Question

I have four sides measuring 5 cm, 8 cm, 9 cm and 7 cm. Which shape is it.

Answer 1

Step-by-step explanation:

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Answer 2

AB= 9

BC= 40

CD= 28

DA= 15

join A and C to form a diagonal AC in quadrilateral ABCD

using pythagorean theorem,

$AB ^(2) + BC ^(2) = AC ^(2)</p><p>9*9 + 40*40 = AC ^(2)</p><p>81 + 1600 = AC ^(2)</p><p>1681 = AC^(2)</p><p>\sqrt{1681} </p><p>1681</p><p> </p><p> = AC</p><p>41 = AC</p><p></p><p>now, using herons formula, u can find the area of triangle ABC an d triangle ADC.</p><p></p><p>in triangle ABC</p><p>\sqrt{s(s-a)(s-b)(s-c)} </p><p>s(s−a)(s−b)(s−c)</p><p> </p><p> = \sqrt{45(45-9)(45-40)(45-41)} </p><p>45(45−9)(45−40)(45−41)</p><p> </p><p> </p><p> = \sqrt{45(36) (5)(4)} </p><p>45(36)(5)(4)</p><p> </p><p> </p><p> = \sqrt{5*3*3*2*3*2*3*5*2*2} </p><p>5∗3∗3∗2∗3∗2∗3∗5∗2∗2</p><p> </p><p> </p><p> = \sqrt{5*5*3*3*2*2*3*3*2*2} </p><p>5∗5∗3∗3∗2∗2∗3∗3∗2∗2</p><p> </p><p> </p><p> = 5*3*2*3*2</p><p> = 5*36</p><p> = 180</p><p></p><p></p><p>in triangle ADC</p><p>\sqrt{s(s-a)(s-b)(s-c)} </p><p>s(s−a)(s−b)(s−c)</p><p> </p><p> = \sqrt{42(42-28)(42-15)(42-41)} </p><p>42(42−28)(42−15)(42−41)</p><p> </p><p> </p><p> = \sqrt{42*14*27*1} </p><p>42∗14∗27∗1</p><p> </p><p> </p><p> = \sqrt{2*3*7*2*7*3*3*3} </p><p>2∗3∗7∗2∗7∗3∗3∗3</p><p> </p><p> </p><p> = \sqrt{2*2*7*7*3*3*3*3} </p><p>2∗2∗7∗7∗3∗3∗3∗3</p><p> </p><p> </p><p> =2*7*3*3*</p><p> = 14*9</p><p> = 126$

area of ABC + area of ADC = area of ABCD

180 + 126 = 306