In Fig. 12.8, PQRS is a parallelogram, HSR isa straight line and HPQ = 90°. If|HQ| = 10 cm and |PQ| = 6 cm, what is thearea of the parallelogram?

In Fig 128 PQRS is a parallelogram HSR isa straight line and HPQ 90 IfHQ 10 cm and PQ 6 cm what is thearea of the parallelogram class=

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Answer

48 cm²

Step-by-step explanation

First, we need to calculate the height of the parallelogram, segment HP. Applying the Pythagorean theorem to triangle HPQ, we get:

[tex]\begin{gathered} HQ^2=HP^2+PQ^2 \\ \text{ Substituting with HQ = 10 cm, and PQ = 6 cm, and solving for HP:} \\ 10^2=HP^2+6^2 \\ 100=HP^2+36 \\ 100-36=HP^2 \\ 64=HP^2 \\ \sqrt{64}=HP \\ HP=8\text{ cm} \end{gathered}[/tex]

The area of a parallelogram is calculated as follows:

[tex]A=base\times height[/tex]

In this case, the height is HP = 8 cm, and the base is PQ = 6 cm. Then the area of parallelogram PQRS is:

[tex]\begin{gathered} A=HP\times PQ \\ A=8\times6 \\ A=48\text{ cm}^2 \end{gathered}[/tex]