Question

Prove that " Between any two parallel lines , the area of all triangles drawn on the same base have same area " ​

Answer 1

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$\large \quad \qquad { \bigstar { \underline { \boxed { \displaystyle { \tt {theorem}}}}}}$

Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

EXAMPLE:-

■ Show that a median of a triangle divides it into two triangles of equal areas.

$\Large\underline{\pink{\underline{\frak{\pmb{ \: given }}}}}$

ABC is a Triangle,

AD is A Median.

$\Large\underline{\pink{\underline{\frak{\pmb{to \: prove}}}}}$

Median of a triangle divides it into two triangles of equal areas.

$\Large\underline{\pink{\underline{\frak{\pmb{proof }}}}}$

Let ABC be a triangle and let AD be one of its medians (see Fig. 9.21) You wish to show that

=> ar (ABD) = ar (ACD).

● Since the formula for area involves altitude, let us draw AN ⟂ BC.

Now,

=> ar(ABD) = ½ x base x altitude (of ABD)

=> ½ x BD x AN

=> ½ x CD x AN ( As BD = CD )

=> ½ x base x altitude (of Triangle ACD)

= ar(ACD).

$\quad \qquad { \bigstar \: \: { \underline \red{ \boxed { \displaystyle { \sf { \: \: \: \: hence \: proved \: \: \: }}}}}}$

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