Advertisements
Advertisements
प्रश्न
Jiya walks 6 km due east and then 8 km due north. How far is she from her starting place?
Advertisements
उत्तर
Let A be the starting point and B be the ending point of Jiya.
Since ∆ABC is right-angled.
∴ (AB)2 = (AC)2 + (BC)2
⇒ (AB)2 = 62 + 82
= 36 + 64
= 100
⇒ AB = 10
Thus, Jiya is 10 km away from her starting place.
APPEARS IN
संबंधित प्रश्न
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
In triangle ABC, AB = AC and BD is perpendicular to AC.
Prove that: BD2 − CD2 = 2CD × AD
In the given figure, BL and CM are medians of a ∆ABC right-angled at A. Prove that 4 (BL2 + CM2) = 5 BC2.
In ∆ ABC, AD ⊥ BC.
Prove that AC2 = AB2 +BC2 − 2BC x BD
The sides of the triangle are given below. Find out which one is the right-angled triangle?
40, 20, 30
The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2
If in a ΔPQR, PR2 = PQ2 + QR2, then the right angle of ∆PQR is at the vertex ________
From the given figure, in ∆ABQ, if AQ = 8 cm, then AB =?
If S is a point on side PQ of a ΔPQR such that PS = QS = RS, then ______.
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 ______.
