Can You Find the Shaded Area? | #math | #geometry
3.6K views · Sep 23, 2026 · Education
Comments · 14
@ロプノールs · 6 days ago
考え方(how to solve) yellowの△を直角二等辺三角形&①:②:√3の△に分けて考える。<br>高さ(h)を求める。斜辺4から高さ=2√2......①。<br>底辺(b)を求める。①:√3=2√2:? ?=2√6 2√2+2√6.....②<br>yellowの△の面積(area)=①×②×1/2。2√2×(2√2+2√6)×1/2=(8+8√3)/2=4+4√3(4で括る)<br>=4(1+√3)。 answer=4(1+√3)
2
@MrPaulc222 · 6 days ago
Height is 2*sqrt(2) (from 4/sqrt(2) on the left triangle).<br>Therefore, PQS area is 4.<br>As PSR is a 30 60 90, SR = 2*sqrt(6)<br>2*sqrt(2)*sqrt(6) = 2*sqrt(12) = 4*sqrt(3) for the right side.<br>Total are 4 + 4*sqrt(3) = approx 10.93 un^2
3
@gregoryzlotnik5071 · 6 days ago
Легкая задача, решил за 5 минут.
1
@libbychang413 · 6 days ago
|PS| = |QS| = 4*√2/2 = 2*√2<br>|PS| = |PR|*sin(30°) = |PR|/2 -> |PR| = 4*√2<br>|SR| = |PR|*cos(30°) =|PR|*√3/2 =2*√2*√3<br>|QR| = 2*√2*(1+√3)<br>area(△PQR) = |QR|*|PS|/2 = 2*√2*(1+√3)*2*√2/2 = 4*(1+√3)
1
@ollitrop46 · 6 days ago
PQS: 4²=x²+x²==16=2x²==8=x²==2√2=x. so 30º-60º-90º PRS sides are 2√2, 2√6, 4√2 Area of PQS: <br>A=½×2√2×2√2==A=√2×2√2==A=4units²==== Area of PRS: A=½×2√2×2√6==A=√2×2√6==A=4√3units²<br>Area of PQR: A=4+4√3==A=4(1+√3)units²
1
@bkarazi · 5 days ago
Hello,<br>Yes, it is very easy to find.<br>According to the laws of physics, in the figure shown; <br>a) The base = 2 * sides ➣ 4 = √‾(2 * √₂̅)² + (2 * √₂̅)² u ➣<br> ➣ PS = PQ = 2 * √₂̅ u ➣ PR = 2 * (2 * √₂̅)= 4 * √₂̅ u.<br>b) SR = √‾(4 * √₂̅ )² - (2 * √₂̅)² = 2 * √‾₆ u.<br> Two ways to find the total area:<br>1) Use QR as the base for one large triangle.<br>2) Add the left-side triangle and the right-side triangle areas.<br>a₁) QPS area = (2 * √₂̅) * (2 * √₂̅) ÷ 2 = (4 sq u).<br>b₂) RSP area = (2 * √‾₆) * (2 * √₂̅) ÷ 2 = (4 * √₃ sq u).<br>c) Total area = (4 sq u) + (4 * √₃ sq u) = 4 + (4 * √₃) sq u ≈ 10.928 sq u.<br>Total area = 10.928 sq u.<br>It should not take more than 40 seconds to solve the question. Especially, knowing: “a) The base = 2 * sides” <br>Thank you,
1
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