Let ABC be an isosceles triangle with AB = AC and ∠BAC=30°. Side AB and the median AD contain points Q and P respectively (point
s P and Q are not at point A) in such a way that PC=PQ. Find m∠PQC.
1 answer:
Answer:
15°
Step-by-step explanation:
Since P is on the median of ΔABC, it is equidistant from points B and C as well as from C and Q. Thus, points B, C, and Q all lie on a circle centered at P. (See the attached diagram.)
The base angles (B and C) of triangle ABC are (180° -30°)/2 = 75°. This means arc QC of the circle centered at P has measure 150°. The diameter of circle P that includes point Q is defined to intersect circle P at R.
Central angle RPC is the difference between arcs QR and QC, so is 180° -150° = 30°. Inscribed angle RQC has half that measure, so is 15°. Angle PQC has the same measure as angle RQC, so is 15°.
Angle PQC is 15°.
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<u>Answer</u>:
The vertices are:
A' = (-3, -3)
B' = (3, -3)
C' = (3, 3)
D' = (-3, 3)
The ratio of area of larger square to smaller square is 9:1
<u>Step-by-step explanation:</u>
Given:
A 2 x 2 square is centered at the origin.
So, the center of the square is (0, 0)
Since it is 2 x 2 square, the side of the square is 2 units.
So, the vertices of the 2 x 2 square are A (-1, -1), B(1, -1), C(1. 1), D(-1, 1)
The above square is dilated by a factor of 3.
Let's name the dilated square A'B'C'D'
To find the coordinates of the vertices of dilated square, we need to multiply each vertices of ABCD by 3.
A(-1, -1) = 3(-1, -1) = A'(-3, -3)
B(1, -1) = 3(1, -1) = B'(3, -3)
C(1, 1) = 3(1, 1) = C'(3, 3)
D(-1, 1) = 3(-1, 1) = D'(-3, 3)
To find the area of the small square
the side of the small square is 2 units
so the are of the small square is = 4 square units
To find the area of the larger square
lets find the side AB of the square using distance formula
=>
=>
=>
=>
=>
=>6
AB =6 units
In a square all the sides will be equal
Now the area of the larger square will be
36 square units
The ratio of larger square to smaller square is
=>36 : 4
=>9 : 1