Look at the given triangles. A student argued that since the perimeter of △XYZ is more than that of △ABC, so area of △XYZ is also greater than that of △ABC. Is he right?

From figure,

Perimeter of △ABC = AB + BC + CA = 5 + 5 + 6 = 16 cm

Perimeter of △XYZ = XY + YZ + ZX = 5 + 5 + 8 = 18 cm.

By formula,

Area of an isosceles triangle = $\dfrac{1}{4}b\sqrt{4a^2 - b^2}$, where a is the length of equal sides and b is the length of base.

Area of △ABC = $\dfrac{1}{4} \times 6 \times \sqrt{4(5)^2 - 6^2}$

$= \dfrac{1}{4} \times 6 \times \sqrt{100 - 36} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{64} \\[1em] = \dfrac{1}{4} \times 6 \times 8 \\[1em] = 12 \text{ cm}^2.$

Area of △XYZ = $\dfrac{1}{4} \times 8 \times \sqrt{4(5)^2 - 8^2}$

$= \dfrac{1}{4} \times 8 \times \sqrt{100 - 64} \\[1em] = \dfrac{1}{4} \times 8 \times \sqrt{36} \\[1em] = \dfrac{1}{4} \times 8 \times 6 \\[1em] = 12 \text{ cm}^2.$

Although, perimeter of △XYZ is more than that of △ABC, but area of both the triangles are equal.

Hence, the student is wrong.