Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर

It is given that, Pranali and Prasad have same speed.
Thus, they cover same distance in 2 hours.
i.e. OA = OB
Let the speed be x km per hour.
According to Pythagoras theorem,
In ∆AOB
\[{AB}^2 = {AO}^2 + {OB}^2 \]
\[ \Rightarrow \left( 15\sqrt{2} \right)^2 = {AO}^2 + {OA}^2 \]
\[ \Rightarrow 450 = 2 {AO}^2 \]
\[ \Rightarrow {AO}^2 = \frac{450}{2}\]
\[ \Rightarrow {AO}^2 = 225\]
\[ \Rightarrow AO = 15 km\]
\[ \Rightarrow BO = 15 km\]
\[\text{Speed} = \frac{Distance}{Time}\]
\[ = \frac{15}{2}\]
\[ = 7 . 5 \text{km per hour}\]
APPEARS IN
संबंधित प्रश्न
Two towers of heights 10 m and 30 m stand on a plane ground. If the distance between their feet is 15 m, find the distance between their tops
ABC is a right triangle right-angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular from C on AB, prove that
(i) cp = ab
`(ii) 1/p^2=1/a^2+1/b^2`
In Figure ABD is a triangle right angled at A and AC ⊥ BD. Show that AC2 = BC × DC

ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4PM2 – 3PR2.

In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.
In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS2 + TQ2 = TP2 + TR2 (As shown in the figure, draw seg AB || side SR and A-T-B)

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)

In ΔABC, Find the sides of the triangle, if:
- AB = ( x - 3 ) cm, BC = ( x + 4 ) cm and AC = ( x + 6 ) cm
- AB = x cm, BC = ( 4x + 4 ) cm and AC = ( 4x + 5) cm
AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.
In triangle ABC, AB = AC and BD is perpendicular to AC.
Prove that: BD2 − CD2 = 2CD × AD
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

If the angles of a triangle are 30°, 60°, and 90°, then shown that the side opposite to 30° is half of the hypotenuse, and the side opposite to 60° is `sqrt(3)/2` times of the hypotenuse.
In the given figure, angle ACP = ∠BDP = 90°, AC = 12 m, BD = 9 m and PA= PB = 15 m. Find:
(i) CP
(ii) PD
(iii) CD

From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.
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 length of sides of a triangle are a, b, c and a2 + b2 = c2, then which type of triangle it is?
In an isosceles triangle PQR, the length of equal sides PQ and PR is 13 cm and base QR is 10 cm. Find the length of perpendicular bisector drawn from vertex P to side QR.
