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Maakalyani Studio 6 years, 2 months ago
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Sia ? 6 years, 2 months ago
Since, sides of coloured triangular wall are 15 m, 11 m and 6 m.
{tex}\therefore{/tex} Semi-perimeter of coloured triangular wall
S ={tex}\frac{15+11+6}{2}{/tex}={tex}\frac{32}{2}{/tex}=16 m
Now, Using Heron’s formula,
Area of coloured triangular wall
={tex}\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}{/tex}
={tex}\sqrt{16\left( 16-15 \right)\left( 16-11 \right)\left( 16-6 \right)}{/tex}
={tex}\sqrt{16\times 1\times 5\times 10}{/tex}
={tex}20\sqrt{2}{{m}^{2}}{/tex}
Hence area painted in blue colour = {tex}20\sqrt{2}{{m}^{2}}{/tex}
Angadh Kamboj 6 years, 2 months ago
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Yogita Ingle 6 years, 2 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 6 years, 2 months ago
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Yogita Ingle 6 years, 2 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 6 years, 2 months ago
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Posted by Ridhima Rana 6 years, 2 months ago
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Sia ? 6 years, 2 months ago
- In {tex}\triangle{/tex}ABC and {tex}\triangle{/tex}BAC
AD = BC [Given]
{tex} \angle {/tex}DAB = {tex} \angle {/tex}CBA [Given]
AB = AB [Common]
{tex}\therefore \triangle ABD \cong \triangle BAC{/tex} [By SAS congruency] - Since
ΔABD≅ΔBAC
{tex}\therefore {/tex} BD=AC[By C.P.C.T.] - Since
ΔABD≅ΔBAC{tex}\triangle \mathrm{ABD} \cong \triangle \mathrm{BAC}{/tex}
{tex}\therefore{/tex}{tex} \angle {/tex}ABD = {tex} \angle {/tex}BAC [By C.P.C.T.]
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Prabhpreet Singh Sardar 6 years, 2 months ago
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Rohan Bamniya 6 years, 2 months ago
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