plz help me in no 3 and 4 .
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preeti5390:
No 4 aditi
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Q.3. solution:
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let the base of parallelogram be 'x' cm
altitude = 1/ 3 rd of base = x / 3 cm
area of parallelogram = 243 cm^2
base × altitude = 243
x × ( x/ 3) = 243
x^2 = 729 => x = √729 => x = 27 cm
Altitude = x/ 3 = 27/ 3 = 9 cm
Answer: altitude = 9 cm
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Q.4 solution:
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let two equal side be ' a, a'
let a = AB = AC = 8 cm ,
and base b = BC = 10 cm
area of isosceles triangle
= [ b ×√( 4a^2 - b^2) ] / 4
= [ 10×√(4 (8)^2 - (10)^2)]/ 4
= 10 × √156 / 4 = (5 × 2√39 )/ 2
= 5√39 cm = 5 × 6.24 = 31.2 cm^2
Answer:
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area of isosceles triangle= 31.2 cm^2
-------------------
let the base of parallelogram be 'x' cm
altitude = 1/ 3 rd of base = x / 3 cm
area of parallelogram = 243 cm^2
base × altitude = 243
x × ( x/ 3) = 243
x^2 = 729 => x = √729 => x = 27 cm
Altitude = x/ 3 = 27/ 3 = 9 cm
Answer: altitude = 9 cm
---------------------------------
Q.4 solution:
-------------------
let two equal side be ' a, a'
let a = AB = AC = 8 cm ,
and base b = BC = 10 cm
area of isosceles triangle
= [ b ×√( 4a^2 - b^2) ] / 4
= [ 10×√(4 (8)^2 - (10)^2)]/ 4
= 10 × √156 / 4 = (5 × 2√39 )/ 2
= 5√39 cm = 5 × 6.24 = 31.2 cm^2
Answer:
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area of isosceles triangle= 31.2 cm^2
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