Answer:
Step-by-step explanation:
The base and height will be the two equal sides of the isosceles right triangle.
⇒ b = h = x cm
[tex]Area \ of \ triangle = \dfrac{1}{2}bh[/tex]
[tex]\dfrac{1}{2}bh = 72 \ cm^{2}[/tex]
[tex]\dfrac{1}{2}*x*x=72\\\\\\\dfrac{1}{2}*x^{2}=72\\\\\\x^{2}=72*2\\\\x^{2}=144\\\\Take \ square \ root,\\\\\sqrt{x^{2}}=\sqrt{144}\\\\x = \sqrt{12*12}\\\\x = 12 cm[/tex]
Hypotenuse² = b² + h²
= 12² + 12²
= 144 + 144
= 288
hypotenuse = √288
= 16.97 cm
Answer:
[tex]\large{\boxed{\sf Hypotenuse = 16.97\ cm }}[/tex]
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 .
[tex]\sf\qquad\longrightarrow Area =\dfrac{1}{2}(base)(height)\\ [/tex]
[tex]\sf\qquad\longrightarrow 72cm^2=\dfrac{1}{2}(x)(x)\\[/tex]
[tex]\sf\qquad\longrightarrow x^2= 144cm^2\\ [/tex]
[tex]\sf\qquad\longrightarrow x =\sqrt{144cm^2}\\ [/tex]
[tex]\sf\qquad\longrightarrow \pink{x = 12cm }[/tex]
Hence we may find hypotenuse using Pythagoras Theorem as ,
[tex]\sf\qquad\longrightarrow h =\sqrt{ p^2+b^2} [/tex]
[tex]\sf\qquad\longrightarrow h =\sqrt{ (12cm)^2+(12cm)^2}\\[/tex]
[tex]\sf\qquad\longrightarrow h =\sqrt{144cm^2+144cm^2}\\[/tex]
[tex]\sf\qquad\longrightarrow h =\sqrt{288cm^2}\\[/tex]
[tex]\sf\qquad\longrightarrow \pink{ hypotenuse= 16.97cm }[/tex]
Hence the hypotenuse is 16.97 cm .
