Advertisements
Advertisements
Question
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
Solution
∆ 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
RELATED QUESTIONS
In a right triangle ABC right-angled at C, P and Q are the points on the sides CA and CB respectively, which divide these sides in the ratio 2 : 1. Prove that
`(i) 9 AQ^2 = 9 AC^2 + 4 BC^2`
`(ii) 9 BP^2 = 9 BC^2 + 4 AC^2`
`(iii) 9 (AQ^2 + BP^2 ) = 13 AB^2`
In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid.
In triangle ABC, angle A = 90o, CA = AB and D is the point on AB produced.
Prove that DC2 - BD2 = 2AB.AD.
M andN are the mid-points of the sides QR and PQ respectively of a PQR, right-angled at Q.
Prove that:
(i) PM2 + RN2 = 5 MN2
(ii) 4 PM2 = 4 PQ2 + QR2
(iii) 4 RN2 = PQ2 + 4 QR2(iv) 4 (PM2 + RN2) = 5 PR2
Prove that (1 + cot A - cosec A ) (1 + tan A + sec A) = 2
In the given figure, PQ = `"RS"/(3)` = 8cm, 3ST = 4QT = 48cm.
SHow that ∠RTP = 90°.
∆ABC is right-angled at C. If AC = 5 cm and BC = 12 cm. find the length of AB.
Find the unknown side in the following triangles
From the given figure, in ∆ABQ, if AQ = 8 cm, then AB =?
Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.
