Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumIn parallelogram $XYZW$,$XY = 24 \, cm$. Altitudes $WP$ and $WQ$ are corresponding to bases $XY$ and $YZ$ respectively. If $WP = 6 \, cm$ and $WQ = 8 \, cm$,then find $YZ$ and the perimeter of parallelogram $XYZW$.
(N/A) The area of a parallelogram is given by the formula: $\text{Area} = \text{base} \times \text{corresponding altitude}$.
For base $XY = 24 \, cm$ and altitude $WP = 6 \, cm$,the area is: $\text{Area} = 24 \times 6 = 144 \, cm^2$.
Since the area of the parallelogram is constant,we can also write: $\text{Area} = YZ \times WQ$.
Given $WQ = 8 \, cm$,we have: $144 = YZ \times 8$.
Therefore,$YZ = 144 / 8 = 18 \, cm$.
The perimeter of a parallelogram is given by $2 \times (\text{sum of adjacent sides})$.
Perimeter $= 2 \times (XY + YZ) = 2 \times (24 + 18) = 2 \times 42 = 84 \, cm$.
For base $XY = 24 \, cm$ and altitude $WP = 6 \, cm$,the area is: $\text{Area} = 24 \times 6 = 144 \, cm^2$.
Since the area of the parallelogram is constant,we can also write: $\text{Area} = YZ \times WQ$.
Given $WQ = 8 \, cm$,we have: $144 = YZ \times 8$.
Therefore,$YZ = 144 / 8 = 18 \, cm$.
The perimeter of a parallelogram is given by $2 \times (\text{sum of adjacent sides})$.
Perimeter $= 2 \times (XY + YZ) = 2 \times (24 + 18) = 2 \times 42 = 84 \, cm$.
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