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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.
i have the same question...
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
No
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
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!!!
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.
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.