Advertisements
Advertisements
प्रश्न
Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, the distance between their tops is
पर्याय
12 m
14 m
13 m
11 m
उत्तर
Given: Two poles of heights 6m and 11m stand vertically upright on a plane ground. Distance between their foot is 12 m.
To find: Distance between their tops.
Let CD be the pole with height 6m.
AB is the pole with height 11m, distance between their foot i.e. DB is 12 m.
Let us assume a point E on the pole AB which is 6m from the base of AB.
Hence
AE = AB − 6 = 11 − 6 = 5 m
Now in right triangle AEC, Applying Pythagoras theorem
AC2 = AE2 + EC2
AC2 = 52 + 122 (since CDEB forms a rectangle and opposite sides of rectangle are equal)
AC2 = 25 + 144
AC2 = 169
`AC= 13cm`
Thus, the distance between their tops is 13m.
Hence correct answer is `C`.
APPEARS IN
संबंधित प्रश्न
In each of the following figures, you find who triangles. Indicate whether the triangles are similar. Give reasons in support of your answer.
Corresponding sides of two triangles are in the ratio 2 : 3. If the area of the smaller triangle is 48 cm2, determine the area of the larger triangle.
A point D is on the side BC of an equilateral triangle ABC such that\[DC = \frac{1}{4}BC\]. Prove that AD2 = 13 CD2.
In ∆ABC, given that AB = AC and BD ⊥ AC. Prove that BC2 = 2 AC. CD
If ABC and DEF are similar triangles such that ∠A = 47° and ∠E = 83°, then ∠C =
In an equilateral triangle ABC if AD ⊥ BC, then AD2 =
In a ∆ABC, perpendicular AD from A and BC meets BC at D. If BD = 8 cm, DC = 2 cm and AD = 4 cm, then
In a ∆ABC, point D is on side AB and point E is on side AC, such that BCED is a trapezium. If DE : BC = 3 : 5, then Area (∆ ADE) : Area (◻BCED) =
In an isosceles triangle ABC if AC = BC and AB2 = 2AC2, then ∠C =
D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, length of DE (in cm) is ______.
