Advertisements
Advertisements
प्रश्न
In the figure below, find the value of 'x'.
Advertisements
उत्तर
In the right-angled triangle EDF,
∠D = 90°.
Hence, side EF is the hypotenuse.
According to Pythagoras' theorem,
l(EF)2 = l(ED)2 + l(DF)2
⇒ (17)2 = (x)2 + (8)2
⇒ 289 = x2 + 64
⇒ x2 = 289 − 64
⇒ x2 = 225
⇒ `root 225`
⇒ x = 15
∴ The value of x is 15.
संबंधित प्रश्न
A man goes 10 m due east and then 24 m due north. Find the distance from the starting point
In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
`"AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2`
In triangle ABC, AB = AC and BD is perpendicular to AC.
Prove that: BD2 − CD2 = 2CD × AD
In triangle PQR, angle Q = 90°, find: PR, if PQ = 8 cm and QR = 6 cm
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 = AD2 + BC x DE + `(1)/(4)"BC"^2`
If in a ΔPQR, PR2 = PQ2 + QR2, then the right angle of ∆PQR is at the vertex ________
A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
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).
In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2
[Hint: Produce AB and DC to meet at E.]
The foot of a ladder is 6 m away from its wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?
