Advertisements
Advertisements
प्रश्न
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
Advertisements
उत्तर
Let the initial position of the man be “O” and his final position be “B”.
By Pythagoras theorem
In the right ∆OAB,
OB2 = OA2 + AB2
= 182 + 242
= 324 + 576 = 900
OB = `sqrt(900)` = 30
The distance of his current position is 30 m
APPEARS IN
संबंधित प्रश्न
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

In the given figure, ∠B = 90°, XY || BC, AB = 12 cm, AY = 8cm and AX : XB = 1 : 2 = AY : YC.
Find the lengths of AC and BC.

Diagonals of rhombus ABCD intersect each other at point O.
Prove that: OA2 + OC2 = 2AD2 - `"BD"^2/2`
Choose the correct alternative:
In right-angled triangle PQR, if hypotenuse PR = 12 and PQ = 6, then what is the measure of ∠P?
In the figure below, find the value of 'x'.

A man goes 10 m due east and then 24 m due north. Find the distance from the straight point.
In the given figure. PQ = PS, P =R = 90°. RS = 20 cm and QR = 21 cm. Find the length of PQ correct to two decimal places.
In a right-angled triangle ABC, if angle B = 90°, BC = 3 cm and AC = 5 cm, then the length of side AB is ______.
Two squares having same perimeter are congruent.
