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  • In fig.8 , the vertices of  are A (4,6), B(1,5) and C(7,2) . A line segment CE is drawn to intersect the sides AB and AC at  D and E res

In fig.8 , the vertices of \Delta ABC are A (4,6), B(1,5) and C(7,2) . A line segment CE is drawn to intersect the sides AB and AC at 

D and E respectively such that \frac{AD}{AB} = \frac{AE }{AC} = \frac{1}{3} . Calculated the area of \Delta ADE and compare it with area of \Delta ABC

 

 

 

 
 
 
 
 

Answers (1)

A = ( x_1,y_1)=(4,6)

B=( x_2,y_2)=(1,5) 

C=( x_3,y_3)=(7,2) .

\frac{AD }{AB} = \frac{AE }{AC} = 1/3

\frac{AD}{AB} = \frac{1}{3}\: \: \: ,\: \: \: \: \: \: \: \: \: \: \: \: \: \frac{AE }{AC}= \frac{1}{3} \\\\ 3AD = AB \: \: \: , \:\: \: \: \: \: \:\: \: \: \: \: \: \: 3 AE = AC \\\\ 3AD = AD + BD \: \: \: \: \: , \: \: \: 3 AE = AE + CE\\\\ 2 AD = BD \: \: \: \: \: \: \: \:\: \: \: , \: \: \: \: \: \: \: \:2 AE = CE \\\\ \frac{AD}{BD} = \frac{1}{2} \: \:\: \: \: ,\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{AE }{CE } = \frac{1}{2}

Thus D divide AB in 1:2       ,     Thus E divides AC in 1:2

 Using section formula 

D = ({\frac{mx_2+ n x_1}{m+n}}, {\frac{my_2+ n y_1}{m+n}}) \: \: \: \: \: \: ,\: \: \: \: \: \: E= ({\frac{mx_2+ n x_1}{m+n}}, {\frac{my_2+ n y_1}{m+n}})

here , x_1 = 4 , x _2 = 1 ; y _ 1 =6 ; y_2 =5, m= 1 ; n = 2\: \: \: \: ,\: \: \: here , x_1 = 4 , x _2 = 7; y _ 1 =6 ; y_2 =2, m= 1 ; n = 2

D = ({\frac{1\times1+2 \times 4}{1+2}}, {\frac{1 \times 5+2 \times 6}{1+2}}) \: \: \: \: \: \: ,\: \: \: \: \: \:E= ({\frac{1\times7+2 \times 4}{1+2}}, {\frac{1 \times 2+2 \times 6}{1+2}})\\\\\\ D = \left ( \frac{1+8}{3}, \frac{5+12}{3} \right ), E = \left ( \frac{7+8}{3}, \frac{2+12}{3} \right )\\\\\\D = \left ( \frac{9}{3},\frac{17}{3} \right )\: \: \: \: ,\: \: \: E = \left (5,\frac{14}{3} \right )\\\\

Now area of \Delta = 1/2 \left [ x _1 ( y_2 - y_3) + x_2 ( y_3 - y_1) + x _3 ( y_1 - y_2)\right ]\\\\ area \: \: of \: \: \Delta ABC = 1/2 \left [ 4 ( 5-2)+ 1 ( 2-6 )++7 ( 6-5 ) \right ] \\\\ = 1/2 [4 \times 3 + 1 \times ( -4)+ 7 \times 1 ]\\\\ =1/2 [ 12-4+7 ] \\\\ 15/2 \: \: \: \: unit ^2

Area \: \: of \: \: \Delta ADE = \frac{1}{2} [ 4 (\frac{17}{3}-\frac{14}{3})] + 3 [ \frac{14}{3}-6] + 5 \left ( 6 - \frac{17}{3} \right )\\\\ 1/2 [ 4 \times\frac{3}{3}+ 3x \frac{-4}{3}+ 5 \times \frac{1}{3} ] \\\\ = 1/2 [ \frac{12-12+5 }{3}]\\\\ area \: \: of \: \: \Delta ADE = 5/6 unit ^2

\frac{area (\Delta ADE )}{area ( \Delta ABC )} = \frac{\frac{5}{6}}{\frac{15}{2}}= \frac{5 \times 2}{6 \times 15} = 1/9

\frac{area (\Delta ADE )}{area ( \Delta ABC )} = 1/9

Posted by

Ravindra Pindel

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