Advertisements
Advertisements
Question
In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL2 + CM2) = 5 BC2
Advertisements
Solution
Given: Δ ABC right angled at A, i.e., A = 90, where BL and CM are the medians.
To Prove: 4(BL2 + CM2) = 5BC2
Proof:
Since BL is the median,
AL = CL = `1/2` AC ...(1)
Similarly, CM is the median
AM = MB = `1/2`AB ...(2)
∴ by Pythagoras theorem,
(Hypotenuse)2 = (Height)2 + (Base)2 ...(3)
In ΔBAC,
(BC)2 = (AB)2 + (AC)2 ...(4)
∠BAC = 90°
In ΔBAL,
(BL)2 = AB2 + AL2 ...(From 1)
BL2 = AB2 + `(("AC")/2)^2`
BL2 = AB2 + `("AC"^2/4)`
Multiply both sides by 4,
4BL2 = 4AB2 + AC2 ...(5)
In ΔMAC,
CM2 = AM2 + AC2 ...(From 2)
CM2 = `(("AB")/2)^2` + AC2
CM2 = `("AB")^2/4` + AC2
Multiply both sides by 4,
4CM2 = AB2 + 4AC2 ...(6)
Adding 5 and 6,
4BL2 + 4CM2 = (4AB2 + AC2) + (AB2 + 4AC2)
4(BL2 + CM2) = 5AB2 + 5AC2
∴ 4(BL2 + CM2) = 5BC2
Hence, proved.
APPEARS IN
RELATED QUESTIONS
Prove that the diagonals of a rectangle ABCD, with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6), are equal and bisect each other.
In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 − 2BC.BD.
In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4PM2 – 3PR2.
In the figure: ∠PSQ = 90o, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.
If the sides of the triangle are in the ratio 1: `sqrt2`: 1, show that is a right-angled triangle.
In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1: 3.
Prove that : 2AC2 = 2AB2 + 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
Choose the correct alternative:
In right-angled triangle PQR, if hypotenuse PR = 12 and PQ = 6, then what is the measure of ∠P?
Find the side of the square whose diagonal is `16sqrt(2)` cm.
In the given figure, angle BAC = 90°, AC = 400 m, and AB = 300 m. Find the length of BC.
The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder?
The sides of the triangle are given below. Find out which one is the right-angled triangle?
11, 12, 15
The sides of the triangle are given below. Find out which one is the right-angled triangle?
1.5, 1.6, 1.7
A ladder 15m long reaches a window which is 9m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to other side of the street to reach a window 12m high. Find the width of the street.
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
The hypotenuse of a right angled triangle of sides 12 cm and 16 cm is __________
In ΔABC, if DE || BC, AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, then value of x is ______.
A right-angled triangle may have all sides equal.
Jiya walks 6 km due east and then 8 km due north. How far is she from her starting place?
