Answer
Verified
420.3k+ views
Hint: Here, first we have to draw the figure. Draw $DL\bot AB$ and $CM\bot AB$ such that CD = ML and AL = MB. Here, we have to apply the formulas:
Perimeter of a trapezium = sum of its sides
Area of a trapezium = $\dfrac{1}{2}\times$sum of the parallel sides $\times$ height
$A{{D}^{2}}=A{{L}^{2}}+D{{L}^{2}}$
Complete step-by-step solution -
First, we have to draw the figure. Draw $DL\bot AB$ and $CM\bot AB$.
Here we have to find the area of the trapezium.
We are given that the trapezium is isosceles, which means the opposite sides are equal. Therefore, we can say that:
AD = BC ….. (1)
We have the perimeter of the trapezium = $16cm$
Here, in trapezium we have $AB\parallel CD$.
We are also given that the shorter parallel side is one-third of the longer parallel side. i.e.
Now, consider the length of CD to be a. Then the length of AB will be 3 times that of CD.
CD = a ….. (2)
AB = 3a …. (3)
Hence, we are given that the lengths of the opposite sides are equal. We can write:
AD + BC = CD + AB
Hence, from equations (1) and equation (2), we can write:
2AD = a + 3a
2AD = 4a
By cross multiplication we get:
AD = $\dfrac{4}{2}a$
Now, by cancellation we obtain:
AD = 2a
Therefore, we will obtain:
AD = BC = 2a
We know that the perimeter of a trapezium is the length of its four sides. i.e.
Perimeter = AB + BC + CD + AD
Now by substituting all the values we obtain:
16 = 3a + 2a + a + 2a
16 = 8a
By cross multiplication, we get:
$\dfrac{16}{8}$ = a
Now, by cancellation we obtain:
2 = a
Therefore the sides of the trapezium are:
CD = 2 cm
AD = 2a
AD = $2\times 2$
AD= 4 cm
BC = 4 cm
AB = 3a
AB= $3\times 2$
AB = 6 cm
Now, consider the right angled triangle, $\Delta ALD$, Hence by Pythagoras theorem, we have
$A{{D}^{2}}=A{{L}^{2}}+D{{L}^{2}}$
First we have to find AL.
From the figure, we can say that:
CD = ML =2 cm
AL = MB
Now, we can write:
AL + ML + MB = 6
2AL + 2 = 6
2AL = 4
By cross multiplication we get:
AL = $\dfrac{4}{2}$
By cancellation we get:
AL = 2 cm
We have AD = 4cm, AL = 2 cm. Now we have to find DL using Pythagoras theorem. i.e. we have:
$\begin{align}
& D{{L}^{2}}=A{{D}^{2}}-A{{L}^{2}} \\
& D{{L}^{2}}={{4}^{2}}-{{2}^{\begin{smallmatrix}
2 \\
\end{smallmatrix}}} \\
& D{{L}^{2}}=16-4 \\
& D{{L}^{2}}=12 \\
\end{align}$
Now, by taking square root on both the sides we get:
$\begin{align}
& \sqrt{D{{L}^{2}}}=\sqrt{12} \\
& DL=\sqrt{3\times 2\times 2} \\
& DL=2\sqrt{3} \\
\end{align}$
Therefore, we will get DL = $2\sqrt{3}$ cm.
Next, we have to find the area of the trapezium, we know that:
Area of the trapezium = $\dfrac{1}{2}\times $sum of the parallel sides $\times $ height
Hence, we have the parallel sides as AB and CD and the height is AL
$\begin{align}
& Area=\dfrac{1}{2}\times (AB+CD)\times AL \\
& Area=\dfrac{1}{2}\times (6+2)\times 2\sqrt{3} \\
& Area=\dfrac{1}{2}\times 8\times 2\sqrt{3} \\
& Area=\dfrac{16\sqrt{3}}{2} \\
\end{align}$
Now by cancellation we obtain:
$Area=8\sqrt{3}c{{m}^{2}}$. Hence the required area of the trapezium is $8\sqrt{3}c{{m}^{2}}$.
Therefore, the correct answer for this question is option (a).
Note: Here we have two right angled triangles $\Delta ALD$ and $\Delta BMC$, DL and CM are the heights where, DL = CM. So instead of $\Delta ALD$ you can also consider $\Delta BMC$ to find height CM.
Perimeter of a trapezium = sum of its sides
Area of a trapezium = $\dfrac{1}{2}\times$sum of the parallel sides $\times$ height
$A{{D}^{2}}=A{{L}^{2}}+D{{L}^{2}}$
Complete step-by-step solution -
First, we have to draw the figure. Draw $DL\bot AB$ and $CM\bot AB$.
Here we have to find the area of the trapezium.
We are given that the trapezium is isosceles, which means the opposite sides are equal. Therefore, we can say that:
AD = BC ….. (1)
We have the perimeter of the trapezium = $16cm$
Here, in trapezium we have $AB\parallel CD$.
We are also given that the shorter parallel side is one-third of the longer parallel side. i.e.
Now, consider the length of CD to be a. Then the length of AB will be 3 times that of CD.
CD = a ….. (2)
AB = 3a …. (3)
Hence, we are given that the lengths of the opposite sides are equal. We can write:
AD + BC = CD + AB
Hence, from equations (1) and equation (2), we can write:
2AD = a + 3a
2AD = 4a
By cross multiplication we get:
AD = $\dfrac{4}{2}a$
Now, by cancellation we obtain:
AD = 2a
Therefore, we will obtain:
AD = BC = 2a
We know that the perimeter of a trapezium is the length of its four sides. i.e.
Perimeter = AB + BC + CD + AD
Now by substituting all the values we obtain:
16 = 3a + 2a + a + 2a
16 = 8a
By cross multiplication, we get:
$\dfrac{16}{8}$ = a
Now, by cancellation we obtain:
2 = a
Therefore the sides of the trapezium are:
CD = 2 cm
AD = 2a
AD = $2\times 2$
AD= 4 cm
BC = 4 cm
AB = 3a
AB= $3\times 2$
AB = 6 cm
Now, consider the right angled triangle, $\Delta ALD$, Hence by Pythagoras theorem, we have
$A{{D}^{2}}=A{{L}^{2}}+D{{L}^{2}}$
First we have to find AL.
From the figure, we can say that:
CD = ML =2 cm
AL = MB
Now, we can write:
AL + ML + MB = 6
2AL + 2 = 6
2AL = 4
By cross multiplication we get:
AL = $\dfrac{4}{2}$
By cancellation we get:
AL = 2 cm
We have AD = 4cm, AL = 2 cm. Now we have to find DL using Pythagoras theorem. i.e. we have:
$\begin{align}
& D{{L}^{2}}=A{{D}^{2}}-A{{L}^{2}} \\
& D{{L}^{2}}={{4}^{2}}-{{2}^{\begin{smallmatrix}
2 \\
\end{smallmatrix}}} \\
& D{{L}^{2}}=16-4 \\
& D{{L}^{2}}=12 \\
\end{align}$
Now, by taking square root on both the sides we get:
$\begin{align}
& \sqrt{D{{L}^{2}}}=\sqrt{12} \\
& DL=\sqrt{3\times 2\times 2} \\
& DL=2\sqrt{3} \\
\end{align}$
Therefore, we will get DL = $2\sqrt{3}$ cm.
Next, we have to find the area of the trapezium, we know that:
Area of the trapezium = $\dfrac{1}{2}\times $sum of the parallel sides $\times $ height
Hence, we have the parallel sides as AB and CD and the height is AL
$\begin{align}
& Area=\dfrac{1}{2}\times (AB+CD)\times AL \\
& Area=\dfrac{1}{2}\times (6+2)\times 2\sqrt{3} \\
& Area=\dfrac{1}{2}\times 8\times 2\sqrt{3} \\
& Area=\dfrac{16\sqrt{3}}{2} \\
\end{align}$
Now by cancellation we obtain:
$Area=8\sqrt{3}c{{m}^{2}}$. Hence the required area of the trapezium is $8\sqrt{3}c{{m}^{2}}$.
Therefore, the correct answer for this question is option (a).
Note: Here we have two right angled triangles $\Delta ALD$ and $\Delta BMC$, DL and CM are the heights where, DL = CM. So instead of $\Delta ALD$ you can also consider $\Delta BMC$ to find height CM.
Recently Updated Pages
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Find the values of other five trigonometric functions class 10 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Two charges are placed at a certain distance apart class 12 physics CBSE
Difference Between Plant Cell and Animal Cell
What organs are located on the left side of your body class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The planet nearest to earth is A Mercury B Venus C class 6 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE