Home
Class 12
MATHS
In a Delta ABC, the median AD is perpend...

In a `Delta ABC`, the median AD is perpendicular to AC. If b = 5 and c = 11, then a =

A

10

B

12

C

14

D

`sqrt(221)`

Text Solution

Verified by Experts

The correct Answer is:
C

Using rule in `Delta ACD`, we get
`(5)/(sin theta)=(CD)/(sin 90^(@))` or `a = 10 cosec theta` ….(1)
Using sine rule in `Delta ADB`, we get
`(11)/(sin(pi-theta))=(BD)/(sin(A-90^(@)))`
`(11)/(sin theta)=-(a)/(2cos A)`
`therefore a=-22 cosec theta cos A`
From (1) and (2), we get
co A = -5/11
Also cos `A=(25+121-a^(2))/(2(5)(11))=-(5)/(11)`
`therefore a^(2)=196`
`therfore a = 14`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS AND PROPERTIES OF TRIANGLE

    CENGAGE PUBLICATION|Exercise Multiple Correct Answers Type|13 Videos

  • SOLUTIONS AND PROPERTIES OF TRIANGLE

    CENGAGE PUBLICATION|Exercise Comprehension Type|6 Videos

  • SET THEORY AND REAL NUMBER SYSTEM

    CENGAGE PUBLICATION|Exercise Archives|1 Videos

  • STATISTICS

    CENGAGE PUBLICATION|Exercise Archives|10 Videos

Similar Questions

Explore conceptually related problems

In triangle ABC, D is the mid point of side BC . If AD is perpendicular to AC . then prove that cosA cosC =(2(c^2-a^2))/(3ca)

Let us consider one vertex and one side through the vertex along x -axis of a triangle . Now the coordinates of the vertices B,C and A of any triangle ABC (0,0) ,(a,0) and (h,k) respectively should be taken . If in triangle ABC , AC =3 ,BC=4 and median AD is perpendicular with median BE , then the area of DeltaABC is -

In the isosceles triangle ABC, AB=AC and BE is perpendicular to AC from B. Prove that BC^(2) = 2ACxxCE

In DeltaABC , AD is the perpendicular bisector of BC (See adjacent figure). Show that DeltaABC is an isosceles triangle in which AB = AC.

If the medium AD is perpendicular to the side AB in the triangle ABC, show that tan A+ 2 tan B=0 .

In Delta ABC ,AD is a perpendicular drawn from A to the side BC. If AD^(2)=BD . CD , then prove that the triangle is a right-angled triangle.

ABC is a triangle and D is the middle point of BC. If AD is perpendicular to AC, prove that, cosA cos C =(2(c^(2)-a^(2))/(3ca))

Delta ABC is an obtus triangle in which /_B = obtus angle. If AD is perpendicular to BC ( or extended BC ) ,then prove that AC^(2) = AB^(2) +BC^(2) + 2BC xx BD .