Home
Class 10
MATHS
In the figure , angle LMN = angle LKN =...

In the figure , ` angle LMN = angle LKN = 90^(@)`
seg MK ` bot ` seg LN .
Complete the following activtiy
to prove R is the midpoint of seg MK .

Promotional Banner

Topper's Solved these Questions

  • PYTHAGORAS THEOREM

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise EXAMPLES FOR PRACTICE ( 3 MARK QUESTIONS )|8 Videos

  • PYTHAGORAS THEOREM

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise EXAMPLES FOR PRACTICE ( 2 MARK QUESTIONS )|2 Videos

  • PYTHAGORAS THEOREM

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise EXAMPLES FOR PRACTICE ( 1 MARK MULTIPLE CHOICE QUESTIONS )|11 Videos

  • PROBABILITY

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise ASSIGNEMENT 5.4|12 Videos

  • QUADRATIC EQUATION

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise CHALLENGIN QUESTIONS|2 Videos

Similar Questions

Explore conceptually related problems

In the figure, /_LMN=/_LKN=90^(@) seg MK bot seg LN . Complete the following activity to prove R is the midpoint of seg MK . Proof: In DeltaLMN , /_LMN=90^(@) seg MR bot hypotenuse LN :. by property of geometric mean, MR^(2)=squarexxRN In DeltaLKN , /_LKN=90^(@) seg KR bot hypotenuse LN :. by property of geometric mean, KR^(2)=LRxxsquare From (1) and (2) , we get MR^(2)=square :. MR=square . :. R is the midpoint of seg MK .

In the figure, /_DFE=90^(@) , seg FG bot side DE , DG=8 , FG=12 then complete the following activity to find the length of seg DE . In DeltaDFE , /_DFE=90^(@) , seg FG bot hypotenuse DE :. by theorem of geometric mean, FG^(2)=squarexxEG :.12^(2)=squarexxEG EG=(12xx12)/(square) :.EG=square DE=DG+GE=8+square=square

In the given figure, angle PQR=90^(@)," seg "QNbot seg PR, PN=9,NR=16 .Find QN.

In the given figure, angle BAC=90^(@) and AD bot BC .then,

(A) Complete any two out of three activites : In the figure , seg SP bot side YK . Seg YT bot side SK . If SP = 6 , YK = 13, YT = 5 and TK = 12 then complete the following activity to find A (Delta SYK ) : A (Delta YTK) The ratio of the areas of the two Deltas is equal to the ratio of the products of their square and the corresponding heights :. (A(Delta SYK))/(A (Delta YTK)) = (square xxSP)/(TK xx TY) = (13 xx 6) /(12 xx square ) = square /square

In the figure, /_QPR=90^(@) , seg PM bot seg QR and Q-M-R , PM=10 . QM=8 , find QR .

In the figure , angle PQR =32^@ seg SN bot ray QP and seg SM bot ray QR . Find angle PQS . state the reason for your answer.

In the given figure, angleP~=angleR , seg PQ ~= seg RQ. Prove that DeltaPQT~=DeltaRQS .