Triangle ABP

An isosceles trapezoid ABCD is given. The length of side AB is 10 cm, the length of CD is 7 cm, and the height drawn to side AB is 4 cm. Point P is the foot of the altitude from A to side AD. Calculate the area of triangle ABP.

Final Answer:

S₂ =  10.4335 cm²

Step-by-step explanation:

$$\begin{gathered} a=10\text{cm} \\ c=7\text{cm} \\ h=4\text{cm} \\ x=\frac{a-c}{2}=\frac{10-7}{2}=\frac{3}{2}=1.5\text{cm} \\ d=\sqrt{x^2+h^2}=\sqrt{1.5^2+4^2}\doteq 4.272\text{cm} \\ {d}_2=d/2=4.272/2\doteq 2.136\text{cm} \\ {e}_2=\sqrt{a^2-d_2^2}=\sqrt{10^2-2.136^2}\doteq 9.7692\text{cm} \\ {S}_2=\frac{d_2\cdot e_2}{2}=\frac{2.136\cdot 9.7692}{2}=10.4335{\text{cm}}^2 \end{gathered}$$

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