Question

please give me answer​

Answer 1

Answer:

19.17suare .cm helpful or not

Answer 2

Answer:

B

Step-by-step explanation:

Area of given quadrilateral is sum of triangles ∆ABC and ∆ACD .

First we will calculate area of ∆ABC

Note that sides of this traingle are 3 , 4 , 5 and

they satisfies Pythogoras Theorem

${5}^{2} = {3}^{2} + {4}^{2}$

So it's right angel triangle so it's area will be

$\frac{1}{2} \times 3 \times 4= 6 \: sq \: cm$

Now ∆ADC is a Isosceles triangle .

To find it's area put a perpendicular (p) AP on base DC ( b = 4 cm ) .

Note that P will equally divide base DC ( Congruent triangles ) . So AP = 2 cm .

Now use Pythogoras Theorem to find length p .

$AD {}^{2} = AP {}^{2} + DP {}^{2} \\ \\ \implies \: {5}^{2} = {p}^{2} + {2}^{2} \\ \\ \implies \: p = \sqrt{25 - 4} = \sqrt{21}$

So area of ∆ADC will be

$\frac{1}{2} bp = \frac{1}{2} \times 4 \times \sqrt{21} = 2 \sqrt{21} \:$

So total area of quadrilateral will be

$= 6 + 2 \sqrt{21} \: sq \: cm \\ \\ \approx \: 6 + 2 \times (1.73 \times 2.64) \\ \\ = 6 + 9.13 \: \: \: \: \: \: \: \\ = 15.13 \: sq \: cm$

Which is close to Option B