Answer:
area of triangle MQR =½×20×10=100Cm.
area of triangle PQR=½×10×10=50.
AREA OF SHADED REGION = AREA OF SQUARE - AREA OF ALL TIANGLES.
400-250=150cm² ans.
Step-by-step explanation:
i)
Area of whole square including circle = [tex](14cm)^2 = 196cm^2[/tex]
Area of only circle = [tex]\pi {( \frac{14}{2}) }^{2} = 49\pi = 49 \times \frac{22}{7} = 154 {cm}^{2}[/tex]
Area of shaded part = area of whole square - area of circle
= [tex]196cm^2 - 154cm^2 = 42cm^2[/tex]
ii)
let [tex]s[/tex] be the side of shaded square.
using Pythagorean theorem,
[tex]{s}^{2} = {12}^{2} + {5}^{2}[/tex]
=> [tex] {s}^{2} = 144 + 25[/tex]
=> [tex] {s}^{2} = 169[/tex]
=> [tex]s = \sqrt{169} [/tex]
=> [tex]s = 13cm[/tex]
Area of shaded square = [tex](13cm)^2 = 169cm^2[/tex]
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Verified answer
Answer:
area of triangle MQR =½×20×10=100Cm.
area of triangle PQR=½×10×10=50.
AREA OF SHADED REGION = AREA OF SQUARE - AREA OF ALL TIANGLES.
400-250=150cm² ans.
Step-by-step explanation:
i)
Area of whole square including circle = [tex](14cm)^2 = 196cm^2[/tex]
Area of only circle = [tex]\pi {( \frac{14}{2}) }^{2} = 49\pi = 49 \times \frac{22}{7} = 154 {cm}^{2}[/tex]
Area of shaded part = area of whole square - area of circle
= [tex]196cm^2 - 154cm^2 = 42cm^2[/tex]
ii)
let [tex]s[/tex] be the side of shaded square.
using Pythagorean theorem,
[tex]{s}^{2} = {12}^{2} + {5}^{2}[/tex]
=> [tex] {s}^{2} = 144 + 25[/tex]
=> [tex] {s}^{2} = 169[/tex]
=> [tex]s = \sqrt{169} [/tex]
=> [tex]s = 13cm[/tex]
Area of shaded square = [tex](13cm)^2 = 169cm^2[/tex]