Question
Find the area of maximum side of square that can be inscribed in a right angled triangle of side 15, 20 and 25
Find the area of maximum side of square that can be inscribed in a right angled triangle of side 15, 20 and 25
cm.
More Height and Distance Questions
- A Navy captain going away from a lighthouse at the speed of 4[(√3) – 1] m/s. He observes that it takes him 1 minute to change the angle of elev...
- From a point on the ground, the angle of elevation of the top of a tower is 30°. After moving 20 m closer to the tower in a straight line, the angle of ele...
- The distance between two parallel poles is 65√3 m. The angle of depression of the top of the second pole when seen from the top of first pole is...
- The angle of elevation of the top of a tower from a point on the ground is 30°. On moving 20 m closer to the tower, the angle of elevation becomes 45°. Fin...
- The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the bui...
- From a point on the ground, the angle of elevation of the top of a tower is 30 degrees. After moving 20 m towards the tower, the angle becomes 60 degrees. ...
- The angle of elevation of an aeroplane from a point on the ground is 60°. After 15 seconds flight the elevation changes to 30°, if the aeroplane is flying ...
- The angle of depression from the top of a light-house of two boats are 60° and 30° towards the west, if the two boats are 60m apart, then the height of the...
- The shadow of a pole 24m long is 56m. Then what will be the approximate shadow of building 40m high at the similar situation?
- From the top of a tower, the angle of depression of a car on the ground is 30°. After the car moves 40 m towards the tower in a straight line, the angle of...
Relevant for Exams:
Hey! Ask a query
Please enter email id
The email must be a valid email address.
Please enter Mobile Number
Please enter valid Mobile Number
Please enter your Doubt
Let the side of the square be a Then, AD = 15 – a and FC = 20 - a Area of Triangle ABC = 1/2 × Base × Height = 1/2 × 15 ×20 = 150 cm² Now, Area of Triangles ADE and EFC + Area of Square BDEF = Area of Triangle ABC ∴ 1/2 ×a × (15 - a) + 1/2 × a × (20 - a) + a² = 150 15a/2 - a²/2 + 10a - a²/2 + a² = 150 (15a+20a)/2 = 150 35a = 300 a = 300/35 = 60/7 cm Area of Square = 3600/49 cm²