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 ______.
विकल्प
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
ABCD is a rectangle whose three vertices are B (4, 0), C(4, 3) and D(0,3). The length of one of its diagonals is
(A) 5
(B) 4
(C) 3
(D) 25
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals
PQR is a triangle right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Identify, with reason, if the following is a Pythagorean triplet.
(4, 9, 12)
In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS2 + TQ2 = TP2 + TR2 (As shown in the figure, draw seg AB || side SR and A-T-B)
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 following Figure ∠ACB= 90° and CD ⊥ AB, prove that CD2 = BD × AD
Find the value of (sin2 33 + sin2 57°)
In ΔABC, AD is perpendicular to BC. Prove that: AB2 + CD2 = AC2 + BD2
Two squares having same perimeter are congruent.
