Question-212829

Question Number 212829 by mr W last updated on 25/Oct/24

Commented by mr W last updated on 25/Oct/24

$${find}\:{the}\:{total}\:{area}\:{of}\:{the}\:{rectangle}. \\ $$

Answered by ajfour last updated on 25/Oct/24

let ((4+A)/(12))=(p/q) y=((h/p))x & y=−((h/(p+q)))(x−p−q) ⇒ ((hx)/p)((q/(p+q)))=h ⇒ x=(((p+q)p)/q) Now (A/4)=(x/p)=(p/q)+1 ⇒ ((A−4)/4)=((4+A)/(12)) 3A−12=4+A A=8 rectangle area = 2(A+4+12) =2×24=48 sq units.

$${let}\:\:\:\frac{\mathrm{4}+{A}}{\mathrm{12}}=\frac{{p}}{{q}} \\ $$$${y}=\left(\frac{{h}}{{p}}\right){x}\:\:\:\:\&\:\:{y}=−\left(\frac{{h}}{{p}+{q}}\right)\left({x}−{p}−{q}\right) \\ $$$$\Rightarrow\:\:\frac{{hx}}{{p}}\left(\frac{{q}}{{p}+{q}}\right)={h} \\ $$$$\Rightarrow\:\:\:{x}=\frac{\left({p}+{q}\right){p}}{{q}} \\ $$$${Now}\:\:\frac{{A}}{\mathrm{4}}=\frac{{x}}{{p}}=\frac{{p}}{{q}}+\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{{A}−\mathrm{4}}{\mathrm{4}}=\frac{\mathrm{4}+{A}}{\mathrm{12}} \\ $$$$\mathrm{3}{A}−\mathrm{12}=\mathrm{4}+{A} \\ $$$${A}=\mathrm{8} \\ $$$${rectangle}\:{area}\:=\:\mathrm{2}\left({A}+\mathrm{4}+\mathrm{12}\right) \\ $$$$\:\:\:\:=\mathrm{2}×\mathrm{24}=\mathrm{48}\:{sq}\:{units}. \\ $$

Answered by A5T last updated on 25/Oct/24

Commented by A5T last updated on 25/Oct/24

A+4=4+B⇒A=B A+C=4+B+12⇒C=16 (A/4)=(C/B)⇒A^2 =4C=64⇒A=8 [rectangle]=8+8+16+12+4=48

$${A}+\mathrm{4}=\mathrm{4}+{B}\Rightarrow{A}={B} \\ $$$${A}+{C}=\mathrm{4}+{B}+\mathrm{12}\Rightarrow{C}=\mathrm{16} \\ $$$$\frac{{A}}{\mathrm{4}}=\frac{{C}}{{B}}\Rightarrow{A}^{\mathrm{2}} =\mathrm{4}{C}=\mathrm{64}\Rightarrow{A}=\mathrm{8} \\ $$$$\left[{rectangle}\right]=\mathrm{8}+\mathrm{8}+\mathrm{16}+\mathrm{12}+\mathrm{4}=\mathrm{48} \\ $$

Answered by Spillover last updated on 25/Oct/24

Answered by Spillover last updated on 25/Oct/24

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