The perimeter of a rhombus ABCD is 40 cm. Find the area of a rhombus if it’s diagonal BD measures 12 cm.
Answer
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Hint: First we will use the formula of the perimeter of the rhombus, that is, \[4s\], where \[s\] is the side of the rhombus and then find the value of side. Then since diagonals are perpendicular with each other, we will use the Pythagorean theorem \[{h^2} = {a^2} + {b^2}\], where \[h\] is the hypotenuse, \[a\] is the height and \[b\] is the base of the right-angled triangle and then multiply the above side by 2 to find the other diagonal. Then use the formula of area of rhombus, \[\dfrac{1}{2}{d_1}{d_2}\], where \[{d_1}\] and \[{d_2}\] are the diagonals in the rhombus ABCD to find the required value.
Complete step by step answer:
We are given that the perimeter of the rhombus ABCD is 40 cm.
We know that the perimeter is the sum of the length of all sides of the rhombus, that is, \[4s\], where \[s\] is the side of the rhombus.
Finding the side of the rhombus using the above formula, we get
\[ \Rightarrow 40 = 4s\]
Dividing the above equation by 4 on both sides, we get
\[
\Rightarrow \dfrac{{40}}{4} = \dfrac{{4s}}{4} \\
\Rightarrow 10 = s \\
\Rightarrow s = 10{\text{ cm}} \\
\]
We are also given that the diagonal BD measure 12 cm, we have
We know that the diagonal of the rhombus intersect each other at O is 90 degrees.
We will use the Pythagorean theorem \[{h^2} = {a^2} + {b^2}\], where \[h\] is the hypotenuse, \[a\] is the height and \[b\] is the base of the right-angled triangle.
Applying the Pythagorean theorem in the triangle COB, we get
\[
\Rightarrow {10^2} = {x^2} + {6^2} \\
\Rightarrow 100 = {x^2} + 36 \\
\]
Subtracting the above equation by 36 on each of the sides, we get
\[
\Rightarrow 100 – 36 = {x^2} + 36 – 36 \\
\Rightarrow 64 = {x^2} \\
\Rightarrow {x^2} = 64 \\
\]
Taking the square root on both sides in the above equation, we get
\[
\Rightarrow x = \sqrt {64} \\
\Rightarrow x = \pm 8 \\
\]
Since the side can never be negative, the negative value of \[x\] is discarded.
Multiplying the above side by 2 to find the complete diagonal, we get
\[ \Rightarrow 2 \cdot 8 = 16{\text{ cm}}\]
Using the formula of area of rhombus, \[\dfrac{1}{2}{d_1}{d_2}\], where \[{d_1}\] and \[{d_2}\] are the diagonals in the rhombus ABCD, we get
\[ \Rightarrow \dfrac{1}{2} \times 16 \times 12 = 96{\text{ c}}{{\text{m}}^2}\]
Thus, the area of the given rhombus is 96 cm\[^2\].
Note: In solving this question, students should note here that both diagonals bisect each other at 90 degrees. Students should remember that the perimeter of the rhombus is the sum of all the sides of the rhombus and the unit of the perimeter has cm, not the square cm.
Complete step by step answer:
We are given that the perimeter of the rhombus ABCD is 40 cm.
We know that the perimeter is the sum of the length of all sides of the rhombus, that is, \[4s\], where \[s\] is the side of the rhombus.
Finding the side of the rhombus using the above formula, we get
\[ \Rightarrow 40 = 4s\]
Dividing the above equation by 4 on both sides, we get
\[
\Rightarrow \dfrac{{40}}{4} = \dfrac{{4s}}{4} \\
\Rightarrow 10 = s \\
\Rightarrow s = 10{\text{ cm}} \\
\]
We are also given that the diagonal BD measure 12 cm, we have
We know that the diagonal of the rhombus intersect each other at O is 90 degrees.
We will use the Pythagorean theorem \[{h^2} = {a^2} + {b^2}\], where \[h\] is the hypotenuse, \[a\] is the height and \[b\] is the base of the right-angled triangle.
Applying the Pythagorean theorem in the triangle COB, we get
\[
\Rightarrow {10^2} = {x^2} + {6^2} \\
\Rightarrow 100 = {x^2} + 36 \\
\]
Subtracting the above equation by 36 on each of the sides, we get
\[
\Rightarrow 100 – 36 = {x^2} + 36 – 36 \\
\Rightarrow 64 = {x^2} \\
\Rightarrow {x^2} = 64 \\
\]
Taking the square root on both sides in the above equation, we get
\[
\Rightarrow x = \sqrt {64} \\
\Rightarrow x = \pm 8 \\
\]
Since the side can never be negative, the negative value of \[x\] is discarded.
Multiplying the above side by 2 to find the complete diagonal, we get
\[ \Rightarrow 2 \cdot 8 = 16{\text{ cm}}\]
Using the formula of area of rhombus, \[\dfrac{1}{2}{d_1}{d_2}\], where \[{d_1}\] and \[{d_2}\] are the diagonals in the rhombus ABCD, we get
\[ \Rightarrow \dfrac{1}{2} \times 16 \times 12 = 96{\text{ c}}{{\text{m}}^2}\]
Thus, the area of the given rhombus is 96 cm\[^2\].
Note: In solving this question, students should note here that both diagonals bisect each other at 90 degrees. Students should remember that the perimeter of the rhombus is the sum of all the sides of the rhombus and the unit of the perimeter has cm, not the square cm.
Last updated date: 16th Aug 2023
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