Advertisements
Advertisements
प्रश्न
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Advertisements
उत्तर

In the given figure, BC represents the unbroken part of the tree. Point C represents the point where the tree broke and CA represents the broken part of the tree. Triangle ABC, thus formed, is right-angled at B.
Applying Pythagoras theorem in ΔABC,
AC2 = BC2 + AB2
AC2 = (5 m)2 + (12 m)2
AC2 = 25 m2 + 144 m2
AC2 = 169 m2
AC = 13 m
Thus, original height of the tree = AC + CB
= 13 m + 5 m
= 18 m
APPEARS IN
संबंधित प्रश्न
Identify, with reason, if the following is a Pythagorean triplet.
(11, 60, 61)
Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was \[15\sqrt{2}\]
km. Find their speed per hour.
Digonals of parallelogram WXYZ intersect at point O. If OY =5, find WY.
In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC = 3 cm. Calculate the length of OC.

In figure AB = BC and AD is perpendicular to CD.
Prove that: AC2 = 2BC. DC.
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 - AB2 = 2BC x ED
In the given figure. PQ = PS, P =R = 90°. RS = 20 cm and QR = 21 cm. Find the length of PQ correct to two decimal places.
In a right-angled triangle PQR, right-angled at Q, S and T are points on PQ and QR respectively such as PT = SR = 13 cm, QT = 5 cm and PS = TR. Find the length of PQ and PS.
From the given figure, in ∆ABQ, if AQ = 8 cm, then AB =?

In ∆PQR, PD ⊥ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b)(a – b) = (c + d)(c – d).
