Advertisements
Advertisements
प्रश्न
The top of a broken tree touches the ground at a distance of 12 m from its base. If the tree is broken at a height of 5 m from the ground then the actual height of the tree is ______.
पर्याय
25 m
13 m
18 m
17 m
Advertisements
उत्तर
The top of a broken tree touches the ground at a distance of 12 m from its base. If the tree is broken at a height of 5 m from the ground then the actual height of the tree is 18 m.
Explanation:
Let BC is the broken part of tree and AB is the unbroken part of tree.
Here, ΔABC is right angled triangle.
∴ (BC)2 = (AB)2 + (AC)2
⇒ (BC)2 = (5)2 + (12)2
⇒ (BC)2 = 25 + 144 = 169
⇒ (BC)2 = 132
⇒ BC = 13 m
∴ Actual height of tree is AB + BC = (5 + 13) m = 18 m.
APPEARS IN
संबंधित प्रश्न
In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AB2 = BC × BD
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals
Find the side and perimeter of a square whose diagonal is 10 cm.
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.
In the figure below, find the value of 'x'.
The sides of the triangle are given below. Find out which one is the right-angled triangle?
1.5, 1.6, 1.7
A right triangle has hypotenuse p cm and one side q cm. If p - q = 1, find the length of third side of the triangle.
From a point O in the interior of aΔABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that: AF2 + BD2 + CE2 = OA2 + OB2 + OC2 - OD2 - OE2 - OF2
A point OI in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that OB2 + OD2 = OC2 + OA2
