Sorry, but data not adequate. However, I'll try. Suppose the triangle is right angled at R. Then PQ = hypotenuse = 8. Further, if the triangle is isosceles, then RQ = 4 times sqrt(2).
PQ ST
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!!!
Since 36=24+1/2.24 and abc and pqr are similar, ab= 10+1/2.10=15.
A triangle has 3 line segments
Is PQ |_ RS
PQ ST
m = pqr/s Multiply both sides by s: ms = pqr Divide both sides by pq: ms/pq = r
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!!!
Since 36=24+1/2.24 and abc and pqr are similar, ab= 10+1/2.10=15.
Since the sides of triangle are equal, the triangles are equilateral. Just for your information, in this question, we do not require the length of sides. It is just additional information. :) The area of equilateral triangle is: (√3)/4 × a², where a is the side of the equilateral triangle. For triangle ABC, area will be = (√3)/4 × a² (Let 'a' is the side of triangle ABC) Since, side of triangle PQR is half that of ABC, it will be = a/2 Therefore, area of triangle PQR = (√3)/4 × (a/2)² = (√3)/16 × a² Take the ratio of areas of triangle ABC and PQR: [(√3)/4 × a²] / [(√3)/16 × a²] = 4:1
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.
A triangle has 3 line segments
Not sure the exact length but the connection is 3oq=2ob , 3/2oq=ob
z=pq
Is PQ |_ RS
In triangle ABC, let P and Q be the midpoints of sides AB and AC, respectively. By the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. Therefore, since PQ connects the midpoints P and Q, it follows that PQ is parallel to side BC of triangle ABC. This establishes that PQ is parallel to BC, as required.
To find the length of PR, you can use the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. In this case, PR must be less than the sum of PQ and QR, so PR < 20 + 22 = 42. Therefore, PR could be any value less than 42.