Triangle ABC is an isosceles right triangle. If AB = AC, then the area

Question

Triangle ABC is an isosceles right triangle. If AB = AC, then the area of a square with a side length equal to twice the length of BC is how many times the area of triangle ABC?

A. 13
B. 14
C. 15
D. 16
E. 17

Archit3110 · Answer

let isoscles triangle side = 2 
so area is 2
and side BC =2√2
side of square ; 4√2 ; area = 32
i.e 32/2 ; 16 times
IMO D

DaniyalAlwani · Answer

Archit3110

Can you please explain how you deduce area of triangle in detail?

ananyamohata15 · Answer

Area of triangle = 1/2 * base * height

Say original triangle ABC has AB (base) and AC (height) = 1cm,
Hence hypotenuse = √2
Area of ABC = 0.5

Now, square PQRS has side as QR which is twice that of hypotenuse of ABC, ie 2√2,
 Area of square = 2√2 * 2√2 = 8, which is 16 times that of ABC

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GunjanMadan1 · Answer

Since this is an isosceles right angled triangle, its base and height has to be equal

Lets Base= Height = b

Hypo= \sqrt{2}b

Area=  \frac{base*height}{2} 
 =b^2/2

Side of Square=  2*Hypo= 2b\sqrt{2}

Area of Square =  4b^2*2= 8b^2= 16*Area of Triangle

suchita2409 · Answer

Side of a square = 2* BC

Area of square =(2*BC)^2
 =4*BC^2
 
The area should be a multiple of 4.
The only option multiple of 4 is 16.
Therefore, D

Triangle ABC is an isosceles right triangle. If AB = AC, then the area

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Triangle ABC is an isosceles right triangle. If AB = AC, then the area

Prep Toolkit

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