Can You Find the Shaded Area? | Geometry Challenge
1.2K views · Sep 28, 2026 · Education
Comments · 19
@ロプノールs · 2 days ago
考え方（how to solve) QR→PR→SP→△ＳＰＲの高さ(h)の順に求める。<br>QR＝３３６/４８＝１４．PR＝４８²+１４²＝√２５００＝５０ SP＝４８²+６４²＝√６４００＝８０<br>△SPRの（h)=５０²ー40²＝√９００＝３０ ８０×３０×１/２＝１２００ answer=1200㎠
2
@sakurayayoi-p2r · 1 day ago
黄色の三角は底辺が50、高さが48、面積は５０＊４８／２＝1200
1
@ollitrop46 · 2 days ago
1) Triangle PQR: Let RQ be x === 48x/2=336 === 2/48[48x/2=336 === x=672/48 === x=14cm === <br>Let PR & RS be y, so QS=x+y. === 48²+14²=y² === 2304+196=y² === 2500=y² ===√2500=√y² === 50cm=y=PR=RS === 2) QS=x+y=14+50=64cm === Triangle PQS' Area: A=½×48×64 === A=24×64 <br>=== A= 1536cm² === 3) Shaded Region's Area PRS: A=1536-336 === A=1200cm² Final Answer.
2
@DonaldSax-q4h · 2 days ago
Required shaded area = 1200cm^2
2
@klementhajrullaj1222 · 2 days ago
A=(64•48)/2-336=32•48-336=1536-336=1200
2
@ollitrop46 · 2 days ago
You don't need the hypotenuse PS to solve the Area of the Shaded Region.
1
@humbertorios2941 · 2 days ago
Boa explicação.
1
@RamanaCv-o2l · 2 days ago
Is this lenghty method necessary?. Area =1/2×base×height=(1/2)×50×48=1200.
1
@debbie-g6x · 1 day ago
Let θ = ∡PRQ ⇒ ∡PRS = 180° - θ and letting a = PR = RS ⇒ RQ = 14 ⇒ a = 50 ⇒ sin(θ) = sin(180° - θ) = 24/25 ⇒ A(ΔPRS) = 1/2 * (25^2 * 2^2) * (24/25) = 1200.
1
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