Home
Class 10
MATHS
In the given figure DEFG is a square and...

In the given figure DEFG is a square and `angleBAC=90^(@)` . Show that `FG^(2) = BG xx FC ` .

Text Solution

Verified by Experts

The correct Answer is:
`FG^(2)=BGxx FC `
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • TRIANGLES

    OSWAL PUBLICATION|Exercise SELF ASSESSMENT -1 |3 Videos
  • TRIANGLES

    OSWAL PUBLICATION|Exercise SELF ASSESSMENT -1 ( FILL IN THE BLANKS) |3 Videos
  • TRIANGLES

    OSWAL PUBLICATION|Exercise CASE - BASED MCQs|15 Videos
  • SURFACE AREAS AND VOLUMES

    OSWAL PUBLICATION|Exercise BOARD CORNER (LONG ANSWER TYPE QUESTIONS)|9 Videos

Similar Questions

Explore conceptually related problems

In the given figure, DEFG is a square and angle BAC= 90^(@) , Prove that (i) Delta AGF~ Delta DBG (ii) Delta AGF~ Delta EFC (iii) DE^(2)=BDxxEC

In the given figure , DEFG is a square and angleBAC= 90^(@) prove that (i) triangleAGF~ triangleDBG (ii) triangleAGF~triangleEFC

Knowledge Check

  • In the given figure, the measure of angleBAC is,

    A
    `56^@`
    B
    `58^@`
    C
    `62^@`
    D
    `48^@`
  • In the figure given below angleBAC=90^(@) and AD bot BC. Then

    A
    `BCxx CD= AC^(2)`
    B
    `AB xx AC =BC^(2)`
    C
    `BDxx CD =AD^(2)`
    D
    `AB xx AC=AD^(2)`
  • In the given figure BC is produced to D and angleBAC=40^(@) and angle ABC=70^(@) . Find the value of

    A
    `30^(@)`
    B
    `40^(@)`
    C
    `70^(@)`
    D
    `110^(@)`
  • Similar Questions

    Explore conceptually related problems

    In the figure, DEFG is a square and angle BAC = 90^(@). Prove that (A) Delta AGF ~ Delta DBG (B) Delta AGF ~ Delta EFC (C) Delta DBG ~ Delta EFC

    In fig.7.84, DEFG is a square and angleBAC = 90° . Prove that: (i) DeltaAGF~DeltaDBG (ii) DeltaAGF~ DeltaEFC (iii) DeltaDBG~ DeltaEFC (iv) DE^(2) = BD xx EC

    In the given figure, AB = AD and angleBAC=angleDAC . Then (ii) BC = ____.

    In the given figure, ABCD is a quadrilateral angleADB=60^(@), angleBAC=70^(@),angleDBC=30^(@) , and

    In the given figure, O is the centre of circle. angleAOC = 120^(@) . Find angleBAC :