You’re Probably Missing the Key Step in This Geometry Problem
1.1K views · Sep 29, 2026 · Education
Comments · 15
@ロプノールs · 1 day ago
考え方（how to solve) QRと同じ長さの円の接線ABをPSと平行（∦）に引くとBSはBRと対照なので４。<br>Tと円の中心（O）を結びTから２の位置に点Cを取り△ACOを作図する。半径（radius)をrとして式を作る。<br>（r-4)²+(r-2)²=r² ｒ²ー８r+16+ｒ²ー４r＋４=ｒ² r²－１２r＋２０=０ (因数分解をする）<br>（r-2)(r-10) ｒ＝２or 10 2=NG. r=10 blueの円の面積＝１０²×π＝１００π answer=100π
1
@MrPaulc222 · 2 days ago
A mirror image of RB to the lower right means adding 4 to obtain a horizontal that is flush with the tangent.<br>M is the midpoint of AS.<br>OM is r - 4<br>MA is r - 2<br>OA is r<br>(r - 4)^2 + (r - 2)^2 = r^2<br>r^2 - 8r + 16 + r^2 - 4r + 4 = r^2<br>r^2 - 12r + 20 = 0<br>r = (12+or-sqrt(144 - 4*20))/2<br>r = (12+or-8)/2<br>r = 10 or r = 2<br>Can't be 2 so must be 10.<br>100pi un^2 or about 314.16
1
@Ramkabharosa · 1 day ago (edited)
4+ 2.|PT| = 4+ 2.|SD| = 2x -4. So |PT| = x-4. By the <br>tangent-secant thm, |PT|² =2.|PS|. So x² -8x+16 = <br>(x-4)² = 2.(x+ x-2) = 4x -4. Hence 0= x² -12x+20 = <br>(x-2).(x-10). So x = 10 because x>2. ∴ area=100π.
2
@sanjeevkumarpamu2206 · 1 day ago
Madam! Initial reasoning is most crucial for me 😮😮😮😮😮
1
@edheakes2663 · 2 days ago
If you know the secant tangent theorem for lengths the answer can be found a little more quickly. But the solution given is impeccable.
1
@vierinkivi · 1 day ago
R pisteen potenssi. (r-2)^2=4(2r-4). r=10
1
@lux-vacui · 1 day ago (edited)
<a href="https://www.youtube.com/watch?v=Gom-jWHK8Aw&t=8">0:08</a> "we have a square". QRSP cannot possibly be a square with that data, it's a rectangle.
1
@klementhajrullaj1222 · 1 day ago
Or, x^2=(x-2)^2+(x-4)^2 ...
1
@h0wards0n · 1 day ago
PQRS must be a rectangle not a square.
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