In a triangle pqr the angles q and r are acute if qs is perpendicular to pr and rt is perpendicular to pq
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Let PQ=x, QR=y and PR=z.
Given: x+y+z=60 (i);
Equate the areas: 12∗xy=12∗QS∗z (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) --> xy=12z (ii);
Aslo x2+y2=z2 (iii);
So, we have:
(i) x+y+z=60;
(ii) xy=12z;
(iii) x2+y2=z2.
From (iii) (x+y)2−2xy=z2 --> as from (i) x+y=60−z and from (ii) xy=12z then (60−z)2−2∗12z=z2 --> 3600−120z+z2−24z=z2 --> 3600=144z --> z=25;
From (i) x+y=35 and from (ii) xy=300 --> solving for x and y --> x=20 and y=15 (as given that x>y).
Next, perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. So, PQS and SQR are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides: AREAarea=S2s2.
So, x2y2=AREAarea --> AREAarea=400225=169
Given: x+y+z=60 (i);
Equate the areas: 12∗xy=12∗QS∗z (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) --> xy=12z (ii);
Aslo x2+y2=z2 (iii);
So, we have:
(i) x+y+z=60;
(ii) xy=12z;
(iii) x2+y2=z2.
From (iii) (x+y)2−2xy=z2 --> as from (i) x+y=60−z and from (ii) xy=12z then (60−z)2−2∗12z=z2 --> 3600−120z+z2−24z=z2 --> 3600=144z --> z=25;
From (i) x+y=35 and from (ii) xy=300 --> solving for x and y --> x=20 and y=15 (as given that x>y).
Next, perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. So, PQS and SQR are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides: AREAarea=S2s2.
So, x2y2=AREAarea --> AREAarea=400225=169
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