both the triangles are similiarjapan1453 wrote:Could you solve this problem. Thanks
OA: A
from therom
ratio of the areas of similiar triangle = ratio of the square of corresponding sides
therefore
ratio = (PR/LN)^2
= (4/12)^2
= 1/9
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
gmat prep coordinate system
Source: Beat The GMAT — Problem Solving |
from therom
ratio of the areas of similiar triangle = ratio of the square of corresponding sides
therefore
ratio = (PR/LN)^2
= (4/12)^2
= 1/9
one more solution to your prob: (w/o using any theorem)
A=0.5XbXh
base= X2-X1
height= Y2-Y1
just substitute the values of X2,X1,Y2,Y1 for both the triangles.
area of bigger triangle = 0.5 X 12 X 12
= 72
area of smaller= 0.5 X 4 X 4
= 8
ratio=
8/72=1/9
A=0.5XbXh
base= X2-X1
height= Y2-Y1
just substitute the values of X2,X1,Y2,Y1 for both the triangles.
area of bigger triangle = 0.5 X 12 X 12
= 72
area of smaller= 0.5 X 4 X 4
= 8
ratio=
8/72=1/9
