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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.