Advertisements
Advertisements
प्रश्न
The foot of a ladder is 6 m away from its wall and its top reaches a window 8 m above the ground. Find the length of the ladder.
Advertisements
उत्तर
Let the length of the ladder be x m.
In right angled ΔABC,
AC2 = AB2 + BC2 ...[By Pythagoras theorem]
⇒ (x)2 = (8)2 + (6)2
⇒ `sqrt((8)^2 + (6)^2)` = `sqrt(64 + 36)` = `sqrt(100)`
⇒ x = 10 m
Hence, the length of the ladder is 10 m.
APPEARS IN
संबंधित प्रश्न
In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
`"AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2`
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Diagonals of rhombus ABCD intersect each other at point O.
Prove that: OA2 + OC2 = 2AD2 - `"BD"^2/2`
In the following Figure ∠ACB= 90° and CD ⊥ AB, prove that CD2 = BD × AD
Triangle ABC is right-angled at vertex A. Calculate the length of BC, if AB = 18 cm and AC = 24 cm.
In the given figure, angle ADB = 90°, AC = AB = 26 cm and BD = DC. If the length of AD = 24 cm; find the length of BC.
In the figure below, find the value of 'x'.
In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i) `"AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2`
(ii) `"AB"^2 = "AD"^2 - "BC"."DM" + (("BC")/2)^2`
(iii) `"AC"^2 + "AB"^2 = 2"AD"^2 + 1/2"BC"^2`
Two squares having same perimeter are congruent.
Points A and B are on the opposite edges of a pond as shown in the following figure. To find the distance between the two points, the surveyor makes a right-angled triangle as shown. Find the distance AB.
