Question

(ii) The area of the largest triangle inscribed in a semicircle of radius 18 cm is​

Answer 1

Answer:

324cm²

Step-by-step explanation:

BY THE ABOVE ATTACHMENT YOU CAN EASILY UNDERSTAND IT

Answer 2

Given :

• The area of the largest triangle inscribed in a semicircle of radius 18 cm

To find :

• Area of required triangle

Solution :

$\bold \blue{Let \: \triangle ABC \: be \: the \: largest \: triangle }\\ \\ \sf \red {Given \: radius \: of \: semi - circle = OA = OB = 18} \\ \\ \underline {\boxed{\sf \purple{Area \: of \: \triangle \: ABC = \frac{1}{2} \times BC \times AO}}} \\ \\ \implies \sf \pink{ \frac{1}{2} \times (OB + OC) \times AO } \\ \\ \implies \sf \green{ \frac{1}{2} \times (18 + 18) \times 18 } \\ \\ \implies \sf \orange{ \frac{1}{2} \times 36 \times 18 } \\ \\ \implies \underline{\boxed{\sf \blue{324 \: cm^2}}} \\ \\ \underline{\sf \red{\therefore \: Hence \: area \: of \: required \: triangle \: is \: 324\: cm^2}}$

Extra information :

• Volume of cylinder = πr²h
• T.S.A of cylinder = 2πrh + 2πr²
• Volume of cone = ⅓ πr²h
• C.S.A of cone = πrl
• T.S.A of cone = πrl + πr²
• Volume of cuboid = l × b × h
• C.S.A of cuboid = 2(l + b)h
• T.S.A of cuboid = 2(lb + bh + lh)