In triangle PQR, M is the midpoint of PQ. Let X be the point on QR such that PX bisects angle QPR, and let the perpendicular bisector of PQ intersect AX at Y. If PQ = 36, PR = 22, QR = 26, and MY = 8, then find the area of triangle PQR
In triangle PQR, M is the midpoint of PQ. Let X be the point on QR such that PX bisects angle QPR, and let the perpendicular bisector of PQ intersect AX at Y. If PQ = 36, PR = 22, QR = 26, and MY = 8, then find the area of triangle PQR
I was trying to draw the figure, and then
you threw me a curve when you said AX.
Where did that come from?
Then I saw that you have the values of all
three sides. You don't need anything else.
Add all three sides, then take half of it.
Call this "s".
36 + 22 + 26
s = —————— = 42
2
Then use the following to get the area. A = sqrt[(s)(s – a)(s – b)(s – c)]
This is called Heron's Formula.
A = sqrt[(42)(42 – 36)(42 – 22)(42 – 26)]
A = sqrt(42 • 6 • 20 • 16)
A = sqrt(80,640)
A = 283.97 square units
That answer looks weird,but it has
to be right ... it's Heron's Formula.
.
First we denote the length of PX as 'x'. By applying the Angle Bisector Theorem, we have:
PR / RQ = PX / XQ
22 / 26 = x / (26 - x)
22 * (26 - x) = 26x
572 - 22x = 26x
48x = 572
x ≈ 11.92
Now, let's find the length of XQ:
XQ = QR - PX ≈ 26 - 11.92 ≈ 14.08
Since M is the midpoint of PQ, we can determine the length of PM:
PM = PQ / 2 = 36 / 2 = 18
Since MY is perpendicular to PQ and MY = 8, we can determine the length of PY:
PY = PM - MY = 18 - 8 = 10
Therefore, AY is equal to PY:
AY = PY = 10
Now, we have all the side lengths of triangle AXY. We can calculate its area using Heron's formula:
s = (XY + AY + AX) / 2 = (14.08 + 10 + 11.92) / 2 ≈ 18
Area_AXY = sqrt(s * (s - XY) * (s - AY) * (s - AX))
Area_AXY = sqrt(18 * (18 - 14.08) * (18 - 10) * (18 - 11.92))
Area_AXY ≈ sqrt(18 * 3.92 * 8 * 6.08) ≈ 24.48
Finally, the area of triangle PQR is twice the area of triangle AXY:
Area_PQR = 2 * Area_AXY ≈ 2 * 24.48 ≈ 48.96 square units
sorry, but none of these are correct...
thanks for trying though....
I still need help
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