Advertisements
Advertisements
Question
In the given figure, AD = 13 cm, BC = 12 cm, AB = 3 cm and angle ACD = angle ABC = 90°. Find the length of DC.
Advertisements
Solution
Given :
∆ ACD = ∠ABC = 90°
and AD = 13 cm, BC = 12 cm, AB = 3 cm
To find : Length of DC.
(i) In right angled ∆ ABC
AB = 3 cm, BC = 12 cm
According to Pythagoras Theorem,
(AC)2 = (AB)2 + (BC)2
(AC)2 = (3)2 + (12)2
(AC) =`sqrt(9+144)=sqrt153` cm
(ii) In right angled ∆ ACD
AD = 13 cm, AC =`sqrt153`
According to Pythagoras Theorem,
DC2 = AB2 − AC2
DC2= 169 − 153
DC = `sqrt16` = 4 cm
∴ Length of DC is 4 cm
APPEARS IN
RELATED QUESTIONS
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC.BD.
In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4PM2 – 3PR2.
In triangle ABC, AB = AC and BD is perpendicular to AC.
Prove that: BD2 − CD2 = 2CD × AD
In a rectangle ABCD,
prove that: AC2 + BD2 = AB2 + BC2 + CD2 + DA2.
In the figure below, find the value of 'x'.
Find the Pythagorean triplet from among the following set of numbers.
2, 4, 5
Find the Pythagorean triplet from among the following set of numbers.
9, 40, 41
Two poles of height 9m and 14m stand on a plane ground. If the distance between their 12m, find the distance between their tops.
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.
