Question

Find the area of the triangle with sides 21 cm, 16 cm and 13 cm. Also, find the perimeter of an equilateral
triangle equal in area to this triangle.​

Answer 1

Explanation:

$AnswEr :</p><p></p><p>☯ Given sides of the equilateral ∆ are 21 cm, 16 cm & 13 cm.</p><p></p><p>{\dag}\:\underline{\frak{Using \ Heron's \ Formula \: \: :}}†Using Heron′s Formula:</p><p></p><p>\star\: \small\boxed{\sf{\purple{Area \: of \: \triangle = \sqrt{s(s - a) (s - b) (s - c)}}}}⋆Areaof△=s(s−a)(s−b)(s−c)</p><p></p><p>⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>\begin{gathered}:\implies\sf S_{(semi perimeter)} = \dfrac{a + b + c}{2} \\\\\\:\implies\sf s = \dfrac{21 + 16 + 13}{12} \\\\\\:\implies\sf s = \cancel\dfrac{50}{2} \\\\\\:\implies\boxed{\sf{\purple{ s = 25}}}\end{gathered}:⟹S(semiperimeter)=2a+b+c:⟹s=1221+16+13:⟹s=250:⟹s=25</p><p></p><p>⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>\begin{gathered}:\implies\sf \sqrt{25(25 - 21) (25 - 16) (25 - 13)}\\\\\\:\implies\sf\sqrt{25 \times 4 \times 9 \times 12} \\\\\\:\implies\sf\sqrt{ 5 \times 5 \times 2 \times 2 \times 3 \times 3 \times 2 \times 3 \times 2} \\\\\\:\implies\sf 60 \sqrt{3}\end{gathered}:⟹25(25−21)(25−16)(25−13):⟹25×4×9×12:⟹5×5×2×2×3×3×2×3×2:⟹603</p><p></p><p>⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>\therefore\: \underline{\sf{Here\: we \: get \: area \: is \: \bf{60 \sqrt{3}}}}∴Herewegetareais603</p><p></p><p>⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>━━━━━━━━━━━━━━━━━</p><p></p><p>☯ Now, finding area of the equilateral ∆ :</p><p></p><p>\star\: \small\boxed{\sf{\pink{ Area \: of \: equilateral \: \triangle = \dfrac{\sqrt{3}}{4} (a)^2}}}⋆Areaofequilateral△=43(a)2</p><p></p><p>\begin{gathered}:\implies\sf 60 \sqrt{3} = \dfrac{\sqrt{3}}{4}(a)^2 \qquad \quad \Bigg[ Equating \: both \: areas \Bigg] \\\\\\:\implies\sf 60 \cancel{\sqrt{3}} = \dfrac{\cancel{\sqrt{3}}}{4}(a)^2 \\\\\\:\implies\sf 60 = \dfrac{(a^2)}{4} \\\\\\:\implies\sf a^2 = 4 \times 60 \\\\\\:\implies\sf a = \sqrt{240} \\\\\\:\implies\boxed{\sf{\pink{a = 4\sqrt{15}}}}\end{gathered}:⟹603=43(a)2[Equatingbothareas]:⟹603=43(a)2:⟹60=4(a2):⟹a2=4×60:⟹a=240:⟹a=415</p><p></p><p>⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>━━━━━━━━━━━━━━━━━</p><p></p><p>⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>☯ Perimeter of equilateral ∆ is equal to the Area of the ∆ :</p><p></p><p>\star\: \small\boxed{\sf{\purple{Perimeter \: of \: \triangle = 3 \times Area }}}⋆Perimeterof△=3×Area</p><p></p><p>\begin{gathered}:\implies\sf Area = 3 \times 4 \sqrt{15} \qquad \quad \Bigg[ Area = 4 \sqrt{15}\Bigg] \\\\\\\:\implies \boxed{\sf{\purple{ Area = 12\sqrt{15}}}}\end{gathered}:⟹Area=3×415[Area=415]⟹Area=1215</p><p></p><p>\therefore\: \underline{\sf{Perimeter \: of \: equilateral \: \triangle \: is \: \bf{12\sqrt{15}}}}∴Perimeterofequilateral△is1215</p><p></p><p>AnswEr :</p><p></p><p>☯ Given sides of the equilateral ∆ are 21 cm, 16 cm & 13 cm.</p><p></p><p>{\dag}\:\underline{\frak{Using \ Heron's \ Formula \: \: :}}† </p><p>Using Heron </p><p>′</p><p> s Formula:</p><p> </p><p> </p><p></p><p>\star\: \small\boxed{\sf{\purple{Area \: of \: \triangle = \sqrt{s(s - a) (s - b) (s - c)}}}}⋆ </p><p>Areaof△= </p><p>s(s−a)(s−b)(s−c)</p><p> </p><p> </p><p> </p><p> </p><p></p><p>⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>\begin{gathered}:\implies\sf S_{(semi perimeter)} = \dfrac{a + b + c}{2} \\\\\\:\implies\sf s = \dfrac{21 + 16 + 13}{12} \\\\\\:\implies\sf s = \cancel\dfrac{50}{2} \\\\\\:\implies\boxed{\sf{\purple{ s = 25}}}\end{gathered} </p><p>:⟹S </p><p>(semiperimeter)</p><p> </p><p> = </p><p>2</p><p>a+b+c</p><p> </p><p> </p><p>:⟹s= </p><p>12</p><p>21+16+13</p><p> </p><p> </p><p>:⟹s= </p><p>2</p><p>50</p><p> </p><p> </p><p> </p><p> </p><p>:⟹ </p><p>s=25</p><p> </p><p> </p><p> </p><p> </p><p></p><p>⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>\begin{gathered}:\implies\sf \sqrt{25(25 - 21) (25 - 16) (25 - 13)}\\\\\\:\implies\sf\sqrt{25 \times 4 \times 9 \times 12} \\\\\\:\implies\sf\sqrt{ 5 \times 5 \times 2 \times 2 \times 3 \times 3 \times 2 \times 3 \times 2} \\\\\\:\implies\sf 60 \sqrt{3}\end{gathered} </p><p>:⟹ </p><p>25(25−21)(25−16)(25−13)</p><p> </p><p> </p><p>:⟹ </p><p>25×4×9×12</p><p> </p><p> </p><p>:⟹ </p><p>5×5×2×2×3×3×2×3×2</p><p> </p><p> </p><p>:⟹60 </p><p>3</p><p> </p><p> </p><p> </p><p> </p><p></p><p>⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>\therefore\: \underline{\sf{Here\: we \: get \: area \: is \: \bf{60 \sqrt{3}}}}∴ </p><p>Herewegetareais60 </p><p>3</p><p> </p><p> </p><p> </p><p> </p><p></p><p>⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>━━━━━━━━━━━━━━━━━</p><p></p><p>☯ Now, finding area of the equilateral ∆ :</p><p></p><p>\star\: \small\boxed{\sf{\pink{ Area \: of \: equilateral \: \triangle = \dfrac{\sqrt{3}}{4} (a)^2}}}⋆ </p><p>Areaofequilateral△= </p><p>4</p><p>3</p><p> </p><p> </p><p> </p><p> (a) </p><p>2</p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies\sf 60 \sqrt{3} = \dfrac{\sqrt{3}}{4}(a)^2 \qquad \quad \Bigg[ Equating \: both \: areas \Bigg] \\\\\\:\implies\sf 60 \cancel{\sqrt{3}} = \dfrac{\cancel{\sqrt{3}}}{4}(a)^2 \\\\\\:\implies\sf 60 = \dfrac{(a^2)}{4} \\\\\\:\implies\sf a^2 = 4 \times 60 \\\\\\:\implies\sf a = \sqrt{240} \\\\\\:\implies\boxed{\sf{\pink{a = 4\sqrt{15}}}}\end{gathered} </p><p>:⟹60 </p><p>3</p><p> </p><p> = </p><p>4</p><p>3</p><p> </p><p> </p><p> </p><p> (a) </p><p>2</p><p> [Equatingbothareas]</p><p>:⟹60 </p><p>3</p><p> </p><p> </p><p> </p><p> = </p><p>4</p><p>3</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> (a) </p><p>2</p><p> </p><p>:⟹60= </p><p>4</p><p>(a </p><p>2</p><p> )</p><p> </p><p> </p><p>:⟹a </p><p>2</p><p> =4×60</p><p>:⟹a= </p><p>240</p><p> </p><p> </p><p>:⟹ </p><p>a=4 </p><p>15</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p>⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>━━━━━━━━━━━━━━━━━</p><p></p><p>⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀</p><p></p><p>☯ Perimeter of equilateral ∆ is equal to the Area of the ∆ :</p><p></p><p>\star\: \small\boxed{\sf{\purple{Perimeter \: of \: \triangle = 3 \times Area }}}⋆ </p><p>Perimeterof△=3×Area</p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies\sf Area = 3 \times 4 \sqrt{15} \qquad \quad \Bigg[ Area = 4 \sqrt{15}\Bigg] \\\\\\\:\implies \boxed{\sf{\purple{ Area = 12\sqrt{15}}}}\end{gathered} </p><p>:⟹Area=3×4 </p><p>15</p><p> </p><p> [Area=4 </p><p>15</p><p> </p><p> ]</p><p>⟹ </p><p>Area=12 </p><p>15</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p>\therefore\: \underline{\sf{Perimeter \: of \: equilateral \: \triangle \: is \: \bf{12\sqrt{15}}}}∴ </p><p>Perimeterofequilateral△is12 </p><p>15</p><p> </p><p> </p><p>$