Advertisements
Advertisements
प्रश्न
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

Advertisements
उत्तर
∆ ABD
∠ACB = ∠ACD = 90°
and AB = 10 cm, BC = 6 cm and AD = 17 cm

To find:
(i) Length of AC
(ii) Length of CD
Proof:
(i) In right-angled triangle ABC
BC = 6 cm, AB = 110 cm
According to Pythagoras Theorem,
AB2 = AC2 + BC2
(10)2 = (AC)2 + (6)2
100 = (AC)2 + 36
AC2 = 100 − 36 = 64 cm
AC2 = 64 cm
∴ AC = `sqrt(8xx8)` = 8 cm
(ii) In right-angle triangle ACD
AD = 17 cm, AC = 8 cm
According to Pythagoras Theorem,
(AD)2 = (AC)2 + (CD)2
(17)2 = (8)2 + (CD)2
289 – 64 = CD2
225 = CD2
CD =`sqrt(15xx15)` = 15 cm
APPEARS IN
संबंधित प्रश्न
In an equilateral triangle ABC, D is a point on side BC such that BD = `1/3BC` . Prove that 9 AD2 = 7 AB2
Identify, with reason, if the following is a Pythagorean triplet.
(24, 70, 74)
If P and Q are the points on side CA and CB respectively of ΔABC, right angled at C, prove that (AQ2 + BP2) = (AB2 + PQ2)
Find the length of diagonal of the square whose side is 8 cm.
The sides of the triangle are given below. Find out which one is the right-angled triangle?
11, 12, 15
A ladder 15m long reaches a window which is 9m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to other side of the street to reach a window 12m high. Find the width of the street.
In a right-angled triangle PQR, right-angled at Q, S and T are points on PQ and QR respectively such as PT = SR = 13 cm, QT = 5 cm and PS = TR. Find the length of PQ and PS.
If length of sides of a triangle are a, b, c and a2 + b2 = c2, then which type of triangle it is?
For going to a city B from city A, there is a route via city C such that AC ⊥ CB, AC = 2x km and CB = 2(x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2
[Hint: Produce AB and DC to meet at E.]
