Question

Given: ΔPSQ, PS = SQ

Answer 1

Answer:

To solve this problem we will use Heron's formula:

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Where $a, \ b \ and \ c$ are the side lengths of the triangle and $s$ is the semiperimeter (half the perimeter of the triangle). We know that:

$Perimeter \ P=\triangle PSQ=PS+PQ+SQ: \\ \\ \triangle PSQ=P=50 \\ \\ Semiperimeter \ s: \\ \\ s=\frac{P}{2}=25$

Also:

$(I) \ PS=SQ \\ \\ (II) \ SQ-PQ = 1 \\ \\ (III) \ PS+PQ+SQ=50 \\ \\ \\ (I) \ into \ (III): \\ \\ SQ+PQ+SQ=50 \\ \\ \therefore (IV) \ 2SQ+PQ=50 \\ \\ From \ (II): \\ \\ PQ=SQ-1 \\ \\ (II) \ into \ (IV): \\ \\ 2SQ+(SQ-1)=50 \\ 3SQ-1=50 \\ 3SQ=51 \\ \\ \boxed{SQ=17} \\ \\ \boxed{PS=17} \\ \\ PQ=SQ-1=17-1 \therefore \boxed{PQ=16}$

Finally:

$A=\sqrt{s(s-a)(s-b)(s-c)} \\ \\ A=\sqrt{s(s-PS)(s-SQ)(s-PQ)} \\ \\ A=\sqrt{s(s-17)(s-17)(s-16)} \\ \\ A=\sqrt{25(25-17)(25-17)(25-16)} \\ \\ \boxed{A=120}$