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Triangle ABC and parallelogram ABEP are on the same base AC in between the same parallel AB and EF. Prove that area $\left( \vartriangle ABC \right)=\dfrac{1}{2}$ ar parallelogram (ABEP).

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Last updated date: 23rd Feb 2024
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IVSAT 2024
Answer
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Hint: We are given that we have a $\left( \vartriangle ABC \right)$ and a parallelogram that lie on the same base we have to show that
$\left( \vartriangle ABC \right)=\dfrac{1}{2}$ar (||gm ABEP)
To do so we do construction we draw BH||AC that meet FE at H
Then we use that diagonal of parallelogram to divide it into two congruent triangles and we use that two parallelograms on the same base between the same parallel are equal in area. Using them we get our required solution.

Complete step by step answer:
We are given that $\left( \vartriangle ABC \right)=\dfrac{1}{2}$ar (||gm ABEP) are on the same base while AC is between the same parallel AB and EF
So, using these data we will first draw our figure to get better understanding of problem
Now as Ab is common to $\vartriangle ABC$and parallelogram ABEF, so, common base in AB. So, we have,
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Now to solve this problem and get our required solution,
Through B we draw BH||AC to meet FE provided at H
Such that ABHC is a parallelogram.
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As we know that diagonal of parallelogram divide it into two congruent,
So, as CB is diagonal at ||ABHC
So, $\vartriangle ABC\cong \vartriangle BHC$
A congruent triangle has some area.
So, $ar\left( \vartriangle ABC \right)=ar\left( \vartriangle BHC \right)..........(i)$
Now as
$ar\left( ||gmABHC \right)=ar\left( \vartriangle ABC \right)+ar\left( BHC \right)$
Why (i), we get
$ar\left( ||gmABHC \right)=ar\left( \vartriangle ABC \right)+ar\left( BHC \right)$
So, we get
$ar\left( \vartriangle ABC \right)=\dfrac{1}{2}\left( ar\left( ||gm\,ABHC \right) \right).......(ii)$
Now
We know that two parallelograms on same base and between same parallel are equal in equal we have that
||gm ABHC and ||gm ABEP are on the same base AB and between the same parallel AB and EF.
So,
$ar\left( ||gmABHC \right)=ae\left( ||gmABEF \right)........(iii)$
Now, using (iii) and (ii)
$ar\left( \vartriangle ABC \right)=\dfrac{1}{2}\left( ||gmABEF \right)$
Hence proved

Note:
Parallelogram divide into two parts by the diagonal as we can see that through diagonal we have $\vartriangle ABC\,\operatorname{and}\vartriangle ACD$ in ||gm ABCD.
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In $\vartriangle ABC\,\operatorname{and}\vartriangle ACD$, AB=CD (opposite sides are equal in parallelogram)
BC=AD (opposite sides are equal common)
AC=AC
So, using SSS rule, we have these two are congruent.
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