Question

Answer ASAP, Will give brainliest!!

Answer 1

Answer:

Step-by-step explanation:

1)As ABCD is isosceles trapezium,

∠ADC= ∠DAB

∠ADC = 115°

2) AD //BC

∠ADC + ∠DCB = 180°   {co interior angles}

115 + ∠DCB = 180

 ∠DCB = 180 - 115

 ∠DCB = 65°

3) As ABCD is isosceles trapezium,

∠CBA = ∠DCB

∠CBA = 65°

4) As ABCD is isosceles trapezium, non parallel sides are congruent.

AB = DC

3x + 4 = 25 in

3x = 25 - 4

3x = 21

x = 21/3

x = 7 in

5) height = 13.5 in

a= 43 in

b= 20 in

Area of trapezium = $\frac{(a+b)*h}{2}$

                   $= \frac{(43 +20)*13.5}{2}\\\\=\frac{63*13.5}{2}\\\\\\= 425.25 in^{2}$

Answer 2

Answer:

First. 115°

Second. 65°

Third. 65°

Fourth. 7

Fifth. 425.25

First

angle DAB = angle ADC (since this is an isosceles trapezoid)

Second

In a trapezoid adjacent angle are supplmentary (that is their sum is 180°)

180-115 is 65°

Third

(Same reason as second)

Fourth

The side 3x+4 is same as the opposite side

So 3x + 4 = 25

on solving you get x = 7 in

Fifth

$area \: = \frac{1}{2} \times length \: of \: the \: perpendicular \: (b1 + b2)$

area = 1/2 × 13.5 (20+43)

area = 1/2 × 13.5 × 63

Thus area is 425.25