Question:
Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is
(a) $5 \sqrt{10} \mathrm{~cm}^{2}$
(b) $50 \mathrm{~cm}^{2}$
(c) $10 \sqrt{3} \mathrm{~cm}^{2}$
(d) $75 \mathrm{~cm}^{2}$
Solution:
(b) 50 cm2
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:
Area of triangle $=\frac{1}{2} \times$ Base $\times$ Height
$=\frac{1}{2} \times 10 \times 10$
$=50 \mathrm{~cm}^{2}$
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