Advertisements
Advertisements
प्रश्न
A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final position of the boy.
Advertisements
उत्तर
Given: Direction of north = 5 m i.e. AC Direction of east = 12 m i.e. AB
To find: BC
According to Pythagoras Theorem,
In right angled Δ ABC
(BC)2 = (AC)2 + (AB)2
(BC)2 = (5)2 + (12)2
(BC)2 = 25 + 144
(BC)2= 169
∴ BC = `sqrt169=sqrt(13xx13)` = 13 m
APPEARS IN
संबंधित प्रश्न
P and Q are the mid-points of the sides CA and CB respectively of a ∆ABC, right angled at C. Prove that:
`(i) 4AQ^2 = 4AC^2 + BC^2`
`(ii) 4BP^2 = 4BC^2 + AC^2`
`(iii) (4AQ^2 + BP^2 ) = 5AB^2`
The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD . Prove that 2AB2 = 2AC2 + BC2.
Which of the following can be the sides of a right triangle?
2.5 cm, 6.5 cm, 6 cm
In the case of right-angled triangles, identify the right angles.
Which of the following can be the sides of a right triangle?
1.5 cm, 2 cm, 2.5 cm
In the case of right-angled triangles, identify the right angles.
In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.
A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.
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.
Find the Pythagorean triplet from among the following set of numbers.
4, 7, 8
Calculate the area of a right-angled triangle whose hypotenuse is 65cm and one side is 16cm.
In an equilateral triangle PQR, prove that PS2 = 3(QS)2.
