Brilliant Geometry puzzle 》》 Rectangle puzzle

Geometry Level 3

brilliant geometry puzzle brilliant geometry puzzle

In the rectangle shown above, the areaa of the yellow and blue regions are given as 9 and 4 respectively. Find the area of red region.

You can use Swaroop's theorem to solve 》 bit.ly/Swaroop-theorem


The answer is 6.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Chew-Seong Cheong
Jan 19, 2020

We note that the blue and yellow triangles are similar. Therefore that areas of the two triangles are proportional to the square of their linear dimensions. That is if the blue triangle has a base of 2 a 2a and height of 2 h 2h , the yellow triangle will have a base of 3 a 3a and height of 3 h 3h . Since 4 a h 2 = 4 \dfrac {4ah}2 = 4 and 9 a h 2 = 9 \dfrac {9ah}2 = 9 , a h = 2 ah =2 .

Let the area of the red region be A A . Then A + 9 = 1 2 ( 3 a ) ( 5 h ) = 15 a h 2 = 15 A = 15 9 = 6 A + 9 = \dfrac 12 (3a)(5h) = \dfrac {15ah}2 = 15 \implies A = 15-9 = \boxed 6 .

İlker Can Erten
Jan 29, 2020

blue triangle and yellow triangle are similar with ratio 2 3 \dfrac{2}{3} so (left side lenght of blue) / (right side lenght of yellow) = 2 3 \frac{2}{3} blue triangle and red one shares same height so area ratio will be 2 3 \dfrac{2}{3}

4 S = 2 3 \dfrac{4}{S}=\dfrac{2}{3}

S = 6 S=6

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...