Math, asked by HɪɢʜᴇʀKᴜsʜᴀʟBᴏʏSᴜʙs, 6 months ago

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Answered by Anonymous
43

\huge\red{AnswEr}

\huge\red{Ste\: by step}

A(APB) = 1/2 * 50*100=2500 triangle area

A(CQD) = 1/2* 70 * 100 = 3500 ...since DQ=PB=100

now ARP and BRQ are similar triangles..

so BR/PR = RQ/AR... RQ= BR/PR * AR = (60/40) * 50 = 150/2

RQ= 150/2

A(BRQ) = 1/2 * RQ * RB = 1/2 * 150/2 * 60 = 2250

A(BRC) = 1/2 * RC*RB=1/2 *(150/2+70) *60 = 4350

A(BQC) = BRC-BRQ=4350-2250=2100

BPDQ = 15600 - 2500 - 3500 - 2100 = 7500

Answered by ꜱɴᴏᴡyǫᴜᴇᴇɴ
53

\huge\star{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}

A(APB) = 1/2 * 50*100=2500 triangle area

A(CQD) = 1/2* 70 * 100 = 3500 ...since DQ=PB=100

now ARP and BRQ are similar triangles..

so BR/PR = RQ/AR... RQ= BR/PR * AR = (60/40) * 50 = 150/2

RQ= 150/2

A(BRQ) = 1/2 * RQ * RB = 1/2 * 150/2 * 60 = 2250

A(BRC) = 1/2 * RC*RB=1/2 *(150/2+70) *60 = 4350

A(BQC) = BRC-BRQ=4350-2250=2100

BPDQ = 15600 - 2500 - 3500 - 2100 = 7500

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