Adv
Rajan@2021
  • 0
Guru

In the figure given below, ABCD is a trapezium in which DC is parallel to AB. If AB=9cm,DC=6cm and BB=12cm., find (i) BP (ii) the ratio of areas of △APB and △DPC.

  • 0

This is the basic and conceptual question from similarity chapter in which we are to find

(i) length of BP

(ii) the ratio of areas of △APB and △DPC.

And we have given a figure of trapezium in which it’s parallel sides are AB and DC

ML Aggarwal Avichal Publication Class 10, Chapter 13.3, question no 14a

Share

1 Answer

  1. From the question it is given that,
    DC is parallel to AB
    AB=9cm,DC=6cm and BB=12cm
    (i) Consider the APB and CPD
    APB=CPD … [because vertically opposite angles are equal]
    PAB=PCD … [because alternate angles are equal]
    So, APBCPD
    Then, BP/PD=AB/CD
    BP/(12BP)=9/6
    6BP=1089BP
    6BP+9BP=108
    15BP=108
    BP=108/15
    Therefore, BP=7.2cm
    (ii) We know that, area of APB/area of CPD=AB2/CD2
    area of APB/area of CPD=92/62
    area of APB/area of CPD=81/36
    By dividing both numerator and denominator by 9, we get,
    area of APB/area of CPD=9/4
    Therefore, the ratio of areas of APB and DPC is 9:4.

    • 0
Leave an answer

Leave an answer

Browse

Choose from here the video type.

Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs".

Captcha Click on image to update the captcha.

Related Questions