Question

pls home ans proper​

Answer 1

$\large{\boxed{\boxed{\sf Let's \: Understand \: Concept \: f1^{st}}}}$

Here, we have give a fig. with the measurements of all its sides. We, have to find its area. After using area of Triangle∆ and area of Trapezium in all the triangles and Trapezium visible in the fig. the we will add all the areas of triangles and Trapezium together and we, will got the area of whole fig. as answer.

$\huge{\underline{\boxed{\sf Answer}}}$

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$\large{\underline{\sf Given:-}}$

▣AP = 30m

▣PQ = 20m

▣PR = 20m

▣RD = 30m

▣PB = 20m

▣RC = 40m

▣EQ = 30m

$\large{\underline{\sf Find:-}}$

▩Area of the given figure

$\large{\underline{\sf Formula \: Used:-}}$

$\boxed{\begin{lgathered} \\ \sf Area \: of \: Triangle = \dfrac{1}{2} \times base \times height \\ \\ \end{lgathered}}$

$\boxed{\begin{lgathered} \\ \sf Area \: of \: Trapezium = \dfrac{1}{2} \times (a + b) \times h \\ \\ \end{lgathered}}$

$\large{\underline{\sf Solution:-}}$

In ∆CDR

$\dashrightarrow\sf Area \: of \: \triangle CDR= \dfrac{1}{2} \times base \times height$

$\dashrightarrow\sf Area \: of \: \triangle CDR= \dfrac{1}{2} \times 40\times 30$

$\dashrightarrow\sf Area \: of \: \triangle CDR= \dfrac{1}{2} \times 1200$

$\dashrightarrow\sf Area \: of \: \triangle CDR= \dfrac{1200}{2}$

$\dashrightarrow\sf Area \: of \: \triangle CDR=600 {m}^{2}$

$\small{\therefore\sf ar( CDR)=600 {m}^{2}}$

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In ∆DQE

$\dashrightarrow\sf Area \: of \: \triangle DQE= \dfrac{1}{2} \times base \times height$

$\dashrightarrow\sf Area \: of \: \triangle DQE= \dfrac{1}{2} \times 50 \times 30$

$\dashrightarrow\sf Area \: of \: \triangle DQE= \dfrac{1}{2} \times 1500$

$\dashrightarrow\sf Area \: of \: \triangle DQE= \dfrac{1500}{2}$

$\dashrightarrow\sf Area \: of \: \triangle DQE= 750 {m}^{2}$

$\small{\therefore\sf ar( DQE)= 750 {m}^{2}}$

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In ∆EQA

$\dashrightarrow\sf Area \: of \: \triangle EQA= \dfrac{1}{2} \times base \times height$

$\dashrightarrow\sf Area \: of \: \triangle EQA= \dfrac{1}{2} \times 30\times 50$

$\dashrightarrow\sf Area \: of \: \triangle EQA= \dfrac{1}{2} \times 1500$

$\dashrightarrow\sf Area \: of \: \triangle EQA= \dfrac{1500}{2}$

$\dashrightarrow\sf Area \: of \: \triangle EQA= 750 {m}^{2}$

$\small{\therefore\sf ar(EQA)= 750 {m}^{2}}$

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In ∆ABP

$\dashrightarrow\sf Area \: of \: \triangle ABP= \dfrac{1}{2} \times base \times height$

$\dashrightarrow\sf Area \: of \: \triangle ABP= \dfrac{1}{2} \times 30 \times 20$

$\dashrightarrow\sf Area \: of \: \triangle ABP= \dfrac{1}{2} \times 600$

$\dashrightarrow\sf Area \: of \: \triangle ABP= \dfrac{600}{2}$

$\dashrightarrow\sf Area \: of \: \triangle ABP= 300 {m}^{2}$

$\small{\therefore\sf ar( ABP)= 300 {m}^{2}}$

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In Trapezium PBCR

$\dashrightarrow\sf Area \: of \: PBCR = \dfrac{1}{2} \times (a + b) \times h$

$\dashrightarrow\sf Area \: of \: PBCR = \dfrac{1}{2} \times (20 + 40) \times40$

$\dashrightarrow\sf Area \: of \: PBCR = \dfrac{1}{2} \times (60) \times40$

$\dashrightarrow\sf Area \: of \: PBCR = \dfrac{1}{2} \times 60\times40$

$\dashrightarrow\sf Area \: of \: PBCR = \dfrac{1}{2} \times 2400$

$\dashrightarrow\sf Area \: of \: PBCR = \dfrac{2400}{2}$

$\dashrightarrow\sf Area \: of \: PBCR = 1200 {m}^{2}$

$\small{\therefore\sf ar(PBCR) = 1200 {m}^{2}}$

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➡Area of the field = ar(CDR) + ar(DQE) + ar(EQA) + ar(ABP) + ar(PBCR)

➡Area of the field = 600 + 750 + 750 + 300 + 1200

➡Area of the field = 3600m²

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Hence,

Area of the given field is 3600m²

Answer 2

Answer

Given:−

• AP = 30m
• PQ = 20m
• PR = 20m
• RD = 30m
• PB = 20m
• RC = 40m
• EQ = 30m

Find:−

Area of the given figure.

Solution:−

In ∆CDR

=> Area of△CDR= 21 ×base×height

=> Area of△CDR= 21×40×30

=> Area of△CDR= 21 ×1200

=> Area of△CDR= 21200

=> Area of△CDR=600m²

=> ∴ar(CDR)=600m²

In ∆DQE

=> Area of△DQE= 21 ×base×height

=> Area of△DQE= 21 ×50×30

=> Area of△DQE= 21 ×1500

=> Area of△DQE= 21500

=> Area of△DQE=750m²

=> ∴ar(DQE)=750m²

In ∆EQA

=>Area of△EQA= 21 ×base×height

=> Area of△EQA= 21 ×30×50

=> Area of△EQA= 21 ×1500

=> Area of△EQA= 21500

=> Area of△EQA=750m²

=> ∴ar(EQA)=750m²

In ∆ABP

=> Area of△ABP= 21 ×base×height

=> Area of△ABP= 21 ×30×20

=> Area of△ABP= 21 ×50

=> Area of△ABP= 2600

=> Area of△ABP=300m²

=> ∴ar(ABP)=300m²

In Trapezium PBCR

=> Area of PBCR= 21 ×(a+b)×h

=> Area of PBCR= 21 ×(20+40)×40

=> Area of PBCR= 21 ×(60)×40

=> Area of PBCR= 21 ×60×40

=> Area of PBCR= 21 ×2400

=> Area of PBCR= 22400

=> Area of PBCR=1200m²

=>∴ar(PBCR)=1200m²

➡Area of the field = ar(CDR) + ar(DQE) + ar(EQA) + ar(ABP) + ar(PBCR)

➡Area of the field = 600 + 750 + 750 + 300 + 1200

➡Area of the field = 3600m².

Hence,

Area of the given field is 3600m².