or If AD and PM are the medians of triangles △ABC and △PQR, respectively where △ABC∼△PQR, prove that PQAB=PMAD. 10. Find the unknown entries a,b,c and d in the following distribution of heights of students in the class. Height (in cm) fcf150−155PM012155−16013PM1160−1651035165−170PM243170−1757dPM4 Solve the following system of equations by elimination method: PM5 and PM6 Page 2 of Set 2
Q. or If AD and PM are the medians of triangles △ABC and △PQR, respectively where △ABC∼△PQR, prove that PQAB=PMAD. 10. Find the unknown entries a,b,c and d in the following distribution of heights of students in the class. Height (in cm) fcf150−155PM012155−16013PM1160−1651035165−170PM243170−1757dPM4 Solve the following system of equations by elimination method: PM5 and PM6 Page 2 of Set 2
Given Triangles ABC and PQR: Given: Triangles ABC and PQR are similar, so their corresponding sides are proportional.
Proportional Sides: Since ABC ∼ PQR, we have:PQAB=QRBC=RPCA
Medians Proportional: Medians of similar triangles are also proportional to the corresponding sides. So:PMAD=PQAB
Height Interval 150−155 cm: Therefore, we have:PQAB=PMAD
Height Interval 155−160 cm: Given cumulative frequency (cf) for the height interval 150−155 cm is 12. So, a=12.
Align Coefficients: Given cumulative frequency (cf) for the height interval 155−160 cm is b. Since the cumulative frequency for the next interval (160−165 cm) is 35, we have:b=35−13=22
Subtract Equations: Multiply the first equation by 2 to align coefficients:2(0.2x+0.3y)=2(1.3)0.4x+0.6y=2.6
Solve for y: Subtract the second equation from the modified first equation:(0.4x+0.6y)−(0.4x+0.5y)=2.6−2.30.1y=0.3
Substitute y into Equation: Solve for y:y=0.10.3=3
Substitute y into Equation: Solve for y:y=0.10.3=3 Substitute y = 3 into the first equation:0.2x+0.3(3)=1.30.2x+0.9=1.30.2x=0.4x=0.20.4=2
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