Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumIn parallelogram $PQRS$,$PQ = 15 \, cm$. Altitudes $SM$ and $SN$ are corresponding to bases $PQ$ and $QR$ respectively. If $SM = 6 \, cm$ and $SN = 10 \, cm$,find $QR$ and the perimeter of $PQRS$.
(N/A) The area of a parallelogram is given by the formula: $\text{Area} = \text{base} \times \text{corresponding altitude}$.
For base $PQ = 15 \, cm$ and altitude $SM = 6 \, cm$,the area is: $\text{Area} = 15 \times 6 = 90 \, cm^2$.
Since the area of the parallelogram is constant,we can also write: $\text{Area} = QR \times SN$.
Substituting the known values: $90 = QR \times 10$.
Therefore,$QR = \frac{90}{10} = 9 \, cm$.
The perimeter of a parallelogram is given by $2 \times (\text{sum of adjacent sides}) = 2 \times (PQ + QR)$.
Perimeter $= 2 \times (15 + 9) = 2 \times 24 = 48 \, cm$.
For base $PQ = 15 \, cm$ and altitude $SM = 6 \, cm$,the area is: $\text{Area} = 15 \times 6 = 90 \, cm^2$.
Since the area of the parallelogram is constant,we can also write: $\text{Area} = QR \times SN$.
Substituting the known values: $90 = QR \times 10$.
Therefore,$QR = \frac{90}{10} = 9 \, cm$.
The perimeter of a parallelogram is given by $2 \times (\text{sum of adjacent sides}) = 2 \times (PQ + QR)$.
Perimeter $= 2 \times (15 + 9) = 2 \times 24 = 48 \, cm$.
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