
Find the perimeter of
(i)$\Delta ABC$
(ii) rectangle $BCDE\;$
A.$\left( i \right)8\dfrac{1}{60}cm\text{ }\left( ii \right)5cm\;$
B. $\left( i \right)5cm\text{ }\left( ii \right)10\dfrac{1}{6}cm$
C. $\left( i \right)8\dfrac{1}{60}cm\text{ }\left( ii \right)10\dfrac{1}{5}cm$
D. $\left( i \right)8cm\text{ }\left( ii \right)23cm$
Answer
508.8k+ views
Hint: We need to compute the perimeter of the triangle $\Delta ABC$ and the rectangle $BCDE\;$in the given figure. We solve the given question using the formulae perimeter of triangle = sum of all sides and the perimeter of the rectangle =$2\left( l+b \right)\;$ to get the desired result.
Complete step by step answer:
We are given a figure and are asked to find out the perimeter of the triangle $\Delta ABC$ and the rectangle $BCDE\;$ in the given figure. We will be solving the given question using the formulae of geometry.
A perimeter is defined as a path that surrounds a geometrical figure. In simple terms, it is a distance around the geometric figure.
A perimeter of a triangle is the total path around the edges of a triangle. It is the sum of all the sides of a triangle.
The perimeter of the triangle is given as follows,
$\Rightarrow perimeter\text{ }of\text{ }triangle=sum\text{ }of\text{ }all\text{ }sides$
A perimeter of a rectangle is the total path of all the sides of a rectangle.
The perimeter of the rectangle is given as follows,
$\Rightarrow perimeter\text{ }of\text{ }rec\tan gle=2\left( l+b \right)$
Here,
l is the length of the rectangle
b is the breadth of the rectangle
According to the question,
We need to find the perimeter of the triangle $\Delta ABC$
From the above, we know that the perimeter of the triangle $\Delta ABC$ is the sum of all sides of the triangle $\Delta ABC$ .
Writing the same, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=sum\text{ }of\text{ }all\text{ }sides$
Writing the sides in the above equation, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=AB+BC+CA$
Substituting the values of the sides, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=4\dfrac{2}{3}+2\dfrac{3}{5}+\dfrac{3}{4}$
From the concept of fractions, we know that the value of the fraction $a\dfrac{b}{c}$ can be written as $a+\dfrac{b}{c}$
Following the same, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=\left( 4+\dfrac{2}{3} \right)+\left( 2+\dfrac{3}{5} \right)+\dfrac{3}{4}$
Simplifying the above equation, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=\dfrac{14}{3}+\dfrac{13}{5}+\dfrac{3}{4}$
Taking the LCM for the above equation, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=\dfrac{280+156+45}{60}$
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=\dfrac{481}{60}$
Converting the above fraction into a mixed fraction, we get,
$\therefore perimeter\text{ }of\text{ }\Delta ABC=8\dfrac{1}{60}cm$
Now,
We need to find the perimeter of the rectangle BCDE.
From the above, we know that the perimeter of the rectangle BCDE is the given by $2\left( l+b \right)$
Writing the same, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( l+b \right)$
From the above figure,
l = $2\dfrac{3}{5}$ cm;
b = $\dfrac{5}{2}$ cm
Substituting the values of the sides, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( 2\dfrac{3}{5}+\dfrac{5}{2} \right)$
From the concept of fractions, we know that the value of the fraction $a\dfrac{b}{c}$ can be written as $a+\dfrac{b}{c}$
Following the same, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( \left( 2+\dfrac{3}{5} \right)+\dfrac{5}{2} \right)$
Simplifying the above equation, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( \dfrac{13}{5}+\dfrac{5}{2} \right)$
Taking the LCM for the above equation, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( \dfrac{26+25}{10} \right)$
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( \dfrac{51}{10} \right)$
Canceling the common factors, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=\dfrac{51}{5}$
Converting the above fraction into a mixed fraction, we get,
$\therefore perimeter\text{ }of\text{ }BCDE=10\dfrac{1}{5}cm$
So, the correct answer is “Option C”.
Note: The given question is directly formula based and any errors or mistakes in writing a formula will result in an incorrect solution. We must know that the value of the mixed fraction $a\dfrac{b}{c}$ can be written as $a+\dfrac{b}{c}$ .
Complete step by step answer:
We are given a figure and are asked to find out the perimeter of the triangle $\Delta ABC$ and the rectangle $BCDE\;$ in the given figure. We will be solving the given question using the formulae of geometry.
A perimeter is defined as a path that surrounds a geometrical figure. In simple terms, it is a distance around the geometric figure.
A perimeter of a triangle is the total path around the edges of a triangle. It is the sum of all the sides of a triangle.
The perimeter of the triangle is given as follows,
$\Rightarrow perimeter\text{ }of\text{ }triangle=sum\text{ }of\text{ }all\text{ }sides$
A perimeter of a rectangle is the total path of all the sides of a rectangle.
The perimeter of the rectangle is given as follows,
$\Rightarrow perimeter\text{ }of\text{ }rec\tan gle=2\left( l+b \right)$
Here,
l is the length of the rectangle
b is the breadth of the rectangle
According to the question,
We need to find the perimeter of the triangle $\Delta ABC$
From the above, we know that the perimeter of the triangle $\Delta ABC$ is the sum of all sides of the triangle $\Delta ABC$ .
Writing the same, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=sum\text{ }of\text{ }all\text{ }sides$
Writing the sides in the above equation, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=AB+BC+CA$
Substituting the values of the sides, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=4\dfrac{2}{3}+2\dfrac{3}{5}+\dfrac{3}{4}$
From the concept of fractions, we know that the value of the fraction $a\dfrac{b}{c}$ can be written as $a+\dfrac{b}{c}$
Following the same, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=\left( 4+\dfrac{2}{3} \right)+\left( 2+\dfrac{3}{5} \right)+\dfrac{3}{4}$
Simplifying the above equation, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=\dfrac{14}{3}+\dfrac{13}{5}+\dfrac{3}{4}$
Taking the LCM for the above equation, we get,
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=\dfrac{280+156+45}{60}$
$\Rightarrow perimeter\text{ }of\text{ }\Delta ABC=\dfrac{481}{60}$
Converting the above fraction into a mixed fraction, we get,
$\therefore perimeter\text{ }of\text{ }\Delta ABC=8\dfrac{1}{60}cm$
Now,
We need to find the perimeter of the rectangle BCDE.
From the above, we know that the perimeter of the rectangle BCDE is the given by $2\left( l+b \right)$
Writing the same, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( l+b \right)$
From the above figure,
l = $2\dfrac{3}{5}$ cm;
b = $\dfrac{5}{2}$ cm
Substituting the values of the sides, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( 2\dfrac{3}{5}+\dfrac{5}{2} \right)$
From the concept of fractions, we know that the value of the fraction $a\dfrac{b}{c}$ can be written as $a+\dfrac{b}{c}$
Following the same, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( \left( 2+\dfrac{3}{5} \right)+\dfrac{5}{2} \right)$
Simplifying the above equation, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( \dfrac{13}{5}+\dfrac{5}{2} \right)$
Taking the LCM for the above equation, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( \dfrac{26+25}{10} \right)$
$\Rightarrow perimeter\text{ }of\text{ }BCDE=2\left( \dfrac{51}{10} \right)$
Canceling the common factors, we get,
$\Rightarrow perimeter\text{ }of\text{ }BCDE=\dfrac{51}{5}$
Converting the above fraction into a mixed fraction, we get,
$\therefore perimeter\text{ }of\text{ }BCDE=10\dfrac{1}{5}cm$
So, the correct answer is “Option C”.
Note: The given question is directly formula based and any errors or mistakes in writing a formula will result in an incorrect solution. We must know that the value of the mixed fraction $a\dfrac{b}{c}$ can be written as $a+\dfrac{b}{c}$ .
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