Answer:
Δ MLP ~ Δ HKG
By S-A-S similarity Postulate.
Step-by-step explanation:
Given:
In Δ MLP
MP = 20 ,
LP = 15
∠ P = 42°
In Δ HKG
HG = 16 ,
KG = 12
∠G = 42°
To Prove:
Δ MLP ~Δ HKG
Proof:
In Δ MLP and Δ HKG
..........( 1 )
..............( 2 )
∴ 
...From 1 and 2
∠P ≅ ∠ G = 42° ...............Given
∴ Δ MLP ~Δ HKG ......{ By S-A-S similarity test}............Proved
If you multiply 214 by 112 you would get the answer 23,968.Believe me i did the math.
5 1/2 - 3/8
Make 5 1/2 into a improper fraction
5 1/2=11/2. Mutiply the whole number with the denominator. 5*2= 10. Add 10 with the numerator. 10+1= 11
11/2-3/8.
Find the common denominator for 11/2 which is 8. Mutiply by 4
11(4)/2(4)= 44/8
44/8-3/8= 41/8= 5 1/8
Answer: 41/8 or 5 1/8
For this case we have:
a = 30 cm
c = 16 cm
We look for the length of the diagonal:
d = x + y
Where,
For x:
a ^ 2 = x ^ 2 + x ^ 2
x = a / root (2) = 30 / root (2) = 21.2132 cm
For y:
c ^ 2 = y ^ 2 + y ^ 2
y = c / root (2) = 16 / root (2) = 11.3137 cm
The diagonal is:
d = x + y
d = 21.2132 + 11.3137
d = 32.5269 cm
Then, the height is:
h = h1 + h2
For h1:
h1 = root (x ^ 2 - (a / 2) ^ 2) = root ((21.2132) ^ 2 - (30/2) ^ 2)
h1 = 15 cm
For h2:
h2 = root (y ^ 2 - (c / 2) ^ 2) = root ((11.3137) ^ 2 - (16/2) ^ 2)
h2 = 8 cm
Finally:
h = h1 + h2
h = 15 + 8
h = 23 cm
Then, the area is:
A = (1/2) * (a + c) * (h)
A = (1/2) * (30 + 16) * (23)
A = 529 cm ^ 2
Answer:
the area of an isosceles trapezoid is:
A = 529 cm ^ 2
