Question on: JAMB Mathematics - 2009
If the hypotenuse of a right-angled isosceles triangle is 2cm , what is the area of the triangle?
A
\(\frac{1}{\sqrt{2}}\) cm2
B
1 cm2
C
\(\sqrt{2}\)cm2
D
2\(sqrt{2}\)cm2
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Correct Option: B
sin45o = \(\frac{y}{2}\)
y = 2 sin45o
y = 2 (\(\frac{1}{\sqrt{2}}\))
y = \(\frac{2}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\)
y = \(\frac{2\sqrt{2}}{2}\)
y = \(\sqrt{2}\)
Since it is an isosceles triangle
x = \(\sqrt{2}\)
Area of \(\Delta\) = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) x \(\sqrt{2}\) x \(\sqrt{2}\)
= \(\frac{1}{2}\) x 2
= 1 cm2
y = 2 sin45o
y = 2 (\(\frac{1}{\sqrt{2}}\))
y = \(\frac{2}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\)
y = \(\frac{2\sqrt{2}}{2}\)
y = \(\sqrt{2}\)
Since it is an isosceles triangle
x = \(\sqrt{2}\)
Area of \(\Delta\) = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) x \(\sqrt{2}\) x \(\sqrt{2}\)
= \(\frac{1}{2}\) x 2
= 1 cm2
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