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प्रश्न
Find the side of a square whose diagonal is `10sqrt2` cm.
उत्तर
Let `square`ABCD be the square and AC be the diagonal of length `10sqrt2` cm.
Let the side of the square be x.
In ΔABC, ∠B = 90°
By Pythagoras Theorem,
AC2 = AB2 + BC2
∴ `(10sqrt2)^2 = x^2 + x^2`
∴ `2x^2 = 100 xx 2`
∴ `2x^2 = 200`
∴ `x^2 = 100`
Taking square root on both sides
∴ x = 10 cm
Hence, the side of the square = 10 cm
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संबंधित प्रश्न
In the above figure `square`ABCD is a rectangle. If AB = 5, AC = 13, then complete the following activity to find BC.
Activity: ΔABC is a `square` triangle.
∴ By Pythagoras theorem
AB2 + BC2 = AC2
∴ 25 + BC2 = `square`
∴ BC2 = `square`
∴ BC = `square`
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If tan θ = `12/5`, then 5 sin θ – 12 cos θ = ?
From the information in the figure, complete the following activity to find the length of the hypotenuse AC.
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∴ ∠BAC = `square`
Side opposite angle 45° = `square/square` × Hypotenuse
∴ `5sqrt(2) = 1/square` × AC
∴ AC = `5sqrt(2) xx square = square`
AB, BC and AC are three sides of a right-angled triangle having lengths 6 cm, 8 cm and 10 cm, respectively. To verify the Pythagoras theorem for this triangle, fill in the boxes:
ΔABC is a right-angled triangle and ∠ABC = 90°.
So, by the Pythagoras theorem,
`square` + `square` = `square`
Substituting 6 cm for AB and 8 cm for BC in L.H.S.
`square` + `square` = `square` + `square`
= `square` + `square`
= `square`
Substituting 10 cm for AC in R.H.S.
`square` = `square`
= `square`
Since, L.H.S. = R.H.S.
Hence, the Pythagoras theorem is verified.
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