Question

If a triangle and a rhombus are on the same base and between the same parallels then the ratio of the areas of the triangle and the rhombus is ?

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Answer 1

Step-by-step explanation:

Given:-

A triangle and a rhombus are on the same base and between the same parallels.

To find:-

Find the ratio of the areas of the triangle and the rhombus ?

Solution :-

See the above attachment

Given that

A triangle and a rhombus are on the same base and between the same parallels.

From the above figure ,

∆ABC and ABEF are the triangle and rhombus and they are on the samne bse AB and the same Parallels AB and EF.

Draw a Parallel line BH to AC from B and extend it to H so that it intersects FE.

AB || CH

AC || BH

It is a Parallelogram

ABHC is a Parallelogram

And we know that

Rhombus is a special type of Parallelogram

BC divides ABHC into two congruent triangles

=> ∆ABC is congruent to ∆ BHC

=> area (∆ABC) = area (∆BCH)

=> area (∆ABC) = 1/2(Area (ABHC))

=> area (∆ABC) = (1/2) × Area (ABEF)

=> Area (ABEF) = 2× Area (∆ABC)

Since ABHC and ABEF are lie on the same Parallels AB||EF

Now the ratio of the areas of the triangle and the rhombus

=> area (∆ABC):area (ABEF)

=> area (∆ABC):2 area(∆ABC)

=> 1:2

Answer:-

The ratio of the areas of the triangle and the rhombus is 1:2

Used formulae:-

• Two pairs of opposite sides are parallel and equal in a Parallelogram

• Two pairs of opposite sides are parallel and equal in a rhombus ,so It is a special type of Parallelogram.

Answer 2

Answer:

The ratio of the areas of the triangle and the rhombus on the same base and between same parallels is 1:2

Step-by-step explanation:

Given:

ΔCDE and rhombus ABCD are on the same base CD and between same parallels AB and CD.

To find:

Ar. ΔCDE : Ar. rhombus ABCD

As we know that:

• If a triangle and a parallelogram are on the same base and between the same parallels then

        $\bf{(Ar. \; of \; triangle) =\dfrac{1}{2} \times \; (Ar. \; of \; ||gm)}$

• Rhombus is a type of parallelogram in which all sides are equal.

Solution:

∵ ΔCDE and rhombus ABCD are on the same base CD and between same parallels AB and CD.

$\therefore \sf{(Ar. \; of \; \triangle CDE ) =\dfrac{1}{2} \times \; (Ar. \; of \;rhombus \; ABCD)}$

$\implies \sf{\dfrac{(Ar. \; of \; \triangle CDE)}{(Ar. \; of \;rhombus \; ABCD)} =\dfrac{1}{2}}$

$\implies \sf{(Ar. \; of \; \triangle CDE) : (Ar. \; of \;rhombus \; ABCD) =1:2}$