Advertisements
Advertisements
प्रश्न
Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m;
find the distance between their tips.
Advertisements
उत्तर
The diagram of the given problem is given below,
We have Pythagoras theorem which states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Here, 11 - 6 = 5m ...( Since DC is perpendicular to BC )
base = 12 cm
Applying Pythagoras theorem we get,
hypotenuse2 = 52 + 122
h2 = 25 + 144
h2 = 169
h = 13
Therefore, the distance between the tips will be 13m.
APPEARS IN
संबंधित प्रश्न
In figure, ∠B of ∆ABC is an acute angle and AD ⊥ BC, prove that AC2 = AB2 + BC2 – 2BC × BD
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`
In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AD2 = BD × CD
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Tick the correct answer and justify: In ΔABC, AB = `6sqrt3` cm, AC = 12 cm and BC = 6 cm.
The angle B is:
In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4PM2 – 3PR2.
In ∆ABC, seg AD ⊥ seg BC, DB = 3CD.
Prove that: 2AB2 = 2AC2 + BC2
ABC is a triangle, right-angled at B. M is a point on BC.
Prove that: AM2 + BC2 = AC2 + BM2
In the following figure, OP, OQ, and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC.
Prove that: AR2 + BP2 + CQ2 = AQ2 + CP2 + BR2
In the given figure, angle ACB = 90° = angle ACD. If AB = 10 m, BC = 6 cm and AD = 17 cm, find :
(i) AC
(ii) CD
Find the length of the hypotenuse of a triangle whose other two sides are 24cm and 7cm.
A man goes 10 m due east and then 24 m due north. Find the distance from the straight point.
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 - AB2 = 2BC x ED
PQR is an isosceles triangle with PQ = PR = 10 cm and QR = 12 cm. Find the length of the perpendicular from P to QR.
Find the unknown side in the following triangles
In a right angled triangle, if length of hypotenuse is 25 cm and height is 7 cm, then what is the length of its base?
In ΔABC, if DE || BC, AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, then value of x is ______.
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.
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.
The foot of a ladder is 6 m away from its wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?
