find tha base of this question
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area of isosceles Δ =( a/4)(√4b²-a²)
base-a
side-b
area of the Δ given=12cm²
⇒ ( a/4)(√4b²-a²)=12
(a/4)(√4(25)-a²)=12
(a/4)(√100-a²)=12
a(√100-a²)=48
(√100-a²)=48/a
squaring both sides
100-a²=2304/a
-a²=(2304/a²)-100
a²=100-(2304/a²)
a⁴=2304-100a²
a⁴-100a²=2304
by solving this equation, you will get your answer...............
base-a
side-b
area of the Δ given=12cm²
⇒ ( a/4)(√4b²-a²)=12
(a/4)(√4(25)-a²)=12
(a/4)(√100-a²)=12
a(√100-a²)=48
(√100-a²)=48/a
squaring both sides
100-a²=2304/a
-a²=(2304/a²)-100
a²=100-(2304/a²)
a⁴=2304-100a²
a⁴-100a²=2304
by solving this equation, you will get your answer...............
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