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प्रश्न
∆ABC is an equilateral triangle. Point P is on base BC such that PC = `1/3`BC, if AB = 6 cm find AP.
उत्तर
∆ABC is an equilateral triangle.
It is given that,
PC = `1/3`BC
PC = `1/3 xx 6`
PC = 2 cm
∴ BP = BC − PC
= 6 − 2
= 4 cm
Since, ∆ABC is an equilateral triangle, OA is the perpendicular bisector of BC.
OC = `"BC"/2`
= `6/2`
= 3 cm
∴ OP = OC − PC
= 3 − 2
= 1 cm ...(1)
According to Pythagoras theorem,
In ∆AOB,
AB2 = AO2 + OC2
∴ (6)2 = AO2 + (3)2
∴ 36 = AO2 + 9
∴ AO2 = 36 − 9
∴ AO2 = 27
∴ AO = `sqrt27`
∴ AO = `3sqrt3` cm ...(2)
In ∆AOP,
AP2 = AO2 + OP2
∴ AP2 = `(3sqrt3)^2 + (1)^2 `
∴ AP2 = 27 + 1
∴ AP2 = 28
∴ AP = `sqrt28`
∴ AP = `2sqrt7` cm
संबंधित प्रश्न
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of the squares of the sides making right angle is 169.
(A)15 (B) 13 (C) 5 (D) 12
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In right angled triangle PQR,
if ∠ Q = 90°, PR = 5,
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Choose the correct alternative:
ΔABC and ΔDEF are equilateral triangles. If ar(ΔABC): ar(ΔDEF) = 1 : 2 and AB = 4, then what is the length of DE?
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AC2 = AD2 + `square`
∴ AD2 = AC2 – CD2 ......(I)
Also, In ∆ABD, by pythagoras theorem,
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By Pythagoras Theorem,
PQ2 + `square` = PR2 ......(I)
∴ PR2 = 92 + 122
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From given figure, in ∆PQR, if ∠QPR = 90°, PM ⊥ QR, PM = 10, QM = 8, then for finding the value of QR, complete the following activity.
Activity: In ∆PQR, if ∠QPR = 90°, PM ⊥ QR, ......[Given]
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∴ PM2 + `square` = PQ2 ......(I)
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∴ PQ2 = `square` + 64
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Here, ∆QPR ~ ∆QMP ~ ∆PMR
∴ ∆QMP ~ ∆PMR
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∴ PM2 = RM × QM
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