plz solve question no. 14
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Answer:
Step-by-step explanation:
IN semicircle QRP
area = 1/2 ×22/7×14×14
=308 cm²
area of TRI. QPR = 1/2 × (14+14) ×14
=56cm²
Now , area of smaller circle = 22/7 × 7 ×7
= 154 cm²
now area of shaded region = 154 + (308-56)
= 154 + 252
= 406 cm²
ash1719l:
area of triangle would be 196 m
sorry
koi nahi, hota hain
Answer:
Step-by-step explanation:
IN semicircle QRP
area = 1/2 ×22/7×14×14
=308 cm²
area of TRI. QPR = 1/2 × (14+14) ×14
=56cm²
Now , area of smaller circle = 22/7 × 7 ×7
= 154 cm²
now area of shaded region = 154 + (308-56)
= 154 + 252
= 406 cm²
Step-by-step explanation:
IN semicircle QRP
area = 1/2 ×22/7×14×14
=308 cm²
area of TRI. QPR = 1/2 × (14+14) ×14
=56cm²
Now , area of smaller circle = 22/7 × 7 ×7
= 154 cm²
now area of shaded region = 154 + (308-56)
= 154 + 252
= 406 cm²
Area of shaded region= 154+(308-196)
i.e 154+112
= 266 cm sp.
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