Find the area of an isoceles triangle whose equal sides are 6cm each and unequal side is 4cm. [2 MARKS]

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Solution

Concept : 0.5 Mark
Application : 0.5 Mark
Calculation : 1 Mark

S = $\frac{P e r i m e t e r}{2}$ = $\frac{4 + 6 + 6}{2}$ = 8 cm

Area = $\sqrt{s (s - a) (s - b) (s - c)}$ = $\sqrt{8 (8 - 4) (8 - 6) (8 - 6)}$ = $\sqrt{1} 28$ = $8 \sqrt{2}$ ${c m}^{2}$

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Find the area of an isoceles triangle whose equal sides are 6cm each and unequal side is 4cm. [2 MARKS]

Concept : 0.5 Mark Application : 0.5 Mark Calculation : 1 Mark S = Perimeter2 = 4+6+62 = 8 cm Area = √s(s−a)(s−b)(s−c) = √8(8−4)(8−6)(8−6) = √128 = 8√2 cm2

Concept : 0.5 Mark
Application : 0.5 Mark
Calculation : 1 Mark

S = $\frac{P e r i m e t e r}{2}$ = $\frac{4 + 6 + 6}{2}$ = 8 cm

Area = $\sqrt{s (s - a) (s - b) (s - c)}$ = $\sqrt{8 (8 - 4) (8 - 6) (8 - 6)}$ = $\sqrt{1} 28$ = $8 \sqrt{2}$ ${c m}^{2}$