Question

Given Equilateral triangle ABC with interior point D if the perpendicular distances from D to the sides of 4, 5 and 6 the sum of all numbers so formed is.?

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Answer 1

$\huge{\underline{\sf{\red{Answer}}}}$

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Given:

∆ABC is an Equilateral Triangle

D is a point inside the triangle

Perpendicular Distances

$\perpendicular A = 4 \:,\perpendicular B =6 \: , \perpendicular C = 5$

Formula

• Area of Triangle = ½ x b x h
• Area of Equilateral Triangle = $\dfrac{\sqrt3 a^2}{4}$

SoluTion:

Area of ABC = ½ x 4 + ½ x a x 5 +½ x a x 6

$\implies \dfrac{a}{2} ( 4+5+6) \\ \implies \dfrac{15a}{2}$

now applying the 2nd Formula

Equilateral Triangle =

$\dfrac{\sqrt3 a^2}{4} \\ \leadsto \dfrac{15\cancel a}{\cancel2}=\dfrac{\sqrt3 \cancel {a^2}}{\cancel4} \\ a = \dfrac{30}{\sqrt3}$

Area of Triangle

Putting value of a in Formula

$\dfrac{\sqrt3}{4} bigg( \dfrac{30}{\sqrt3}\bigg)^2 \\ \implies \dfrac{\sqrt3}{\cancel 4} \times \dfrac{\cancel{900}}{\cancel3}$

$\large{\boxed{Area = 75\sqrt3}}}$

Answer 2

${\large\bf{\mid{\overline{\underline{Given:-}}}\mid}}$

• perpendicular distance From Point D to the sides of Equilateral ∆ are 4, 5 and 6 cm...

${\large\bf{\mid{\overline{\underline{Correct\:Question}}}\mid}}$

• we have to Find Area of Equilateral ∆ .

$\Large\underline{\underline{\sf{Solution}:}}$

$\large\underline\textbf{Refer To Image First.}$

$\LARGE\bold\star\underline{\underline\textbf{Formula\:used}}$

• Area of ∆ = 1/2 × Base × Height
• Area of Equilateral ∆ = √3/4 (a)² where a = sides of Equilateral ∆ ..

From image we can see that, we have 3 ∆'s inside Equilateral ∆ABC .

→ Area ∆ADB = 1/2 × a × 4 = 2a

→ Area ∆BDC = 1/2 × a × 6 = 3a

→ Area ∆ADC = 1/2 × a × 5 = 2.5a

Area ∆ABC = Area ∆ADB + Area ∆BDC + Area ∆ADC

Area ∆ABC = 2a + 3a + 2.5a = 7.5a ------ Equation (1)

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Now, Area of ∆ABC also = (√3/4)a² -----Equation(2)

$\red{Putting \: both \: Equal \: we \: get,,}$

$7.5a = \frac{ \sqrt{3} }{4} {a}^{2} \\ \\ \implies \: a \: = \frac{7.5 \times 4}{ \sqrt{3} } \\ \\ \implies \: a = \pink{10 \sqrt{3}}$

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$\textbf{Hence, Area of ABC} = \\ \\ \implies \: \green{ \frac{ \sqrt{3} }{4} ( {10 \sqrt{3}) }^{2}} \\ \\ \implies \orange{ \frac{300 \sqrt{3} }{4} } \\ \\ \implies \: \red{75 \sqrt{3}} \: {cm}^{2}$

$\large\underline\textbf{Hope it Helps You.}$