2006-12-24 22:02:59 · 8 answers · asked by SS 2 in Science & Mathematics ➔ Mathematics
Given:
ABCD is a square that is 1 m on a side.
P is a point on AB.
Q is a point on BC.
< = Angle
Find the area A, of triangle PQB.
AB = 1 m
Therefore P is the midpoint on AB.
PB = AB - AP = 1 - 1/2 = 1/2
tan(
tan(
The tangent subtraction of angles formula is:
tan(a - b) = [(tan a) - (tan b)] / [1 + (tan a)(tan b)]
tan(
= (1 - 1/2) / [1 + 1(1/2)] = (1/2) / (3/2) = 1/3
But
tan(
BQ = BC - QC = 1 - 1/3 = 2/3
So the area A, of triangle PQB = (1/2)(base)(height)
A = (1/2)(PB)(BQ) = (1/2)(1/2)(2/3) = 1/6 m²
2006-12-26 10:49:25 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
Since the right triangle has its second angle equal to 45 degrees, triangle PBQ is a 45-45-90 triangle. This is otherwise known as an isosceles triangle. Since AP = (1/2) m, so will BP = (1/2) since the side of the square ABCD is equal to 1. Since the triangle is isosceles, the legs of the triangle are congruent (BP=BQ= (1/2) m).
The area of a triangle is found by using the formula A = (1/2)bh.
Therefore, A = (1/2)(1/2)(1/2). Thus, the area of triangle PBQ is (1/8) square meters.
2006-12-24 22:45:42 · answer #2 · answered by coachandybrown 2 · 0⤊ 0⤋
Presume the area PQB is required.
Area PQB = PQ * QB/2 = 1/2 * 2/3 * 1/2 = 1/6 m.
nago
2006-12-25 12:05:36 · answer #3 · answered by Anonymous · 0⤊ 0⤋
90
2006-12-25 01:32:18 · answer #4 · answered by Chetan 1 · 0⤊ 0⤋
1/8
2006-12-24 22:15:46 · answer #5 · answered by surya o 2 · 0⤊ 0⤋
it quite is rather very very difficult! i ought to photograph the sq. ABCD E, F, G, H, mid factors of AB, BC, CD, DA respectively i won't photograph what are the line segments interior the sq. which intersects at P, Q, R, S. the only line segments which at the instant are not the aspects of the two squares are the diagonals yet they intersect on the middle of the sq.. So the place are P, Q, R, S ?
2016-11-23 16:23:08 · answer #6 · answered by ? 4 · 0⤊ 0⤋
PD=√1^2+.5^2=1.118
angle ADP=arctan .5/1=26.565
angle QDC=90-45-26.565=18.345
QD=1/cos 18.345=1.054
PQ^2=QD^2+PD^2-2*PD*QD*cos 45
PQ^2=1.054^2+1.118^2-2-1.054*1.118*√2/2
PQ^2=.6944
PQ=.833 m
2006-12-25 05:52:24 · answer #7 · answered by mu_do_in 3 · 0⤊ 0⤋
Area of the triangle PQB = 0.5*BP*BQ
PB =1-0.5=0.5
BQ =1-CQ
CQ/CD=tan(d)
d=90-[45+angle(PDA)]
tan(angle PDA) = AP/AD =0.5/1=0.5
angle PDA = arctan(0.5) =26.56 degrees
Now, backward evaluation:
d = 90-[45+angle(PDA)] = 18.43 degrees.
tan(d)=CQ/CD=CQ/1=CQ so CQ =tan d = 1/3
BQ =1-CQ = 2/3
Finally,
Area of the triangle PQB = 0.5*BP*BQ
=0.5*0.5*2/3=1/6.
2006-12-25 00:55:36 · answer #8 · answered by mulla sadra 3 · 0⤊ 0⤋