Given in the question, QR = 26 cm, PM = 6 cm, MR = 8 cm
According to question, in ? PMR
(Pythagoras Theorem)
PR = ?PM2 + MR2 = ?36 + 64 = ? 100 = 10 cm
According to question, in ? PQR
(Pythagoras Theorem)
PQ = ?QR2 - PR2 = ?262 - 102 = ?576 - 100 = ? 476 = 24
? area of triangle ?PQR = Base length x Height / 2
? area of triangle ?PQR = PR x PQ / 2
? area of triangle ?PQR = 10 x 24 / 2 = 10 x 12
? area of triangle ?PQR = 120
In ||gm ABCD , we have
Since diagonals of parallelogram(||gm ) bisect each other,
? M will be the mid-point of each of the diagonal AC and BD
? In ?ABC AB2 + BC2 = 2(AM2 + MB2) ......... ( 1 ) [Appolonius Theorem]
In ?ADC AD2 + CD2 = 2(AM2 + DM2) ......... ( 2 )
= 2(AM2 + MB2) [? DM = BM]
Adding equations ( 1 ) and ( 2 )
AB2 + BC2 + CD2 + DA2 = 2(AM2 + MB2 + 2(AM2 + MB2 = 4AM2 + 4MB2
= (2AM)2 + (2MB)2= AC2+ BD2. { ? AM = MC , MB = MD }
Then | ar(?DFE) |
ar(?CFB) |
We know that In center of every triangle ( acute , obtuse and right angle ) lies in its interior.
Hence required answer will be any triangle .
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