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in Perimeter and Area of Plane Figures by (23.5k points)
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In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45º. The area of the triangle is:

(a) 25 cm

(b) \(\frac{25}{2}\sqrt2\) cm2 

(c) 25 √2 cm2 

(d) 25 √3 cm2

1 Answer

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Best answer

(c) 25√2 cm2.

ΔABC is an isosceles triangle with AB = AC = 10 cm. ∠A = 45° 

∴ Area of ΔABC

\(\frac{1}{2}\) x 10 x 10 x sin 45°

[Using Δ = \(\frac{1}{2}\) bc sin A]

\(\frac{50}{\sqrt2}\) = \(\frac{50}{\sqrt2}\) x \(\frac{\sqrt2}{\sqrt2}\) = 25√2 cm2.

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