Question

TTL 7. dleL ABCDHL AB = 9 l, BC = 40 4 CD = 28 Al, DA = 15 AAl >i ZB = 90°​

Answer 1

Answer:

306m

Step-by-step explanation:

Step-by-step explanation:

In quad. ABCD,

AB = 9m, BC = 40m, CD= 28m, AD = 15m

\angle ABC∠ABC = 90 °

Now in ABC,

AC^2 = AB^2 + BC^2AC

2

=AB

2

+BC

2

= (9)^2 + (40)^2(9)

2

+(40)

2

= 81 + 1600

= 1681

AC = \sqrt 1681

1

681 = 41 m

∴Area of quad.ABCD = Area of ΔABC+Area of ΔACD

Ar(ABC) = \frac {1}{2} \times base \times altitude

2

1

×base×altitude (Refer Figure)

= \frac {1}{2} \times 9 \times 40

2

1

×9×40

= 180 m^2m

2

""(1)

Ar(ACD) = Applying Heron's Formula

Semiperimeter, s = \frac {a+b+c}{2}

2

a+b+c

= \frac {15+28+41}{2}

2

15+28+41

(Side are 15m, 28m, 41 m)

= 42 m

Area = \sqrt {s(s-a)(s-b)(s-c)}

s(s−a)(s−b)(s−c)

= \sqrt {42(42-15)(42-28)(42-41)}

42(42−15)(42−28)(42−41)

= \sqrt {42 \times 27 \times 14 \times 1}

42×27×14×1

= 126 m^2m

2

"""(2)

As Area of quad.ABCD = Area of ΔABC+Area of ΔACD

= 180 + 126 (From (1) & (2))

Answer 2

Answer:

I don't know the answer maybe someone can answer this

Step-by-step explanation:

:)))))