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Suppose the circle meets QR at A, RP at B and PQ at C. PQ = PR (given) so PC + CQ = PB + BR. But PC and PB are tangents to the circle from point P, so PC = PB. Therefore CQ = BR Now CQ and AQ are tangents to the circle from point Q, so CQ = AQ and BR and AR are tangents to the circle from point R, so BR = AR Therefore AQ = AR, that is, A is the midpoint of QR.
Draw a right angled triangle OPQ with OP the base, PQ the altitude and OQ the hypotenuse. Draw a second right angled triangle OQR with OQ the base, QR the altitude and OR the hypotenuse. Drop a perpendicular from R to intercept OP at T. The line RT crosses OQ at U. Draw a perpendicular from Q to intercept RT at S. Let angle POQ be A and angle QOR be B. Angle OUT = angle QUR therefore angle URQ = A sin (A + B) = TR/OR = (TS + SR)/OR = (PQ + SR)/OR = (PQ/OQ x OQ/OR) + (SR/QR x QR/OR) = sin A cos B + cos A sin B.
Bar code stores numbers only,but QR code can be store be with any multimedia.
QPR is congruent to SPR PR is perpendicular to QPS PQ =~ QR PT =~ RT
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Given the points P, Q and R for a triangle, the inequalities are:PQ + QR > RPQR + RP > PQ andRP + PQ > QR(the sum of two sides of a triangle is greater that the third).If P = (xp, yp) and Q = (xq, yq) then PQ = sqrt{(xq - xp)^2 + (yq - yp)^2}
i have the same question...
Suppose the circle meets QR at A, RP at B and PQ at C. PQ = PR (given) so PC + CQ = PB + BR. But PC and PB are tangents to the circle from point P, so PC = PB. Therefore CQ = BR Now CQ and AQ are tangents to the circle from point Q, so CQ = AQ and BR and AR are tangents to the circle from point R, so BR = AR Therefore AQ = AR, that is, A is the midpoint of QR.
Draw a right angled triangle OPQ with OP the base, PQ the altitude and OQ the hypotenuse. Draw a second right angled triangle OQR with OQ the base, QR the altitude and OR the hypotenuse. Drop a perpendicular from R to intercept OP at T. The line RT crosses OQ at U. Draw a perpendicular from Q to intercept RT at S. Let angle POQ be A and angle QOR be B. Angle OUT = angle QUR therefore angle URQ = A sin (A + B) = TR/OR = (TS + SR)/OR = (PQ + SR)/OR = (PQ/OQ x OQ/OR) + (SR/QR x QR/OR) = sin A cos B + cos A sin B.
PR = 110 because... P = 10, Q = 2... therefore R = 11
The regular configuration of a QR barcode is 3 squares in the corners of the code and pixelated squares in between them. One could also make custom QR barcodes.
Here is the answer to your query. Consider two ∆ABC and ∆PQR. In these two triangles ∠B = ∠Q = 90�, AB = PQ and AC = PR. We can prove the R.H.S congruence rule i.e. to prove ∆ABC ≅ ∆PQR We need the help of SSS congruence rule. We have AB = PQ, and AC = PR So, to prove ∆ABC ≅ ∆PQR in SSS congruence rule we just need to show BC = QR Now, using Pythagoras theorems in ∆ABC and ∆PQR Now, in ∆ABC and ∆PQR AB = PQ, BC = QR, AC = PR ∴ ∆ABC ≅ ∆PQR [Using SSS congruence rule] So, we have AB = PQ, AC = PR, ∠B = ∠Q = 90� and we have proved ∆ABC ≅ ∆PQR. This is proof of R.H.S. congruence rule. Hope! This will help you. Cheers!!!
Bar code stores numbers only,but QR code can be store be with any multimedia.
A QR code (QR = Quick Response) is a two-dimensional barcode.
This isn't a question, even. It's a statement about a line segment, QR.