Cut off Trapezium
The area of the red triangle is 25 and the area of the orange triangle is 49. What is the area of the trapezium A B C D ?
The answer is 144.
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3 solutions
Nice solution. It's interesting to note that, with a being the area of the red triangle and b the area of the orange triangle, the area of A B C D is ( a + b ) 2 .
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Thank you. A v e r y n i c e observation. It seems all ways true since it was a general proof.
Please change sqare roots to square roots . Thanks
Does it matter that the shape is a trapezium and not a trapezoid?
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what's the difference?
Trapezium is UK English and trapezoid US English.
Slightly different approach with same nomenclature
The triangles are similar so the ratio of the squares of their corresponding sides is equal to the ratio of their area.
X/Y =5/7 Y = (7/5)X
Area (Aq) of a quadrilateral equal the average of the bases time the height. Aq = (1/2)(X + Y)(Hx + Hy)
We know the heights of the triangles as a function of their bases since we know their areas.
Hx = 2(Ax)/X = 50/X Hy=2(Ay)/Y = 98/Y Hx + Hy = (50Y + 98X)/(XY)
Aq = (1/2)(X + Y)(50Y + 98X)/(XY) = (1/2)(X + Y)(70X + 98X)/(XY) = 84(X + Y)/Y = 84(1 + X/Y) = 84(1 + 5/7) = 84(12/7) = 144
If the area of the red region is P and the orange region is Q, then the area of each of the white regions in the trapezium can be shown to be sqrt(P x Q).
So the area of the whole trapezium is just P + Q + 2*sqrt(P x Q) = (sqrt(P) + sqrt(Q))^2
How can we deduce that. Is there any theorem
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You can deduce that using elementary geometry.
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It is much advisable if you make your solution capable of accomodating various levels of intelligence by demonstrating the "deduction using elementary geometry".
Is there any proof or psotulate for this formaula?? {sqrt(P x Q)}.
Let's call the unlabeled vertex in the middle E. Let's call the lengths AB=x, DC=y. Call the length of the altitude of triangle ABE (drawn to AB) h and that of triangle CDE (drawn to DC) k. We know 1/2 xh=25. 1/2 yk = 49. Consider the area of triangle ADE as the area of triangle ADB minus the area of triangle ABE namely - (h+k)x-xh=xk. Conversely it can be represented as ADC minus CDE or (h+k)y-yk=yh. So yh=xk. Multiplying the first two equalities together gives 1/4 yhxk=(1/2 yh)^2=25 49. Giving 1/2 yh = 1/2 xk = 35. We can use the same analysis to show that triangle BCE is also equal in area to 1/2 yh = 35 (in fact it is congruent to ADE). We can now sum the four regions to find the area of the trapezoid: 25+35+35+49=144.
L e t X = A B , Y = D C , H x , H y b e r e s p e c t i v e h e i g h t s . ∴ 2 1 ∗ X ∗ H x = 2 5 a n d 2 1 ∗ Y ∗ H y = 4 9 . The triangles are similar. So the ratio of there linear dimensions will be equal to the ratio of the square roots of their areas. ∴ Y X = H y H x = 7 5 . ∴ Y X + Y = H y H x + H y = 7 1 2 . ∴ Y ∗ H y ( X + Y ) ( H x + H y ) = 7 2 1 2 2 . 2 ∗ 4 9 2 ∗ A B C D = 4 9 1 4 4 A r e a A B C D = 1 4 4 .