PlzzZzz Answer to the below question then I will mark it branliest answer ....solve 26 , 30 , 29
Answers
Answer:
Step-by-step explanation:
30. Draw a line BC parallel to AD
Draw a perpendicular line BE on DF
ABCD is a parallelogram.
BC=AD=25cm
CD=AB=60cm
CF=77-CD=17cm
For Δ BCF
perimeter of triangle=(25+26+17)/2
=68/2=34
By Heron's Formula of triangle=√s(s-a)(s-b)(s-c)
=√34(34-25)(34-26)(34-17)
=204cm²
Now area of ΔBCF=1/2xbase x height
=1/2 BExCF204cm²
=1/2x BEx17BE
=408/17
=24cm
Area of Trapezium =1/2(AB+DF)xBE
=1/2(60+77)x24
=1644cm²
Answer:
30.q
Step-by-step explanation:
Area of trapezium = 1/2 h(a + b)
draw ABCD trapezium
AB = 60, CD = 77
draw perpendiculars AN and BM from A and B i.e. height = h
Let distance DN = x and CM = y
Distance MN = 60
x + y = 17
consider triangle ADN and CMB
x² + h² = 25²
y² + h² = 26²
subtract
(y² – x²) = 26² – 25²
(y + x)( y – x) = 26² – 25²
17(y – x) = (26 + 25)(26 – 25) = 51
y – x = 3
y + x = 17
from this 2y = 20
y = 10
from triangle BCM
h² + 10² = 26²
h² = 26² – 10² = (26 + 10)(26 – 10)
h² = 36 × 16
h = 6 × 4 = 24
A = 24/2 (60 + 77) = 1644 cm²
