Question

using the measures given in the finger, find the area of ABCD​

Answer 1

Answer:

taking ∆BDC

using Pythagoras theorem

${BD}^{2} = {BC}^{2} + {CD}^{2} \\ BD = \sqrt{ {BC}^{2} + {CD}^{2} } \\ BD = \sqrt{ {5}^{2} + {12}^{2} } \\ BD = \sqrt{25 + 144} \\ BD = \sqrt{169} \\ BD = 13$

then, base of ∆ABD

we know that

area of triangle ∆ABD

$\frac{1}{2} \times base \times height \\ = \frac{1}{2} \times 13 \times 6 \\ = 13 \times 3 \\ = 39 {cm}^{2}$

area of ∆BCD

$\frac{1}{2} \times 5 \times 12 \\ = 5 \times 6 \\ = 30 {cm}^{2}$

area of given figure = area of ∆ABD+∆BCD

=39+30

$\huge\green{\mathbb{\boxed{69 {cm}^{2} }}}$

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