Advertisements
Advertisements
प्रश्न
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 = AE2 + CD2 + BF2
Advertisements
उत्तर
In right triangles ODB and ODC, we have
OB2 = OD2 + BD2
OC2 = OD2 + CD2
∴ OB2 - OC2 = (OD2 + BD2) - (OD2 + CD2)
⇒ OB2 - OC2 = BD2 - CD2 ....(i)
Similarly, we have
OC2 - OA2 = CE2 - AE2 ....(ii)
OA2 - OB2 = AF2 - BF2 ....(iii)
Adding (i), (ii) and (iii), we get
(OB2 - OC2) + (OC2 - OA2) + (OA2 - OB2) = (BD2 - CD2) + (CE2 - AE2) + (AF2 - BF2)
⇒ (BD2 + CE2 + AF2) - (AE2 + CD2 + BF2) = 0
⇒ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
APPEARS IN
संबंधित प्रश्न
In Figure ABD is a triangle right angled at A and AC ⊥ BD. Show that AC2 = BC × DC
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
In a trapezium ABCD, seg AB || seg DC seg BD ⊥ seg AD, seg AC ⊥ seg BC, If AD = 15, BC = 15 and AB = 25. Find A(▢ABCD)
A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.
The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and ∠ABD = ∠BCD = 90o. Calculate the length of AB.
In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1: 3.
Prove that : 2AC2 = 2AB2 + BC2
In equilateral Δ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.
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
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?
Find the Pythagorean triplet from among the following set of numbers.
2, 6, 7
The sides of the triangle are given below. Find out which one is the right-angled triangle?
40, 20, 30
Calculate the area of a right-angled triangle whose hypotenuse is 65cm and one side is 16cm.
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2(AD2 + CD2)
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.
In a square PQRS of side 5 cm, A, B, C and D are points on sides PQ, QR, RS and SP respectively such as PA = PD = RB = RC = 2 cm. Prove that ABCD is a rectangle. Also, find the area and perimeter of the rectangle.
An isosceles triangle has equal sides each 13 cm and a base 24 cm in length. Find its height
From given figure, In ∆ABC, If AC = 12 cm. then AB =?
Activity: From given figure, In ∆ABC, ∠ABC = 90°, ∠ACB = 30°
∴ ∠BAC = `square`
∴ ∆ABC is 30° – 60° – 90° triangle
∴ In ∆ABC by property of 30° – 60° – 90° triangle.
∴ AB = `1/2` AC and `square` = `sqrt(3)/2` AC
∴ `square` = `1/2 xx 12` and BC = `sqrt(3)/2 xx 12`
∴ `square` = 6 and BC = `6sqrt(3)`
For going to a city B from city A, there is a route via city C such that AC ⊥ CB, AC = 2x km and CB = 2(x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
Two trees 7 m and 4 m high stand upright on a ground. If their bases (roots) are 4 m apart, then the distance between their tops is ______.
