Math, asked by jharashi35, 2 days ago

Find the perimeter and area of the quadrilateral ABCD in which AB = 42cm, BC = 21cm, CD = 29cm, DA = 34cm and angle CBD = 90°.

Answers

Answered by Yoursenorita
3

Given

Sides of a quadrilateral are AB = 42 cm, BC = 21 cm, CD = 29 cm DA = 34 cm and diagonal BD = 20 cm

Area of quadrilateral = area of ∆ADB + area of ∆BCD

Now, area of ∆ABD Perimeter of ∆ABD

We know that,

 \\  = ⇒ S =  \frac{1}{2} (++) \\  \\   ⇒ \frac{1}{2} (34 + 42 + 20) = 96 \\  \\ ⇒  \frac{96}{2}  = 48cm \:  \\  \\ Area  \:  \: of \:  \:  ΔABD  \\  \\ ⇒  \sqrt{s(s - )(s - )(s - )} \\  \\ ⇒  \sqrt{48(48 - 42)(48 - 20)(48 - 34)}  \\  \\ ⇒ 48(14)(6)(28) \\  \\ ⇒ 336 {cm}^{2}  \\  \\ Also  \:  \: for  \:  \: area  \:  \: of \:  \:  ΔBCD, \\  \\ </em></p><p><em>[tex] \\  = ⇒ S =  \frac{1}{2} (++) \\  \\   ⇒ \frac{1}{2} (34 + 42 + 20) = 96 \\  \\ ⇒  \frac{96}{2}  = 48cm \:  \\  \\ Area  \:  \: of \:  \:  ΔABD  \\  \\ ⇒  \sqrt{s(s - )(s - )(s - )} \\  \\ ⇒  \sqrt{48(48 - 42)(48 - 20)(48 - 34)}  \\  \\ ⇒ 48(14)(6)(28) \\  \\ ⇒ 336 {cm}^{2}  \\  \\ Also  \:  \: for  \:  \: area  \:  \: of \:  \:  ΔBCD, \\  \\ Perimeter  \:  \: of  \:  \: ΔBCD \\  \\ 2s =  \: bc + cd + bd \\  \\ ⇒  \frac{1}{2} (29 + 21 + 20)  + 35cm \\  \\ By  \:  \: using  \:  \: heron’s  \:  \: formulae \\  \\ </em></p><p><em>[tex] \\  = ⇒ S =  \frac{1}{2} (++) \\  \\   ⇒ \frac{1}{2} (34 + 42 + 20) = 96 \\  \\ ⇒  \frac{96}{2}  = 48cm \:  \\  \\ Area  \:  \: of \:  \:  ΔABD  \\  \\ ⇒  \sqrt{s(s - )(s - )(s - )} \\  \\ ⇒  \sqrt{48(48 - 42)(48 - 20)(48 - 34)}  \\  \\ ⇒ 48(14)(6)(28) \\  \\ ⇒ 336 {cm}^{2}  \\  \\ Also  \:  \: for  \:  \: area  \:  \: of \:  \:  ΔBCD, \\  \\ Perimeter  \:  \: of  \:  \: ΔBCD \\  \\ 2s =  \: bc + cd + bd \\  \\ ⇒  \frac{1}{2} (29 + 21 + 20)  + 35cm \\  \\ By  \:  \: using  \:  \: heron’s  \:  \: formulae \\  \\ Area of ΔBCD  \\  \\  =  \sqrt{s(s - bc)(s - cd)(s - db)}  \\  \\  =  \sqrt{35(35 - 21)(53 - 29)(35 - 20)}  \\  \\  =  \sqrt{210 \times 210}  \\  \\  = 210 {cm}^{2}  \\  \\ ∴ Area \:  \:  of  \:  \: quadrilateral   \: ABCD  \\  \\ = 336 + 210 = 546 {cm}^{2}  \\  \\

Answered by ItzStarling
0

Answer:

Also for area of ΔBCD,

Perimeter of ΔBCDAlso for area of ΔBCD,

Perimeter of ΔBCDAlso for area of ΔBCD,

Perimeter of ΔBCDAlso for area of ΔBCD,

Perimeter of ΔBCDAlso for area of ΔBCD,

Perimeter of ΔBCDAlso for area of ΔBCD,

Perimeter of ΔBCDAlso for area of ΔBCD,

Perimeter of ΔBCDAlso for area of ΔBCD,

Perimeter of ΔBCDAlso for area of ΔBCD,

Perimeter of ΔBCD

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