In the figure below, PQRS is a parallelogram of perimeter 24 and area 28. What is the perimeter of rectangle QOST? Include an explanation of how you solved the problem.
+2440 Before we begin solving, let's insert \(\overline{PU}\) such that the segments extends perpendicularly from \(\overline{OS}\) , and \(U\) lies on \(\overline{QT}\).
We know from the given information that parallelogram \(PQRS\) has a perimeter of 24. Of course, opposite sides of a parallelogram are congruent, so \(\overline{RS}\cong\overline{QP}\), which means that \(RS=QP=5\). Now, we can solve for the remaining sides of the parallelogram (\(\overline{PS}\) and \(\overline{QR}\))
| \(5+5+PS+QR=24\) | Simplify the left-hand side of the equation. |
| \(10+PS+QR=24\) | Subtract 10 on both sides of the equation. |
| \(PS+QR=14\) | Since \(PS=QR\) because of the property of a parallelogram, we can actually solve this equation. |
| \(PS+PS=14\) | |
| \(2PS=14\) | Divide by 2 on both sides. |
| \(PS=7\) | |
| \(PS=QR=7\) |
To recap, we now know that \(RS=QP=5\) and \(QR=PS=7\). Now, let's worry about that segment I drew in since the beginning of the problem.
\(\overline{PU}\) is an example of an altitude (perpendicula height that extends from a vertex). The area of the parallelogram is given to be \(28units^2\). Let's review the formula for the area of a parallelogram. It is the following:
\(A=bh\)
b = length of the base
h = length of the altitude (\(\overline{PU}\))
Although it is true that we do not know the length of the altitude, we know the length of the base (7) and the total area of the parallelogram (28). We can now solve for the length of the altitude.
| \(A=bh\) | Substitute in the known values and solve for the unknown, in this case h. |
| \(28=7h\) | Divide by 7 on both sides of the equation. |
| \(4=h\) | |
We know that \(PU=OQ=ST=4\). Now, we must solve for the final missing lengths. They are \(\overline{RT}\) and \(\overline{PO}\). We can use Pythagorean's Theorem for this as we know the lengths of the one of the legs and the hypotenuse.
| \(a^2+b^2=c^2\) | This is the equation for the Pythagorean Theorem. Solve for the unknown. |
| \(RT^2+4^2=5^2\) | Simplify. |
| \(RT^2+16=25\) | Subtract 16 on both sides. |
| \(RT^2=9\) | Take the square root of both sides to eliminate the exponent of 2. |
| \(RT=\sqrt{9}=3\) | Of course, we can only have positive lengths in the context of geometry, so you should immediately reject a negative side length. |
| \(RT=OP=3\) |
We now know all the sides of the parallelogram, and we can solve.
| \(OP+PS+ST+TR+RQ+QO=P\) | P, in this case, stands for perimeter, in units, as no specific unit is specified in the problem. |
| \(\textcolor{red}{3+7}+4+\textcolor{red}{3+7}+4=P\) | Since this is all addition, the addition can be computed in any order. In red, I suggest doing that first as it results in 10, which is a friendly number computationally speaking. |
| \(10+4+10+4=P\) | |
| \(28=P\) | |
Therefore, the perimeter of rectangle \(QOST\) is \(28units\)
+73 hi, qr and sr are congruent so qp is 5 too, then 24- 5+5=14 which is the legth for ps plus qr, den divide by 2 so ps and qr are both equal 7. Parallelogram area is bh (base times height) therefore, use 28(the area) divdes by 7(the base) to finde out the height 28/7 is 4. so the height is 4 which is eqvivalent to the height of oq and st. So you know st you know the hypontenuse, use the formula c=square root of a sqaure +b sqaure. Plug it in. 5= sqaure root of x2+4square
25=Xsquare+ 16
9=x2
3=x
there for rt and op are both 3 so 3x2+4x2+7x2+28
Hope this is right PEACE OUT
