Advertisements
Advertisements
प्रश्न
There are two paths that one can choose to go from Sarah’s house to James's house. One way is to take C street, and the other way requires to take B street and then A street. How much shorter is the direct path along C street?

Advertisements
उत्तर
Distance between Sarah’s House and James’s House using “C street”.
AC2 = AB2 + BC2
= 22 + 1.52
= 4 + 2.25
= 6.25
AC = `sqrt(6.25)`
AC = 2.5 miles
Distance covered by using “A Street” and “B Street”
= (2 + 1.5) miles
= 3.5 miles
Difference in distance = 3.5 miles – 2.5 miles = 1 mile
APPEARS IN
संबंधित प्रश्न
In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AD2 = BD × CD

In triangle ABC, AB = AC = x, BC = 10 cm and the area of the triangle is 60 cm2.
Find x.
ABC is a triangle, right-angled at B. M is a point on BC.
Prove that: AM2 + BC2 = AC2 + BM2
In Fig. 3, ∠ACB = 90° and CD ⊥ AB, prove that CD2 = BD x AD.

In the given figure, angle BAC = 90°, AC = 400 m, and AB = 300 m. Find the length of BC.

Find the distance between the helicopter and the ship
In the given figure, ∠T and ∠B are right angles. If the length of AT, BC and AS (in centimeters) are 15, 16, and 17 respectively, then the length of TC (in centimeters) is ______.

The perimeters of two similar triangles ABC and PQR are 60 cm and 36 cm respectively. If PQ = 9 cm, then AB equals ______.
In figure, PQR is a right triangle right angled at Q and QS ⊥ PR. If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.
In the adjoining figure, a tangent is drawn to a circle of radius 4 cm and centre C, at the point S. Find the length of the tangent ST, if CT = 10 cm.

