Question
A trapezium with $3$ equal sides and one side double the equal side can be divided into __________ equilateral triangles of _______ area.
✓
Answer
A trapezium with $3$ equal sides and one side double the equal side can be divided into $3$ equilateral triangles of equal area.
Solution:
Let $ABCD$ be a trapezium, in which
$AD = DC = BC = a ($say$)$
and $AB = 2a$
Draw medians through the vertices $D$ and $C$ on the side $AB.$
$\therefore AE = EB = a$
Now, in parallelogram $ADCE,$ we have
$AD = EC = a$ and $AE = CD = a [ \because$ opposite sides in a parallelogram are equal]
In $\triangle\text{ADE}$ and $\triangle\text{DEC},$ AD = EC
$AE = CD$
and $DE = DE$
By sss, $\triangle\text{ADE}=\triangle\text{DEC}$
By triangle rule, $\triangle\text{ADE}\cong\triangle\text{DEC}$ Thus, $\triangle\text{ADE}$ and $\triangle\text{DEC}$ are equilateral triangles having equal sides. Similarly, in parallelogram $DEBC. $ we can show that $\triangle\text{DEC}\cong\triangle\text{ECB}.$ Hence, the trapezium can be divided into $3$ equilateral triangles of equal area.
Solution:
Let $ABCD$ be a trapezium, in which
$AD = DC = BC = a ($say$)$
and $AB = 2a$
Draw medians through the vertices $D$ and $C$ on the side $AB.$
$\therefore AE = EB = a$
Now, in parallelogram $ADCE,$ we have
$AD = EC = a$ and $AE = CD = a [ \because$ opposite sides in a parallelogram are equal]
In $\triangle\text{ADE}$ and $\triangle\text{DEC},$ AD = EC
$AE = CD$
and $DE = DE$
By sss, $\triangle\text{ADE}=\triangle\text{DEC}$
By triangle rule, $\triangle\text{ADE}\cong\triangle\text{DEC}$ Thus, $\triangle\text{ADE}$ and $\triangle\text{DEC}$ are equilateral triangles having equal sides. Similarly, in parallelogram $DEBC. $ we can show that $\triangle\text{DEC}\cong\triangle\text{ECB}.$ Hence, the trapezium can be divided into $3$ equilateral triangles of equal area.
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