the area of a triangle is 1/2 *b * h
you know that the base is 7.
the height is between 3 and 4. (tricky...look at how the triangle is tilted if you draw a line in the middle of the triangle)
so the area has to be less than 14 and the only choice is 12.5
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
Geometry GMAT Prep
Source: Beat The GMAT — Problem Solving |
Here is the alternative approach for the previous one.
find lenghts of each sides of traingles.
QP = 5
PR = 5
QR = sqrt(50)
since (QR)^2 = (QP) ^ 2 + (PR) ^ 2
Its an right angle triangle
A = 1/2 * B * H = 1/2* QP * PR = 1/2* 5 * 5 = 25/2 = 12.5
find lenghts of each sides of traingles.
QP = 5
PR = 5
QR = sqrt(50)
since (QR)^2 = (QP) ^ 2 + (PR) ^ 2
Its an right angle triangle
A = 1/2 * B * H = 1/2* QP * PR = 1/2* 5 * 5 = 25/2 = 12.5
why does the area have to be less than 14? Because the height's between 3 and 4?wawatan wrote:the area of a triangle is 1/2 *b * h
you know that the base is 7.
the height is between 3 and 4. (tricky...look at how the triangle is tilted if you draw a line in the middle of the triangle)
so the area has to be less than 14 and the only choice is 12.5
find lenghts of each sides of traingles.
But how do you get that QR = sqrt(50) without knowing its a right triangle??
And if it is a right triangle why is the height PR (5)?? I just dont see that in the picture
But how do you get that QR = sqrt(50) without knowing its a right triangle??
And if it is a right triangle why is the height PR (5)?? I just dont see that in the picture
-
AleksandrM
- Legendary Member
- Posts: 566
- Location: Philadelphia
- Thanked: 31 times
- GMAT Score:640
distance1 = sqroot[(0 - 3)^2 + (4 - 0)^2]
distance2 = sqroot[(0 - 4)^2 + (4 - 7)^2]
distance 1 = sqroot25 (base)
distance 2 = sqroot25 (height)
5 x 5 / 2 = 12.5
distance2 = sqroot[(0 - 4)^2 + (4 - 7)^2]
distance 1 = sqroot25 (base)
distance 2 = sqroot25 (height)
5 x 5 / 2 = 12.5
-
AleksandrM
- Legendary Member
- Posts: 566
- Location: Philadelphia
- Thanked: 31 times
- GMAT Score:640
You can prove that this is a right triangle.
If you look at the region underneath QP, you can see that it is a right triangle with hypotenuse 5 and base 4 and height 3. If you drop a line from R, then you will have another right triangle underneth PR, with hypotenuse 5 and base 3 and height 4. If you "close" the two triangles, they will form a box with base 4 and height 3. When you open them up, they form another right triangle, with base and height of 5 and hypotenuse of 5sqroot2 (this is also the value you would get if you used the distance formula). Notice also that the subject triangle is a special triangle in the form x : x : xsqroot2.
If you look at the region underneath QP, you can see that it is a right triangle with hypotenuse 5 and base 4 and height 3. If you drop a line from R, then you will have another right triangle underneth PR, with hypotenuse 5 and base 3 and height 4. If you "close" the two triangles, they will form a box with base 4 and height 3. When you open them up, they form another right triangle, with base and height of 5 and hypotenuse of 5sqroot2 (this is also the value you would get if you used the distance formula). Notice also that the subject triangle is a special triangle in the form x : x : xsqroot2.
