We havemlength of each side of ∆ABC = a = 10 cmArea of ∆ABC = 34a2 = 34×102 = 253 cm2 = 43 30 cm2Consider the ∆BDC,Let a = 7 cm; b = 5 cm; c = 10 cmNow, semi - perimeter, s = a+b+c2 = 7 + 5 + 102 = 11 cmNow, area of ∆BDC = ss-as-bs-c =1111-711-511-10=11×4×6×1=266 cm2=16 25 cm2Area of shaded region = area of ∆ABC - area of ∆BDC=43 30 - 16 25=27 05 cm2

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We havem length of each side of ∆ ABC = a = 10 cm Area of ∆ ABC = \frac{\sqrt{3}}{4} a^{2} = \frac{\sqrt{3}}{4} \times 10^{2} = 25 \sqrt{3} {cm}^{2} = 43.30 {cm}^{2} Consider the ∆ BDC, Let a = 7 cm; b = 5 cm; c = 10 cm Now, semi - perimeter, s = \frac{a + b + c}{2} = \frac{7 + 5 + 10}{2} = 11 cm Now, area of ∆ BDC = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{11 (11 - 7) (11 - 5) (11 - 10)} = \sqrt{11 \times 4 \times 6 \times 1} = 2 \sqrt{66} {cm}^{2} = 16.25 {cm}^{2} Area of shaded region = area of ∆ ABC - area of ∆ BDC = 43.30 - 16.25 = 27.05 {cm}^{2}

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