Advertisements
Advertisements
प्रश्न
Points A and B are on the opposite edges of a pond as shown in the following figure. To find the distance between the two points, the surveyor makes a right-angled triangle as shown. Find the distance AB.
Advertisements
उत्तर
Since, ΔACD is a right angled triangle.
In right angled ΔADC, by Pythagoras theorem,
(AC)2 = (AD)2 + (CD)2
⇒ (AC)2 + (30)2 + (40)2 ...[∵ AD = 30 cm and CD = 40 cm, given]
⇒ (AC)2 = 900 + 1600
⇒ (AC)2 = 2500
⇒ AC = `sqrt(2500)`
∴ AC = 50 m
Now, AB = AC – BC = 50 – 12 = 38 m
Hence, the distance AB is 38 m.
APPEARS IN
संबंधित प्रश्न
In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Tick the correct answer and justify: In ΔABC, AB = `6sqrt3` cm, AC = 12 cm and BC = 6 cm.
The angle B is:
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
Choose the correct alternative:
In right-angled triangle PQR, if hypotenuse PR = 12 and PQ = 6, then what is the measure of ∠P?
In the given figure, angle BAC = 90°, AC = 400 m, and AB = 300 m. Find the length of BC.
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
In the given figure, angle ACB = 90° = angle ACD. If AB = 10 m, BC = 6 cm and AD = 17 cm, find :
(i) AC
(ii) CD
Find the Pythagorean triplet from among the following set of numbers.
4, 5, 6
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.
In triangle ABC, line I, is a perpendicular bisector of BC.
If BC = 12 cm, SM = 8 cm, find CS
