Question

please help
no irrelevant answers​

Answer 1

1st Questions answer in attachment ra..

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2nd answer diagram last pic lo una

For ∆ABC, consider

AB2 + BC2 = 32 + 42 = 25 = 52

⇒ 52 = AC2

Since ∆ABC obeys the Pythagoras theorem, we can say ∆ABC is right-angled at B.

Therefore, the area of ΔABC = 1/2 × base × height

= 1/2 × 3 cm × 4 cm = 6 cm2

Area of ΔABC = 6 cm2

Now, In ∆ADC

we have a = 5 cm, b = 4 cm and c = 5 cm

Semi Perimeter: s = Perimeter/2

s = (a + b + c)/2

s = (5 + 4 + 5)/2

s = 14/2

s = 7 cm

By using Heron’s formula,

Area of ΔADC = √[s(s - a)(s - b)(s - c)]

= √[7(7 - 5)(7 - 4)(7 - 5)]

= √[7 × 2 × 3 × 2]

= 2√21 cm2

Area of ΔADC = 9.2 cm2 (approx.)

Area of the quadrilateral ABCD = Area of ΔADC + Area of ΔABC

= 9.2 cm2 + 6 cm2

= 15.2 cm2

Thus, the area of the quadrilateral ABCD is 15.2 cm2.