Answer:
x=5 y=-6
Step-by-step explanation:
3x+y=9 ------eqn1
y=4-2x -------eqn2
using substitution
put y=4-2x into eqn1
3x+4-2x=9
x=5
put x=5 into eqn2
y=4-2(5)
y=4-10
y=-6
A. Area of ABCD - Area of DGA = Area of DEFG
s^2 - 1/2bh = s^2
(5)^2 - 1/2(4)(3) = (3)^2
25 - 1/2(12) = 9
25 - 24 = 9
1 not equal to 9
B. Area of ABCD - Area of GHIA = Area of DGA
s^2 - s^2 = 1/2bh
(5)^2 - (4)^2 = 1/2(4)(3)
25 - 16 = 1/2(12)
9 not equal to 6
C. Area of ABCD + Area of DGA = Area of GHIA
s^2 + 1/2bh = s^2
(5)^2 + 1/2(4)(3) = (4)^2
25 + 1/2(12) = 16
25 + 6 = 16
31 not equal to 16
D. Area of DEFG + Area of GHIA = Area of ABCD
s^2 + s^2 = s^2
(3)^2 + (4)^2 = (5)^2
9 + 16 = 25
25 = 25
The answer is D.
Answer:
Are you in Connexus ?
Step-by-step explanation:
In order to form triangle PQT and quadrilateral TQRS, point T must lie on line PS which is 16 cm. long.
If the ratio of PT to TS is 5:3 and the total length of PS is 16, then PT must be 10 and TS must be 6 (10 + 6 =16) and 10:6 is the same ratio as 5:3. Another way to think about it is 5/3 = 10/6.
Now you have all the lengths that you need to find the areas of the quadrilateral and the triangle.
Make sure you draw a diagram of it!!
