3(14-18)/2
= 2+8-3(-4)/2
= 2+8+12/2
= 2 + 8 +6
= 10 + 6
= 16
Posted by Shashikala Singh 5 years, 11 months ago
- 1 answers
Yogita Ingle 5 years, 11 months ago
3m - 8n = 27
3m = 27 + 8n
m = (27 + 8n)/3
Taking different values of n, we can calculate the corresponding values of m
For example, when n = 0, m=9. when n=3, m=17 and when n=6, m=25
There can be several sets of other solutions
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Rishabh Gupta 5 years, 11 months ago
0Thank You