Advertisements
Advertisements
प्रश्न
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
उत्तर
Let OA be the wall and AB be the ladder.
Therefore, by Pythagoras theorem,
AB2 = OA + BO
(10 m)2 = (8m)2 + OB2
100 m2 = 64 m2 + OB2
OB2 = 36 m2
OB = 6 m
Therefore the distance of the foot of the ladder from the base of the wall is 6 m
APPEARS IN
संबंधित प्रश्न
In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC = 3 cm. Calculate the length of OC.
Diagonals of rhombus ABCD intersect each other at point O.
Prove that: OA2 + OC2 = 2AD2 - `"BD"^2/2`
Find the side of the square whose diagonal is `16sqrt(2)` cm.
AD is perpendicular to the side BC of an equilateral ΔABC. Prove that 4AD2 = 3AB2.
In the given figure, PQ = `"RS"/(3)` = 8cm, 3ST = 4QT = 48cm.
SHow that ∠RTP = 90°.
In a square PQRS of side 5 cm, A, B, C and D are points on sides PQ, QR, RS and SP respectively such as PA = PD = RB = RC = 2 cm. Prove that ABCD is a rectangle. Also, find the area and perimeter of the rectangle.
If ‘l‘ and ‘m’ are the legs and ‘n’ is the hypotenuse of a right angled triangle then, l2 = ________
Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.