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Sagot :
Answer:
[tex]
[tex] \large{\boxed{\sf Hypotenuse = 16.97\ cm }}[/tex]
Hypotenuse=16.97 cm
Step-by-step explanation:
Here it is given that the area of a right isosceles ∆ is 72 cm² . Let us assume that each equal side is x . Therefore the height and the base of the ∆ will be same that is x .
p
[tex] \begin{gathered}\sf\qquad\longrightarrow [/tex]
Area =\dfrac{1}{2}(base)(height)\\\end{gathered}⟶Area=21(base)(height)
[tex] \begin{gathered}\sf\qquad\longrightarrow[/tex]
[tex] 72cm^2=\dfrac{1}{2}(x)(x)\\\end{gathered} [/tex]
⟶72cm2=21(x)(x)
[tex] \begin{gathered}\sf\qquad\longrightarrow x^2=[/tex]
144cm^2\\\end{gathered}⟶x2=144cm2
[tex] \begin{gathered}\sf\qquad\longrightarrow x[/tex]
[tex] =\sqrt{144cm^2}\\\end{gathered}⟶x=144cm2 [/tex]
[tex]\sf\qquad\longrightarrow \pink{x = 12cm }⟶x=12cm [/tex]
Hence we may find hypotenuse using Pythagoras Theorem as ,
[tex] \sf\qquad\longrightarrow h =\sqrt{ p^2+b^2}⟶h=p2+b2 [/tex]
Here p = b = 12cm ,
[tex] \begin{gathered}\sf\qquad\longrightarrow h =\sqrt{ (12cm)^2+(12cm)^2}\\\end{gathered} [/tex]
⟶h=(12cm)2+(12cm)2
[tex] \begin{gathered}\sf\qquad\longrightarrow h [/tex]
[tex] =\sqrt{144cm^2+144cm^2}\\\end{gathered}[/tex]
⟶h=144cm2+144cm2
[tex] \begin{gathered}\sf\qquad\longrightarrow h[/tex]
=\sqrt{288cm^2}\\\end{gathered}⟶h=288cm2
[tex] \sf\qquad\longrightarrow \pink{ hypotenuse=[/tex]
16.97cm }⟶hypotenuse=16.97cm
Hence the hypotenuse is 16.97 cm .
[/tex]
Answer:
$864
Step-by-step explanation:
100-10 = 90/100= 0.9 (multiplier)
1200*0.9= 1080
100-20 = 80/100= 0.8 (2nd multiplier)
1080*0.8= 864
hope this helps
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