Answer:
i would say no
Step-by-step explanation:
i would say no because quadrilateral FGIH is equal to 120 and quadrilateral SRTQ is equal to 134 and also their areas are indifferent FGIH's being 875 and SRTQ's being 1120
please let me know if this was wrong
hope this helps :) have a nice day !!
It will take Dave 7 weeks to earn the right amount of money.
4. SOLVE FOR X:
Using the Alternate Interior Angles Theorem, we know that the 67 degree angle is congruent with the (12x - 5) degree angle. With this information, all I have to do is set the two equal to each other and solve for x.
67 = 12x - 5
67 + 5 = 12x - 5 + 5
72/12 = 12x/12
6 = x
x = 6
SOLVE FOR Y:
Using the Vertical Angles theorem, we know that angle y must be congruent to the 67 degree angle.
y = 67 degrees.
5. SOLVE FOR Y:
Alternate exterior angles: 6(x - 12) = 120
6x - 72 + 72 = 120 + 72
6x/6 = 192/6
x = 32
SOLVE FOR Y:
6((32) - 12) + y = 180
192 - 72 + y = 180
120 + y - 120 = 180 - 120
y = 60
Answer:
Yes
Step-by-step explanation:
The formula for area of a triangle is A = (1/2)bh,
For the first triangle we can leave it in general terms, so it's area is
A = (1/2)bh, depending on what b and h are, but it doesn't matter here...
The second triangle has base that is twice the other triangles base. Bases being multiples of each other is the definition of being proportional so the bases are proportional, an the area of the second triangle is
A = (1/2)(2b)h, which simplifies to
A = bh
Comparing the 2 areas, you can see that one has a multiplier of (1/2), so their areas are proportional
