Solve this if you can..
Qno.7
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PQR is right angled at P
area of PQR = 1/2 × base × height
= 1/2 × PQ× PR
=1/2 × 5 × 12
= 30 cm^2
Now use PR as base
height = PS
Area = 1/2 × PR × PS
30 = 1/2 × 13 × PS
PS = 30×2/13
PS = 60/13 cm
Hence area is 30cm^2 and PS is 60/13 cm
area of PQR = 1/2 × base × height
= 1/2 × PQ× PR
=1/2 × 5 × 12
= 30 cm^2
Now use PR as base
height = PS
Area = 1/2 × PR × PS
30 = 1/2 × 13 × PS
PS = 30×2/13
PS = 60/13 cm
Hence area is 30cm^2 and PS is 60/13 cm
Answered by
1
PQ = 5 cm
PR = 12 cm
QR = 13 cm
Let s be the semiperimeter (Perimeter/2) of ∆PQR.
Therefore, s = (5+12+13)/2 = 30/2 = 15
Applying Heron's formula,
√{s(s-PQ)(s-PR)(s-QR)}
=√{15(15-5)(15-12)(15-13)}
=√( 15 • 10 • 3 • 2 )
=√900
=30 cm²
Therefore the required area is 30 cm².
Now,
We know,
Area of ∆ = 1/2 • base • height
or, 30 = 1/2 • 13 • PS
or, PS • 13/2 = 30
or, PS = 30 • 2/13 = 60/13 = 4(whole) 8/13 cm. Ans
PR = 12 cm
QR = 13 cm
Let s be the semiperimeter (Perimeter/2) of ∆PQR.
Therefore, s = (5+12+13)/2 = 30/2 = 15
Applying Heron's formula,
√{s(s-PQ)(s-PR)(s-QR)}
=√{15(15-5)(15-12)(15-13)}
=√( 15 • 10 • 3 • 2 )
=√900
=30 cm²
Therefore the required area is 30 cm².
Now,
We know,
Area of ∆ = 1/2 • base • height
or, 30 = 1/2 • 13 • PS
or, PS • 13/2 = 30
or, PS = 30 • 2/13 = 60/13 = 4(whole) 8/13 cm. Ans
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