A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time a tower casts the shadow 40 m long on the ground. Find the height of the tower.

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

According to question,
?B = ?C = 55° , ?D = 25°
We can say ,
AB = AC ( ? ?B = ?C = 55° )
In Triangle ABC,
?A + ?B + ?C = 180°
? ?A + 55° + 55° = 180°
? ?A + 110° = 180°
? ?A = 180° - 110°
? ?A = 70° ..........................(1)
As per given figure,
?ACD + ?ACB = 180° ( ?ACB = ?C = 55°)
? ?ACD + 55° = 180°
? ?ACD = 180° - 55°
? ?ACD = 125° ....................... (2)
Now in Triangle ACD,
?CAD + ?ACD + ?CDA = 180°
? ?CAD + 125° + 25° = 180°
? ?CAD + 150° = 180°
? ?CAD = 30° ...........................(3)
( In a ?, greater angle has longer side opposite to it )
From the equation (1) , (2) and (3);
?B < ?A and ?CAD > ?D ;
? BC > CA and CA < CD

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 }