Question

The side of a triangle are in A.P and its area is 3/5th the area of the equilateral triangle of same perimeter . The sides of triangle are in Ratio :-
a)1:2:√7
b)2:3:5
c)1:6:7
d)3:5:7​

Answer 1

${ \boxed{ \color{blue}{ \bold{Question :–}}}}$

The side of a triangle are in A.P and its area is 3/5th the area of the equilateral triangle of same perimeter . The sides of triangle are in Ratio :-

a)1:2:√7

b)2:3:5

c)1:6:7

d)3:5:7

${ \boxed{ \color{blue}{ \bold{ Answer :–}}}}$

• 3 : 5 : 7

${ \boxed{ \color{blue}{ \bold{ Solution :–}}}}$

$\sf{let \: the \: side \: of \: a \: triangle \: be \: (a - d),(a),(a + d)}$

$\sf{Perimeter \: of \: the \: triangle \: = a - d + a + a + d = 3(a)}$

$\sf{ \therefore \: Each \: side \: of \: a \: equilateral \: triangle \: = \frac{3a}{2} }$

$\sf{ \therefore \: Area \: of \: equilateral \: triangle \: = \frac{ \sqrt{3} }{4} \: {a}^{2} }$

• Area of given triangle ∆

$\circ \: \: \: { \boxed{ \color{red}{ \sqrt{s(s - (a - d)(s - a)( s-a + d) } }}}$

• where s = $\rm{ \frac{3a}{2} }$

$\sf{⇒ \sqrt{ \frac{3a}{2} \left[ \frac{3a}{2} - a + a \right] \left[ \frac{3a}{2} - a \right] \left[ \frac{3a}{2} - a - d \right]}}$

$\sf{⇒ \sqrt{ \frac{3a}{2} \left[ \frac{1}{2} - a + a \right] \left[ \frac{1}{2} a \right] \left[ \frac{1}{2} a - d \right]}}$

$\sf{⇒ \sqrt{ \frac{3}{1} \: {a}^{2} \left[ \frac{1}{4} \: {a}^{2} - {d}^{2} \right]}}$

• Given $\sf{ \sqrt{ \frac{3}{4} \: {a}^{2} - \frac{3}{4} \: {a}^{2} \: {d}^{2} }} = \frac{3}{5} \times \frac{ \sqrt{3} }{4} \: {a}^{2}$

$: \implies\sf{ = \frac{3}{16} {a}^{4} - \frac{3}{4} {a}^{2} {d}^{2} = \left[ \frac{3 \sqrt{3} }{20} \: {a}^{2} \right] }$

$: \implies\sf{ = \frac{3}{16} {a}^{4} - \frac{3}{4} {a}^{2} {d}^{2} = \frac{27}{400} {a}^{4} }$

$: \implies\sf{ = \frac{3}{16} {a}^{4} - \frac{27}{400} {a}^{4} = \frac{3}{4} {a}^{2} {d}^{2}}$

$: \implies\sf{ = \frac{1}{16} {a}^{2} - \frac{9}{400} {a}^{4} = \frac{1}{4} {d}^{2}}$

$: \implies\sf{ = \frac{25 {a}^{2} - 9 {a}^{2} }{400} = \frac{1}{4} \: {d}^{2} }$

$: \implies\sf{ = \frac{16 \: {a}^{2} }{400} = \frac{1}{4} \: {a}^{2} }$

$: \implies\sf{ = \frac{ {a}^{2} }{ {d}^{2} } = \frac{400}{64} }$

$: \implies\sf{ = \frac{ {a}}{ {d}} = \frac{ \sqrt{400} }{ \sqrt{64} } }$

$: \implies\sf{ = \frac{ {a}}{ {d}} = \frac{20}{8} }$

$: \implies\sf{ = \frac{ {a}}{ {d}} = k}$

• calculate the ratio

$\leadsto \: \sf{ (a - d) \ratio(a) \ratio(a + d)}$

$\leadsto \: \sf{ (20k - 8k) \ratio(20k) \ratio(20k + 8k)}$

$\leadsto \: \sf{ (12k) \ratio(20k) \ratio(28k)}$

$\leadsto \: \sf{3 \ratio \: 5 \ratio \: 7 }$

$\sf \pink{hence \: option \:( d) \: is \: correct}$

Answer 2

Step-by-step explanation:

${ \boxed{ \color{blue}{ \bold{Question :–}}}}$

The side of a triangle are in A.P and its area is 3/5th the area of the equilateral triangle of same perimeter . The sides of triangle are in Ratio :-

a)1:2:√7

b)2:3:5

c)1:6:7

d)3:5:7

${ \boxed{ \color{blue}{ \bold{ Answer :–}}}}$

3 : 5 : 7

${ \boxed{ \color{blue}{ \bold{ Solution :–}}}}$

$\sf{let \: the \: side \: of \: a \: triangle \: be \: (a - d),(a),(a + d)}$

$\sf{Perimeter \: of \: the \: triangle \: = a - d + a + a + d = 3(a)}$

$\sf{ \therefore \: Each \: side \: of \: a \: equilateral \: triangle \: = \frac{3a}{2} }$

$\sf{ \therefore \: Area \: of \: equilateral \: triangle \: = \frac{ \sqrt{3} }{4} \: {a}^{2} }$

Area of given triangle ∆

$\circ \: \: \: { \boxed{ \color{red}{ \sqrt{s(s - (a - d)(s - a)( s-a + d) } }}}$

where s = $\rm{ \frac{3a}{2} }$

$\sf{⇒ \sqrt{ \frac{3a}{2} \left[ \frac{3a}{2} - a + a \right] \left[ \frac{3a}{2} - a \right] \left[ \frac{3a}{2} - a - d \right]}}$

$\sf{⇒ \sqrt{ \frac{3a}{2} \left[ \frac{1}{2} - a + a \right] \left[ \frac{1}{2} a \right] \left[ \frac{1}{2} a - d \right]}}$

$\sf{⇒ \sqrt{ \frac{3}{1} \: {a}^{2} \left[ \frac{1}{4} \: {a}^{2} - {d}^{2} \right]}}$

Given $\sf{ \sqrt{ \frac{3}{4} \: {a}^{2} - \frac{3}{4} \: {a}^{2} \: {d}^{2} }} = \frac{3}{5} \times \frac{ \sqrt{3} }{4} \: {a}^{2}$

$: \implies\sf{ = \frac{3}{16} {a}^{4} - \frac{3}{4} {a}^{2} {d}^{2} = \left[ \frac{3 \sqrt{3} }{20} \: {a}^{2} \right] }$

$: \implies\sf{ = \frac{3}{16} {a}^{4} - \frac{3}{4} {a}^{2} {d}^{2} = \frac{27}{400} {a}^{4} }$

$: \implies\sf{ = \frac{3}{16} {a}^{4} - \frac{27}{400} {a}^{4} = \frac{3}{4} {a}^{2} {d}^{2}}$

$: \implies\sf{ = \frac{1}{16} {a}^{2} - \frac{9}{400} {a}^{4} = \frac{1}{4} {d}^{2}}$

$: \implies\sf{ = \frac{25 {a}^{2} - 9 {a}^{2} }{400} = \frac{1}{4} \: {d}^{2} }$

$: \implies\sf{ = \frac{16 \: {a}^{2} }{400} = \frac{1}{4} \: {a}^{2} }$

$: \implies\sf{ = \frac{ {a}^{2} }{ {d}^{2} } = \frac{400}{64} }$

$: \implies\sf{ = \frac{ {a}}{ {d}} = \frac{ \sqrt{400} }{ \sqrt{64} } }$

$: \implies\sf{ = \frac{ {a}}{ {d}} = \frac{20}{8} }$

$: \implies\sf{ = \frac{ {a}}{ {d}} = k}$

calculate the ratio

$\leadsto \: \sf{ (a - d) \ratio(a) \ratio(a + d)}$

$\leadsto \: \sf{ (20k - 8k) \ratio(20k) \ratio(20k + 8k)}$

$\leadsto \: \sf{ (12k) \ratio(20k) \ratio(28k)}$

$\leadsto \: \sf{3 \ratio \: 5 \ratio \: 7 }$

$\sf \pink{hence \: option \:( d) \: is \: correct}$.

muskan