
The area of a parallelogram with base 24cm is equal to the area of a triangle of sides 10cm 24cm and 26cm. find the altitude of the parallelogram.
Answer
587.1k+ views
Hint: First find the area of a triangle using Heron’s formula with the given sides of the triangle and area of a parallelogram with known values, then equate the areas of both the parallelogram and the triangle.
Complete step by step solution: According to the question:
Given the area of triangle = area of the parallelogram
So,
$120 = 24h$
$\Rightarrow h =\dfrac{120}{24} =5 \text{cm}$
Hence the altitude of the parallelogram (h) = 5cm
Area of triangle by heron’s formula = $\sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)} $
Where S = Semi-perimeter of the triangle.
And a, b and c are the sides of the given triangle.
The perimeter of triangle 2S = a + b + c
2S = (10 +24+26)
S = 30 cm
Now the area of triangle =$\sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)} $
\[ = \sqrt {30(30 - 10)(30 - 24)(30 - 26)} \]
\[ = \sqrt {30 \times 20 \times 6 \times 4} \]
\[ = \sqrt {6 \times 5 \times 5 \times 4 \times 4 \times 6} \]
\[ = 6 \times 4 \times 5\]
\[ = 120c{m^2}\]
Area of parallelogram = bh
Where b = base and = height
b is given = 24cm.
Therefore, the area of parallelogram \[ = 24 \times h\]
Now, equating the area of parallelogram and triangle in the next step.
24h = 120
h = 5 cm
Hence, the required altitude is 5 cm.
Note:
Heron’s formula is used to find the area of a triangle when all three sides are given. Heron's formula is a special case of Brahmagupta's formula for the area of a cyclic quadrilateral. Heron's formula and Brahmagupta's formula are both special cases of Bretschneider's formula for the area of a quadrilateral. Heron's formula can be obtained from Brahmagupta's formula or Bretschneider's formula by setting one of the sides of the quadrilateral to zero.
Complete step by step solution: According to the question:
Given the area of triangle = area of the parallelogram
So,
$120 = 24h$
$\Rightarrow h =\dfrac{120}{24} =5 \text{cm}$
Hence the altitude of the parallelogram (h) = 5cm
Area of triangle by heron’s formula = $\sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)} $
Where S = Semi-perimeter of the triangle.
And a, b and c are the sides of the given triangle.
The perimeter of triangle 2S = a + b + c
2S = (10 +24+26)
S = 30 cm
Now the area of triangle =$\sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)} $
\[ = \sqrt {30(30 - 10)(30 - 24)(30 - 26)} \]
\[ = \sqrt {30 \times 20 \times 6 \times 4} \]
\[ = \sqrt {6 \times 5 \times 5 \times 4 \times 4 \times 6} \]
\[ = 6 \times 4 \times 5\]
\[ = 120c{m^2}\]
Area of parallelogram = bh
Where b = base and = height
b is given = 24cm.
Therefore, the area of parallelogram \[ = 24 \times h\]
Now, equating the area of parallelogram and triangle in the next step.
24h = 120
h = 5 cm
Hence, the required altitude is 5 cm.
Note:
Heron’s formula is used to find the area of a triangle when all three sides are given. Heron's formula is a special case of Brahmagupta's formula for the area of a cyclic quadrilateral. Heron's formula and Brahmagupta's formula are both special cases of Bretschneider's formula for the area of a quadrilateral. Heron's formula can be obtained from Brahmagupta's formula or Bretschneider's formula by setting one of the sides of the quadrilateral to zero.
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