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Ann Riya Dsouza 5 years, 9 months ago
Nikhil Pandey 5 years, 9 months ago
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Maakalyani Studio 5 years, 9 months ago
Posted by Lakshita Teotia 5 years, 9 months ago
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Sia ? 5 years, 9 months ago
Since, sides of coloured triangular wall are 15 m, 11 m and 6 m.
∴ Semi-perimeter of coloured triangular wall
S =15+11+62=322=16 m
Now, Using Heron’s formula,
Area of coloured triangular wall
=√s(s−a)(s−b)(s−c)
=√16(16−15)(16−11)(16−6)
=√16×1×5×10
=20√2m2
Hence area painted in blue colour = 20√2m2
Angadh Kamboj 5 years, 9 months ago
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Tishya Jha 5 years, 9 months ago
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Asmita Mukane Mukane 5 years, 9 months ago
Asmita Mukane Mukane 5 years, 9 months ago
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Aditya Tripathi 5 years, 9 months ago
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Prince Jindal 5 years, 9 months ago
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Tishya Jha 5 years, 9 months ago
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Yogita Ingle 5 years, 9 months ago
Given, ∠ DAB + EBA = 180°. CA and CB are bisectors of ∠ DAB ∠ EBA respectively.
∴ ∠ DAC + ∠ CAB = 1/2 (∠ DAB).....(1)
⇒ ∠ EBC + ∠ CBA = 1/2 (∠ EBA)....(2)
⇒ ∠ DAB + ∠ EBA = 180°
⇒ 2 (∠ CAB) + 2 (∠ CBA) = 180° [using (1) and (2)]
⇒ ∠ CAB + ∠ CBA = 90°
In Δ ABC,
∠ CAB + ∠ CBA + ∠ ABC = 180° (Angle Sum property)
⇒ 90° + ∠ ABC = 180°
⇒ ∠ ABC = 180° - 90°⇒ ∠ ABC = 90°
Posted by Shivang Bhatia 5 years, 9 months ago
- 1 answers
Yogita Ingle 5 years, 9 months ago
(a+b+c)3 - a3-b3-c3 = [(a+b)+c]3 – a3-b3-c3
= (a+b)3+c3+3c(a+b)(a+b+c)-a3-b3-c3
= a3+b3+3ab(a+b)+c3+3c(a+b)(a+b+c)-a3-b3-c3
= a3+b3+c3+3ab(a+b)+3c(a+b)(a+b+c)-a3-b3-c3
= 3ab(a+b)+3c(a+b)(a+b+c) 3(a+b)[ab+c(a+b+c)]
= 3(a+b)[ab+ac+bc+c2] 3(a+b)[a(b+c)+c(b+c)]
= 3(a+b)(b+c)(c+a)
Posted by Aryan Shivhare 5 years, 9 months ago
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Posted by Ridhima Rana 5 years, 9 months ago
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Sia ? 5 years, 9 months ago
- In △ABC and △BAC
AD = BC [Given]
∠DAB = ∠CBA [Given]
AB = AB [Common]
∴△ABD≅△BAC [By SAS congruency] - Since
ΔABD≅ΔBAC
∴ BD=AC[By C.P.C.T.] - Since
ΔABD≅ΔBAC△ABD≅△BAC
∴∠ABD = ∠BAC [By C.P.C.T.]
Posted by Anabil Goswami 5 years, 9 months ago
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Prabhpreet Singh Sardar 5 years, 9 months ago
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Rohan Bamniya 5 years, 9 months ago
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